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Calculus III

Tunc Geveci

Spring 2011

Contents

11 Vectors 1

11.1 Cartesian Coordinates in 3D and Surfaces . . . . . . . . . . . . . . . . . . . . . . 1

11.2 Vectors in Two and Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . 6

11.3 The Dot Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

11.4 The Cross Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

12 Functions of Several Variables 35

12.1 Tangent Vectors and Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

12.2 Acceleration and Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

12.3 Real-Valued Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . 61

12.4 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

12.5 Linear Approximations and the Differential . . . . . . . . . . . . . . . . . . . . . 74

12.6 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

12.7 Directional Derivatives and the Gradient . . . . . . . . . . . . . . . . . . . . . . . 94

12.8 Local Maxima and Minima . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

12.9 Absolute Extrema and Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . 121

13 Multiple Integrals 131

13.1 Double Integrals over Rectangles . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

13.2 Double Integrals over Non-Rectangular Regions . . . . . . . . . . . . . . . . . . . 138

13.3 Double Integrals in Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 144

13.4 Applications of Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

13.5 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

13.6 Triple Integrals in Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . 167

13.7 Change of Variables in Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . 179

14 Vector Analysis 187

14.1 Vector Fields, Divergence and Curl . . . . . . . . . . . . . . . . . . . . . . . . . . 187

14.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

14.3 Line Integrals of Conservative Vector Fields . . . . . . . . . . . . . . . . . . . . . 210

14.4 Parametrized Surfaces and Tangent Planes . . . . . . . . . . . . . . . . . . . . . 220

14.5 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239

14.6 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259

14.7 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274

14.8 Gauss’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

K Answers to Some Problems 285

L Basic Differentiation and Integration formulas 309

iii

iv CONTENTS

Preface

This is the third volume of my calculus series, Calculus I, Calculus II and Calculus III. This

series is designed for the usual three semester calculus sequence that the majority of science and

engineering majors in the United States are required to take. Some majors may be required to

take only the first two parts of the sequence.

Calculus I covers the usual topics of the first semester: Limits, continuity, the deriv-

ative, the integral and special functions such exponential functions, logarithms,

and inverse trigonometric functions. Calculus II covers the material of the second

semester: Further techniques and applications of the integral, improper integrals,

linear and separable first-order differential equations, infinite series, parametrized

curves and polar coordinates. Calculus III covers topics in multivariable calculus:

Vectors, vector-valued functions, directional derivatives, local linear approxima-

tions, multiple integrals, line integrals, surface integrals, and the theorems of Green,

Gauss and Stokes.

An important feature of my book is its focus on the fundamental concepts, essential

functions and formulas of calculus. Students should not lose sight of the basic concepts

and tools of calculus by being bombarded with functions and differentiation or antidifferentia-

tion formulas that are not significant. I have written the examples and designed the exercises

accordingly. I believe that "less is more". That approach enables one to demonstrate to the

students the beauty and utility of calculus, without cluttering it with ugly expressions. Another

important feature of my book is the use of visualization as an integral part of the expo-

sition. I believe that the most significant contribution of technology to the teaching of a basic

course such as calculus has been the effortless production of graphics of good quality. Numerical

experiments are also helpful in explaining the basic ideas of calculus, and I have included such

data.

Remarks on some icons: I have indicated the end of a proof by ¥, the end of an example by¤ and the end of a remark by ♦.

Supplements: An instructors’ solution manual that contains the solutions of all the prob-

lems is available as a PDF file that can be sent to an instructor who has adopted the book. The

student who purchases the book can access the students’ solutions manual that contains the

solutions of odd numbered problems via www.cognella.com.

Acknowledgments: ScientificWorkPlace enabled me to type the text and the mathematical

formulas easily in a seamless manner. Adobe Acrobat Pro has enabled me to convert the

LaTeX files to pdf files. Mathematica has enabled me to import high quality graphics to my

documents. I am grateful to the producers and marketers of such software without which I

would not have had the patience to write and rewrite the material in these volumes. I would

also like to acknowledge my gratitude to two wonderful mathematicians who have influenced

me most by demonstrating the beauty of Mathematics and teaching me to write clearly and

precisely: Errett Bishop and Stefan Warschawski.

v

vi PREFACE

Last, but not the least, I am grateful to Simla for her encouragement and patience while I spent

hours in front a computer screen.

Tunc Geveci ([email protected])

San Diego, January 2011

Chapter 11

Vectors

In this chapter we will introduce the concept of a vector and discuss the relevant algebraic

operations. We will also discuss the dot product that is related to angles between vectors

and the cross product that produces a vector that is orthogonal to a pair of vectors. These

concepts and operations will be needed when we develop the calculus of functions of several

variables.

11.1 Cartesian Coordinates in 3D and Surfaces

Cartesian Coordinates in Three Dimensions

Our starting point is the familiar Cartesian coordinate plane. Let’s designate the axes as

the x and y axes. Picture the xy-plane as a plane in the three-dimensional space. The third

axis is placed so that it is perpendicular to the xy-plane and its origin coincides with the origin

of the xy-plane. The positive direction is determined by the right-hand rule. Let us label the

third axis as the z-axis.

x

y

z

-2

-4

2

4

-2

-4

2

4

-2

-4

2

4

Figure 1

We will associate an ordered triple (x, y, z) with each point P in space as follows: If P is thepoints at which the three axes intersect, we will call P the origin and denote it by O. The

ordered triple (0, 0, 0) is associated with O. Let P be a point other than the origin. Consider

the line that passes through P and is perpendicular to the xy-plane. Let Q be the intersection

1

2 CHAPTER 11. VECTORS

of that line with the xy-plane. We will associate with Q the triple (x, y, 0), where x and y aredetermined as Cartesian coordinates in the xy-plane. Consider the plane that passes through

P and is parallel to the xy-plane. If z is the point at which that plane intersects the third

axis, we will associate the ordered triple (x, y, z) with the point P . We are speaking of an“ordered triple”, since the order of the numbers x,y and z matters. We will identify P with

the triple (x, y, z), and refer to “the point (x, y, z)”, just as we identify a point in the Cartesiancoordinate plane with the corresponding order pair of numbers. Thus, we have described the

Cartesian coordinate system in the three-dimensional space. The system is also referred to

as a rectangular coordinate system.

The set of all ordered triples (x, y, z) of real real numbers will be denoted by R3. Thus, R3 canbe identified with the three-dimensional space that is equipped with the Cartesian coordinate

system, just as the set of all ordered pairs (x, y) if real numbers can be denoted as R2 andidentified with the set of points in the Cartesian coordinate plane.

The xy-plane consists points of the form (x, y, 0), the xz-plane consists of points of the form(x, 0, z), and the yz-plane consists of points of the form (0, y, z). We will refer to these planesas the coordinate planes.

The first octant consists of points (x, y, z) such that x ≥ 0, y ≥ 0 and z ≥ 0.

Definition 1 The (Euclidean) distance between P1 = (x1, y1, z1) and P2 = (x2, y2, z2) is

dist (P1, P2) =

q(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2.

Surfaces

Definition 2 Let F (x, y, z) be an expression in the variables x, y, z. and let C be a constant.

The set of points (x, y, z) ∈ R3 such that F (x, y, z) = C is a surface in R3. We say that thesurface is the graph of the equation F (x, y, z) = C.

Example 1 Let a, b, c, d be given constants. The graph of the equation

ax+ by + cz = d

is a plane.

For example, the equation

x+ y + z = 1

describes a plane that intersects the coordinate axes at the points (1, 0, 0), (0, 1, 0) and (0, 0, 1).

1

z

1

x y1

Figure 2

The equation z = 2 describes a plane that is parallel to the the xy-plane.

11.1. CARTESIAN COORDINATES IN 3D AND SURFACES 3

Figure 3

Definition 3 Given a point P0 = (x0, y0, z0) and a positive number r, the set of points P =(x, y, z) such that

(x− x0)2 + (y − y0)2 + (z − z0)2 = r2

is the sphere of radius r centered at P0.

Indeed, P is such a point if

dist (P,P0) =

q(x− x0)2 + (y − y0)2 + (z − z0)2 = r.

For example, the sphere of radius 2 centered at (3, 3, 1) is the graph of the equation

(x− 3)2 + (y − 3)2 + (z − 1)2 = 4.

Figure 4 displays that sphere.

Figure 4

Example 2 Figure 5 shows the surface that is the graph of the equation

z = x2 + y2.

This surface is referred to as a paraboloid.


20-2

5

z

y

0

x

0

-22

Figure 5

Since z = x2 + y2 ≥ 0 for each (x, y) ∈ R2, the surface is above the xy-plane. If (x, y) = (0, 0),then z = 0 so that (0, 0, 0) is on the surface. If c > 0, the intersection of the surface with theplane z = c is a circle whose projection onto the xy-plane is the circle

x2 + y2 = c2

that is centered at (0, 0) and has radius c. As c increases, these concentric circles expand.The intersection of the surface with a plane of the form x = c is a parabola whose projectiononto the yz-plane is the graph of the equation

z = c2 + y2.

Similarly, the intersection of the surface with a plane of the form y = c is a parabola whose

projection onto the xz-plane is the graph of the equation

z = x+ c2.

¤

Example 3 Figure 6 shows the surface that is described by the equation

x2 − y2 + z2 = 1Such a surface is referred to as a hyperboloid of one sheet.

5

-55

0z

x

0

y

5

0

-5-5

Figure 6

Since

x2 + z2 = 1 + y2,

the intersection of the surface with a plane of the form y = c is a circle that projects onto the

xz-plane as the circle

x2 + y2 = 1 + c2

that is centered at (0, 0) and has radius√1 + c2.

11.1. CARTESIAN COORDINATES IN 3D AND SURFACES 5

The intersection of the surface with a plane of the z = c is a hyperbola that projects onto thexy-plane as the graph of the equation

x2 − y2 = 1− c2.Similarly, the intersection of the surface with a plane of the form x = c is a a hyperbola that

projects onto the yz-plane as the graph of the equation

−y2 + z2 = 1− c2.¤

Example 4 Figure 7 shows the surface that is described by the equation

x2 − y2 − z2 = 1Such a surface is referred to as a hyperboloid of two sheets.

Figure 7

Since

y2 + z2 = x2 − 1,the surface does not intersect a plane of the form x = c if −1 < c < 1. If c < −1 or c > 1,the intersection of the surface with a plane of the form x = c is a circle that projects onto the

yz-plane as the circle

y2 + z2 = c2 − 1that is centered at (0, 0) and has radius

√c2 − 1.

The intersection of the surface with a plane of the z = c is a hyperbola that projects onto thexy-plane as the graph of the equation

x2 − y2 = 1 + c2.Similarly, the intersection of the surface with a plane of the form y = c is a a hyperbola that

projects onto the xz-plane as the graph of the equation

x2 − z2 = 1 + c2.¤

Problems

[C] In problems 1-10, make use of a graphing device to plot the surface that is the graph of thegiven equation. In each case identify the curves that are the intersections of the surface with

planes of the form

x = constant, y = constant, z = constant.


1.

2x+ y + 4z = 8

2.

z = 4x− 3y

3.

z = 4x2 + 9y2

4.

x = y2 + 4z2

5.

x2 + 2y2 + 4z2 = 4

6.

4x2 + y2 + 9z2 = 4

7.

x2 − 9y2 − 4z2 = 1

8.

4y2 − x2 − 2z2 = 1

9.

y2 − x2 + z2 = 1

10.

x2 + y2 − z2 = 4

11.2 Vectors in Two and Three Dimensions

In this section we will introduce the concept of a vector. Roughly speaking, a vector is an

entity such as velocity or force that has a magnitude and direction. Most probably you have

encountered vectors intuitively in a Physics course already. Now we will try to express the idea

more precisely.

Two Dimensional Vectors

Algebraically, a two-dimensional vector is simply an ordered pair

v = (v1, v2) .

Even though we use the same the notation for a vector and a point in the plane, the context

will clarify whether the ordered pair refers to a vector or a point. As you will soon see, the

distinction will be possible due to the way we visualize vectors and define certain operations

involving vectors.

We will denote vectors by boldface letters in print. The vector v can be denoted as −→v in writing.The number v1 is the first component v and v2 is the second component of v. Two vectors

v = (v1, v2) and w = (w1, w2) are declared to be equal if and only if

v1 = w1 and v2 = w2.

In this case we write v = w. Thus, two vectors are equal iff their corresponding components

are equal. We define the length of v = (v1, v2) as

||v|| =qv21 + v

22.

The 0-vector 0 is the ordered pair (0, 0). If v = (v1, v2) is not the 0-vector, we can associatewith v a directed line segment that begins at a point P1 = (x1, y1) and ends at a pointP2 = (x2, y2) such that

x2 − x1 = v1 and y2 − y1 = v2.We denote such a directed line segment as

−−−→P1P2 and visualize it as an arrow along the line

segment−−−→P1P2 with its tip at P2. We will refer to

−−−→P1P2 as a representation of the vector

v, or as the vector v located at P1. There are infinitely many representations of the vector

v, but any two representations of v have the same length ||v|| and direction. The0-vector cannot be represented by a directed line segment.

11.2. VECTORS IN TWO AND THREE DIMENSIONS 7

x

y

P1

P2

Figure 1: Representations of a vector have the same length and direction

Conversely, given the points P1 = (x1, y1) and P2 = (x2, y2), the vector

v = (x2 − x1, y2 − y1)

is the vector that is associated with the directed line segment−−−→P1P2. Usually, we will

simply refer to v as the vector−−−→P1P2.

We can associate the origin with the 0-vector. Every nonzero vector can be represented uniquely

by a directed line segment−−→OP that is located at the origin. We will refer to

−−→OP as the position

vector of the point P . Thus, if P = (a, b) then the position vector of P is designated by

the same ordered pair (a, b). The context will clarify whether (a, b) refers to the point P or theposition vector of P .

x

y

P a, b

OP

Figure 2

Example 1

a) Let P1 = (3, 2) and P2 = (2, 4). Determine the vector v that is associated with−−−→P1P2.

b) Determine Q2 such that−−−→Q1Q2 is the representation of v that is located at Q1 = (−1, 1).

Sketch the directed line segments−−−→P1P2.and

−−−→Q1Q2.

Solution

a) We have

v = (2− 3, 4− 2) = (−1, 2) .

b) If Q2 = (x2, y2) the

x2 − (−1) = −1 and y2 − 1 = 2,so that x2 = −2 and y2 = 3. Therefore, Q2 = (−2, 3). ¤


1 2 3x

1

2

3

4y

P1

P2

Q1

Q2

Figure 3

Aside from the fact that we visualize an ordered pair as a point and or a vector differently, the

distinction results from the fact that we endow vectors with two algebraic operations.

Definition 1 Given vectors v = (v1, v2) and w = (w1, w2) we define the sum v+w by setting

v +w = (v1 + w1, v2 + w2) .

Thus, we add vectors by adding the corresponding components.

Within the context of vectors, real numbers are referred to as scalars.

Definition 2 The scalar multiplication of the vector v = (v1, v2) by the scalar c is

denoted as cv and we set

cv = (cv1, cv2) .

Thus, we multiply each component of v by the real number c.

Note that

||cv|| = |c| ||v|| .Indeed, if v = (v1, v2) then

||cv|| =q(cv1)

2+ (cv2)

2=

qc2 (v21 + v

22)

= |c|qv21 + v

22 = |c| ||v|| .

Geometrically, the addition of the vectors v and w can be represented by the parallelogram

picture: If we represent v and w by arrows emanating from the same point, the sum v + wcan be represented by the arrow along the diagonal of the parallelogram that is determined by

v and w.

x

y

v

w

v w

Figure 4: The parallelogram picture for the sum of v and w


We can also represent v +w by the triangle picture: If we have any representation of v, we

can attach w to the tip of that representation. The sum is represented by the arrow from the

point at which v is located to the tip of the arrow that represents w.

x

y

v

wv w

Figure 5: The triangle picture for v +w

Example 2 Let v = (1, 2) and w = (−2, 3).a) Determine v+wb) Sketch the parallelogram picture for the addition of v and w.

Solution

a) We have

v +w = (1− 2, 2 + 3) = (−1, 5)b) Figure 6 illustrates the parallelogram picture for v +w. ¤

2 1 1x

1

2

3

4

5

y

v

w

v w

Figure 6

Remark 1 If the vectors v and w are located at the point P0 and represent forces that act on

object at P0, the sum v +w is the resultant force that acts on that object. If v represents the

velocity of an object in still water and w is the velocity of a current, the sum v +w is the net

velocity of the object.♦

If c = 0, the scalar multiple cv = 0 for any vector v. The scalar multiple is also the 0-vector ifv = 0. If v 6= 0 and c > 0, we can represent cv by an arrow along a representation of v whoselength is c ||v||. If c < 0 then cv can be represented by an arrow along the line determined byv that points in the opposite direction. The length of the arrow is |c| ||v|| = −c ||v|| .


x

y

v

cv

Figure 7: cv if c > 0

x

y

v

cv

Figure 8: cv if c < 0

Example 3 Let v = (1, 2).

a) Determine 3v and −3v and sketch their representations.b) Confirm that ||3v|| = ||−3v|| = 3 ||v||.Solution

a) We have

3v = 3 (1, 2) = (3, 6) and − 3v = −3 (1, 2) = (−3,−6) .

3 2 1 1 2 3x

6

4

2

2

4

6

y

v

3v

3v

Figure 9

b) We have

||3v|| =p32 + 62 =

√45 = 3

√5,

and

3 ||v|| = 3p11 + 22 = 3

√5.

Thus, ||3v|| = 3 ||v|| .


We have

||−3v|| =q(−3)2 + (−62) =

p33 + 62 =

√45 = 3

√5 = 3 ||v|| .

¤

If v = (v1, v2) we set

−v = (−1)v = (−v1,−v2) .Thus, v − v = 0. We can represent −v by an arrow that points in the opposite direction thatis determined by (a representation of) v and has the same length.

If v = (v1, v2) and w = (w1, w2) then

v−w = v+ (−w) = (v1 − w1, v2 − w2) .

We can represent v−w by an arrow from the tip of w to the tip v.

x

y

v

w

v w

v w

w

Figure 10: Representations of v−w

Example 4 Let v = (4, 5) and w = (3, 2). Determine v−w and sketch a representation of the

operation.

Solution

We have

v−w = (4− 3, 5− 2) = (1, 3) .Figure 11 illustrates v −w. ¤

1 2 3 4x

1

2

3

4

5

y

v

w

v w

Figure 11


Definition 3 If v and w are vectors, c and d are real numbers (scalars), the vector cv+ dw is

a linear combination of v and w.

Assume that c and d are real numbers (scalars), and u,v,w are vectors. The following rules

apply are valid for addition and scalar multiplication:

v+w = w+ v,

u+ (v +w) = (u+ v) +w,

v+ 0 = v,

v+ (−v) = 0,

c (v +w) = cv+ cw,

(c+ d)v = cv+ dv,

(cd)v = c (dv) ,

1v = v.

You can confirm the validity of these rules easily (exercise).

Definition 4 A unit vector is a vector of unit length.

Thus, u = (u1, u2) is a unit vector iff

||u|| =qu21 + u

22 = 1.

Definition 5 The normalization of the nonzero vector v or the unit vector along v is

u =1

||v||v.

Note that u is in the same direction as v and has unit length:

||u|| =¯¯1

||v||v¯¯=

1

||v|| ||v|| = 1.

The normalization of v can be written as

v

||v|| .

Example 5 Let v = (2,−3). Determine the unit vector along v.Solution

We have

||v|| =q22 + (−3)2 =

√13.

Therefore, the unit vector along v is

u =1

||v||v =1√13(2,−3) =

µ2√13,− 3√

13

¶.

¤


0.5 1 1.5 2x

0.5

1

1.5

2

2.5

3

y

v

u

Figure 12

Standard Basis Vectors

We set

i = (1, 0) and j = (0, 1) .

Thus, we can represent i as the position vector of the point (1, 0) and j as the position vector ofthe point (0, 1). The vector i points in the positive direction of the horizontal axis and j pointsin the positive direction of the vertical axis. We will refer to i and j as the standard basis

vectors.

i

j

1

1

Figure 13: Standard basis vectors

If v = (v1, v2) thenv = v1 (1, 0) + v2 (0, 1) = v1i+ v2j.

Thus, we can express any vector as a sum of scalar multiples of the standard basis vectors i and

j. We will refer to the expression v1i+ v2j for the vector v as the ijrepresentation of v. We willfavor the ijrepresentation when we wish to emphasize a geometric representation of a vector.

If v = v1i+ v2j and w = w1i+ w2j then

v +w = (v1 + w1) i+ (v2 + w2) j,

and

cv = cv1i+ cv2j

for any real number c.

Example 6 Let v = (4, 1) and w = (−2, 3).


a) Express v and w in terms of the standard basis vectors.

b) Perform the operations to determine 3v, v+w and v−w by making use the expressions for

v and w in terms of the standard basis vectors.

Solution

a) We have

v = (4, 1) = 4i+ j and w = (−2, 3) = −2i+ 3j.b)

3v = 3 (4i+ j) = 12i+ 3j,

v +w = (4i+ j) + (−2i+ 3j) = 2i+ 4j,v −w = (4i+ j)− (−2i+ 3j) = 6i− 2j.

¤

Three Dimensional Vectors

The concept of a three dimensional vector is a straightforward analog of the concept of a two

dimensional vector, and the operations are similar.

Algebraically, a 3D-vector v is an ordered triple (v1, v2, v3) where the components v1, v2 and v3are real numbers (scalars). If v = (v1, v2, v3) and w = (w1, w2, w3) we declare that v = w iff

the corresponding components are equal:

v1 = w1, v2 = w2, v3 = w3.

The 0-vector 0 is (0, 0, 0) .The length of v = (v1, v2, v3) is

||v|| =qv21 + v

22 + v

23.

A (geometric) representation of v = (v1, v2, v3) is a directed line segment−−−→P1P2, where P1 =

(x1, y1, z1), P2 = (x2, y2, z2) are points in R3 such that

v1 = x2 − x1, v2 = y2 − y1 and v3 = z2 − z1.

We may refer to−−−→P1P2 as the vector v that is located at P1 or as v that is attached to P1.

Representations of the same vector have the same length and direction.

z

yx

Figure 14: Representations of a vector in 3D

If P = (x, y, z) is a point in R3, the position vector of P is the directed line segment−−→OP .

The sum of v = (v1, v2, v3) and w = (w1, w2, w3) is the vector

v+w = (v1 + w1, v2 + w2, v3 + w3) .


We can visualize addition via the parallelogram picture or the triangle picture, as in the 2D

case.

If c is a real number (scalar), the scalar multiple cv of v = (v1, v2v3) is the vector

cv = (cv1, cv2, cv3) .

Thus cv is along v, points in the same direction as v if c > 0 and in the opposite direction ifc < 0. We have

||cv|| = |c| ||v|| .Addition and scalar multiplication obey the rules that were listed for two dimensional vectors.

Just as the two dimensional case, the normalization of the nonzero vector v is

u =1

||v||v,

which can be written asv

||v|| .

Example 7 Let v = (2,−2, 3) and w = (−1, 1, 2). Determine v + w, 4v + 3w and the

normalization of v.

Solution

v+w = (2,−2, 3) + (−1, 1, 2) = (1,−1, 5) .

vw

v w

1

0

1

2

x1

0

1

2

y

1

2

3

4

5

z

Figure 15

We also have

2v = 2 (2,−2, 3) = (4,−4, 6) ,

4v + 3w = 4 (2,−2, 3) + 3 (−1, 1, 2)= (8,−8, 12) + (−3, 3, 6) = (5,−5, 18) .

The length of v is

||v|| =q22 + (−2)2 + 32 =

√17.


Therefore, the normalization of v is

u =1

||v||v =1√17(2,−2, 3) =

µ2√17,− 2√

17,3√17

¶.

In the 3D case, the standard basis vectors are

i = (1, 0, 0) , j = (0, 1, 0) and k = (0, 0, 1) .

If we label the Cartesian coordinate axes as the x, y and z axes, then i is a unit vector in the

direction of the positive x-axis, j is a unit vector in the direction of the positive y-axis, and k is

a unit vector in the direction of the positive z-axis.

i

jk

Figure 16: Standard basis vectors in 3D

If v = (v1, v2, v3) then

v = v1i+ v2j+ v3k.

We may refer to the above expression for v as the ijkrepresentation of v.

If v = v1i+ v2j + v3k, w = w1i+ w2j + w3k.and c ∈ R then

v+w = (v1 + w1) i+ (v2 + w2) j+ (v3 + w3)k

and

cv = cv1i+ cv2j+ cv3k.

Example 8 Let

v = 2i− 3j+5k and w = −3i+ 2j− 4k.Determine −3v and v −w.

Solution

−3v = −3 (2i− 3j+5k) = −6i+ 9j− 15k,

v −w = (2i− 3j+5k)− (−3i+ 2j− 4k) = 5i− 5j+9k.¤


Problems

In problems 1-4,

a) Determine the vector v that is associated with the directed line segment−−−→P1P2.

b) Determine Q2 such that−−−→Q1Q2 is the representation of v that is located at Q1. Sketch the

directed line segments−−−→P1P2.and

−−−→Q1Q2.

1.

P1 = (1, 2) , P2 = (3, 5) , Q1 = (2, 1)

2.

P1 = (−2, 3) , P2 = (−4, 2) , Q1 = (1, 3)3.

P1 = (2,−3) , P2 = (4, 2) , Q1 = (3, 2)4.

P1 = (−2,−1) , P2 = (−4, 2) , Q1 = (−1, 2)In problems 5-8

a) Determine v+w,b) Sketch the parallelogram picture for the addition of v and w (represent v and w by directed

line segments attached to the origin):

5. v = (2, 1), w = (3, 4)6. v = (2,−3), w = (3, 2)

7. v = (−2,−1), w = (2, 4)8. v = (2, 3), w = (−1,−5)

In problems 9 and 10

a) Determine v+w,b) Sketch the triangle picture for the addition of v and w.(represent v by a directed line segment

attached to the origin):

9. v = (2, 4), w = (−1, 2) 10. v = (−4, 2), w = (3, 1)


a) Determine v−w,b) Sketch the triangle picture for v−w.(represent v and w by directed line segments attached

to the origin):

11. v = (2, 4), w = (−2, 2) 12. v = (−4, 2), w = (3, 1)

In problems 13 and 14, determine the given linear combination of v and w.

13. 2v − 3w if v = (3,−1) and w = (−2, 5)14. −4v + 5w if v = (2, 4) and w = (1,−4)In problems 15 and 16

a) Determine u, the unit vector along v (the normalization of v),

b) Sketch u and v.(represent v and u by directed line segments attached to the origin):


15. v = (3, 4) 16. v = (−2, 2)

In problems 17-20

a) Express the vectors v and w in terms of the standard basis vectors,

b) Determine the given linear combination of v and w by using their representations in terms

of the standard basis vectors.

17. v = (3, 2) , w = (−2, 4) , 2v − 3w18. v = (−4, 1) , w = (4, 3) , 2v +w19. v = (−2, 3, 6) , w = (4,−2, 1) , −v+ 4w20. v = (−1,−3, 5) , w = (7, 2,−2) , 3v− 2w

11.3 The Dot Product

In this section we will introduce the dot product. This is an operation that assigns a real number

to pairs of vectors and enables us to measure the angle between them.

The Definition of the Dot Product

Definition 1 If v = (v1, v2) and w = (w1, w2) are two dimensional vectors, the dot productv ·w is the real number that is defined as

v ·w = v1w1 + v2w2.If v = (v1, v2,v3) and w = (w1, w2, w3) are three dimensional vectors, the dot product v ·w is

the real number that is defined as

v ·w = v1w1 + v2w2 + v3w3.Note that we can express the length of a vector in terms of the product:

||v|| = √v · v

Example 1 Determine v ·w if

a) v = (2, 5) and w = (4, 3),b) v = 2i− 3j+ k and w = −i+ 2j+ 4k.Solution

a)

v ·w = (2, 5) · (4, 3) = 8 + 15 = 23.b)

v ·w = (2i− 3j+ k) · (−i+ 2j+ 4k) = −2− 6 + 4 = −4.¤The basic properties on the dot product: If u, v and w are vectors in Rn (n = 2 or n = 3)and c is a real number (scalar) then

v ·w = w · v,u· (v+w) = u · v+ u ·w,(cv) ·w = c (v ·w) = v· (cw) ,0 · v = 0.

These properties can be verified easily (exercise).

11.3. THE DOT PRODUCT 19

Theorem 1 (The Cauchy-Schwarz Inequality) For each v and w in Rn (n = 2 or n = 3)we have

|v ·w| ≤ ||v|| ||w||Proof

Let t ∈ R. Set

f (t) = ||v− tw||2 = (v− tw) · (v− tw)= v · v− tv ·w− tw · v+ t2w ·w= ||v||2 − 2tv ·w + t2 ||w||2= ||w||2 t2 − 2v ·wt+ ||v||2

Since

f (t) = ||v− tw||2 ≥ 0 for each t ∈ R,the discriminant of the quadratic expression ||w||2 t2 − 2v ·wt+ ||v||2 is nonpositive. Thus,

(−2v ·w)2 − 4 ||w||2 ||v||2 ≤ 0,

so that

4 (v ·w)2 ≤ 4 ||w||2 ||v||2 ⇒ (v ·w)2 ≤ ||v||2 ||w||2 ⇔ |v ·w| ≤ ||v|| ||w|| ,

as claimed. ¥

The following fact is referred to as the triangle inequality for the obvious reason: The length

of one side of a triangle is less than the sum of the lengths of other two sides.

x

y

v

v w

w

Figure 1: ||v+w|| < ||v||+ ||w||

Theorem 2 (The Triangle Inequality) For each v and w in Rn ( n = 2 or n = 3) wehave

||v+w|| ≤ ||v||+ ||w|| .Proof

We have

||v +w||2 = (v+w) · (v+w)= v · v + v ·w +w · v +w ·w= ||v||2 + 2v ·w+ ||w||2


By the Cauchy-Schwarz Inequality,

2v ·w ≤ 2 ||v|| ||w|| .

Therefore,

||v+w||2 ≤ ||v||2 + 2 ||v|| ||w||+ ||w||2 = (||v||+ ||w||)2 .Thus,

||v+w|| ≤ ||v||+ ||w||¥

Assume that v and w are nonzero vectors in Rn (n = 2 or n = 3). By the Cauchy-Schwarzinequality,

|v ·w| ≤ ||v|| ||w|| ,so that

− ||v|| ||w|| ≤ v ·w ≤ ||v|| ||w|| .Thus

−1 ≤ v ·w||v|| ||w|| ≤ 1.

Therefore there exists a unique angle θ ∈ [0,π] (in radians) such that

cos (θ) =v ·w

||v|| ||w|| ⇔ θ = arccos

µv ·w

||v|| ||w||¶.

x

y

v

w

Θ

Figure 2: The angle between two vectors

Definition 2 If v and w are nonzero vectors in Rn (n = 2 or n = 3).the angle θ ∈ [0,π] suchthat

cos (θ) =v ·w

||v|| ||w|| ⇔ θ = arccos

µv ·w

||v|| ||w||¶

is the angle between the vectors v and w.

Remark 1 It can be shown that the above definition is consistent with the law of cosines:

||v−w||2 = ||v||2 − 2 ||v|| ||w|| cos (θ) + ||w||2 .


x

y

v

w

Θ

v w

Figure 3

Indeed,

||v −w||2 = (v−w) · (v−w) = ||v||2 − 2v ·w+ ||w||2 .

Thus,

||v||2 − 2v ·w + ||w||2 = ||v−w||2= ||v||2 − 2 ||v|| ||w|| cos (θ) + ||w||2 .

Therefore,

v ·w = ||v|| ||w|| cos (θ) .

♦

Example 2 Let

v = (1, 2) and w =³√3 + 2, 2

√3− 1

´.

Determine the angle between v and w.

Solution

We have

||v|| =√5, ||w|| =

r³√3 + 2

´2+³2√3− 1

´2= 2√5,

and

v ·w = (1, 2) ·³√3 + 2, 2

√3− 1

´=√3 + 2 + 4

√3− 2 = 5

√3.

Therefore,

θ = arccos

µv ·w

||v|| ||w||¶= arccos

Ã5√3¡√

5¢ ¡2√5¢! = arccosÃ√3

2

!=

π

6.

Figure 4 illustrates the angle θ. ¤


1 2 3x

1

2

y

v

w

Θ

Figure 4

Definition 3 The vectors v and w in Rn (n = 2 or n = 3)are orthogonal (or perpendicular)if v ·w = 0.

Example 3 Let

v = (2, 1) and w = (−1, 2) .Show that v and w are orthogonal.

Solution

We have

v ·w = (2, 1) · (−1, 2) = −2 + 2 = 0.Therefore v and w are orthogonal.

1 1 2x

1

2

y

v

w

Figure 5: Orthogonal vectors

.¤

Remark 2 The 0-vector is orthogonal to any vector since 0 · v = 0 for any v ∈ Rn. ♦

Remark 3 The standard basis vectors are mutually orthogonal. Therefore, if v = v1i + v2jthen

v · i =(v1i+ v2j) · i = v1 (i · i) + v2 (i · j) = v1 ||i||2 + 0 = v1,and

v · j =(v1i+ v2j) · j = v1 (i · j) + v2 (j · j) = 0 + v2 ||j||2 = v2.Thus,

v = (v · i) i+ (v · j) j.Similarly, if v ∈ R3 then

v = (v · i) i+ (v · j) j+ (v · k)k,where i, j and k are the standard basis vectors in R3. ♦


Let v be a nonzero vector in R2 and let u be the unit vector along v. Thus,

u =1

||v||v.

Let α be the angle between v and the positive x-axis and let β be the angle between v and the

positive y-axis. We have

cos (α) =v · i

||v|| ||i|| =v · i||v|| = u · i,

and

cos (β) =v · j

||v|| ||j|| =v · j||v|| = u · j

Thus

u = (u · i) i+ (u · j) j =cos (α) i+ cos (β) j.Similarly, if v is a nonzero vector in R3 and u is the unit vector along v, then

u = (u · i) i+ (u · j) j =cos (α) i+ cos (β) j+ cos (γ)k,

where α, β and γ are the angles between v and the positive directions of the x, y and z axes,

respectively.

Definition 4 Given a nonzero vector v ∈ R2 the direction cosines of v are the components ofthe unit vector along v. The direction cosines of a nonzero vector v in R3are the componentsof the unit vector along v.

Example 4 Let v =(2, 4, 3). Find the direction cosines of v.

Solution

All we need to do is to determine the unit vector u along v. The length of v is

||v|| =p22 + 42 + 32 =

√29

Therefore,

u =1

||v||v =1√29(2, 4, 3) =

µ2√29,4√29,3√29

¶.

Thus, the direction cosines of v are

2√29,

4√29and

3√29.

¤

Projections

Let w be a nonzero vector. In a number of applications it is useful to express a given vector v

as

v = v1 + v2,

where v1 is along w, i..e, a scalar multiple of w, and v2 is orthogonal to w.


v

w

v1 Pwv

v2

Figure 6

Thus,

v = αw+ v2,

where v2 ·w = 0. Therefore,(v− αw) ·w = 0.

Thus,

v ·w− α ||w||2 = 0⇒ α =v ·w||w||2 .

Therefore,

v =

Ãv ·w||w||2

!w + v2,

where Ãv ·w||w||2

!w

is along w and v2 is orthogonal to w. Note thatÃv ·w||w||2

!w =

µv ·w||w||

¶w

||w|| =µv · w

||w||¶

w

||w|| .

The vector

u =w

||w||is the unit vector along w. Thus, we can express v as

v = (v · u)u+ v2,where u is the unit vector along w and v2 is orthogonal to w.

Definition 5 If w is a nonzero vector, the projection of v along w is

Pwv = (v · u)u,where

u =w

||w||is the unit vector along w. The component of v along w is

compwv = v · u,so that

Pwv = (compwv)u


By the discussion that preceded Definition 4, the vector v −Pwv is orthogonal to w.

Remark 4 If θ is the angle between v and w, then

compwv = v · u =v ·w||w|| =

||v|| ||w|| cos (θ)||w|| = ||v|| cos (θ) ,

and

Pwv = (v · u)u = ||v|| cos (θ)u.♦

Remark 5 We have noted that a vector v ∈ R3 can be expressed as

v = (v · i) i+ (v · j) j+ (v · k)k.

Thus,

v = (compiv) i+¡compjv

¢j+ (compkv)k

♦

Example 5 Let v =(1, 2) and w = (2, 1).a) Determine the component of v along w and the projection of v along w.

b) Express v as Pwv+ v2 where v2 is orhogonal to Pwv

Solution

a) The unit vector along w is

u =1

||w||w =1√5(2, 1) .

Therefore,

compwv = v · u =(1, 2) ·µ1√5(2, 1)

¶=

1√5(2 + 2) =

4√5,

and

Pwv = (v · u)u = 4√5

µ1√5(2, 1)

¶=4

5(2, 1) =

µ8

5,4

5

¶.

b)

v2 = v−Pwv =(1, 2)−µ8

5,4

5

¶=

µ−35,6

5

¶¤

1 2

0.5

1

1.5

2

v

w

v1 Pwv

v2

Figure 7


The idea of projection is related to the definition of work in Physics. Assume that an object

that is moving along a line is subjected to the constant force (vector) F. Let w be the vector

that represents the displacement of the object. The work done by F is defined as

W = F ·w = ||F|| ||w|| cos (θ) = (compwF) ||w||(in appropriate units).

Example 6 Let F = 2i + 3j be the force that is acting on an object whose displacement isrepresented by the vector w = 4i− 2j. Compute the work done by F.Solution

The work done by F is simply

F ·w = (2i+ 3j) · (4i− 2j) = 8− 6 = 2.

¤

1 2 3 4

1

2

3

2

1

F

w

Figure 8

Problems

In problems 1-6,

a) Determine ||v||, ||w|| and the dot product v ·w,b) Determine θ, the angle between v and w. Are the vectors orthogonal to each other?

1.

v = (1, 0) , w = (1, 1)

2.

v = (1, 0) , w =³−1,√3´

3.

v = (1, 1) , w =

Ã√3 + 1

2,

√3− 12

!4.

v = i+ j, w = −i+ j

11.4. THE CROSS PRODUCT 27

5.

v = (0, 1, 1) , w =

Ã0,

√3 + 1

2,1−√32

!6.

v = i+ j, w = i− jIn problems 7-10,

a) Determine ||v||, ||w|| and the dot product v ·w,b) Determine cos (θ), where θ is the angle between v and w,

c) Make use of your computational utility to obtain an approximate value of θ (display 6 signif-

icant digits).

7.

v = (1, 2) , w = (1,−1)8.

v = 3i+ 2j, w = −2i+ 4j

9.

v = (−2,−1) , w = (−1, 3)10.

v = i+ 3j− k, w = 2i− j+ k

In problems 11-14

a) Determine the unit vector u along v (the normalizaton of v).

b) Determine the direction cosines of v.

11.

v = (2, 3)

12.

v = −i+2j

13.

v = (−3, 4)14.

v = i− 2j+ k

In problems 15-18, determine

a) the unit vector u along w,

b) compwv, the component of v along w,

c) Pwv, the projection of v along w,

d) v2 such that v = Pwv + v2 and v2 is orthogonal to w.

15.

v = (3, 4) , w = (6, 2)

16.

v = (−2, 1) , w = (2, 1)

17.

v = i− 2j, w = 2i− j18.

v = (−2,−6) , w = (−3,−4)

11.4 The Cross Product

In the previous section we introduced the dot product that assigns a scalar to a pair of vectors

and enables us to calculate the angle between them. In this section we will introduce the cross

product that assigns to a pair of vectors another vector that is orthogonal to both vectors.

The cross product has many physical and geometric applications. In this section we will de-

scribe a plane in terms of a vector that is orthogonal to that plane by making use of the cross

product. We will also introduce the scalar triple product that corresponds to the volume of

the parallelepiped spanned by three vectors.


The Definition of the Cross Product

Suppose that we wish to construct the cross product of the vectors v and w in R3 as anothervector in R3 that is denoted by v×w that has the following properties:

1. v×w is orthogonal to v and w

2. ||v×w|| is the area of the parallelogram that is spanned by v and w, i.e., ||v|| ||w|| sin (θ),where θ is the angle between v and w. In particular, v×w = 0 if v and w are parallel vectors.

3. The direction of v × w is determined by the right-hand rule: If you imagine that a right-

handed screw turns from v to w, the vector v×w points upwards. Thus, w× v = −v×w, sothat the cross product is anticommutative.

v

w

v x w

Figure 1

4. The cross product is distributive: If α is a scalar and u,v,w are vectors in R3 then

(αv)×w = α (v ×w) = v × (αw) ,v× (u+w) = v × u+ v× w.

In particular,

i× j = k, j× k = i, k× i = j,j× i = −k, k× j = −i, i× k = −j.

The above properties lead to the calculation of the product of vectors v = v1i+ v2j+ v3k andw = w1i+ w2j+ w3k:

v ×w = (v1i+ v2j+ v3k)× (w1i+ w2j+ w3k)= v1w2k− v1w3j− v2w1k+ v2w3i+ v3w1j− v3w2i= (v2w3 − v3w2) i− (v1w3 − v3w1) j+ (v1w2 − v2w1)k.=

¯v2 v3w2 w3

¯i−

¯v1 v3w1 w3

¯j+

¯v1 v2w1 w2

¯k

=

¯¯ i j k

v1 v2 v3w1 w2 w3

¯¯ .

The above expression can be remembered easily by introducing the "symbolic determinant"¯¯ i j k

v1 v2 v3w1 w2 w3

¯¯ .

This is a three-by-three array. The first row consists of the standard basis vectors i, j and k, the

second row is made up of the components of v and the third row is made up of the components


of w. The rule for the evaluation of this symbolic determinant is as follows:¯¯ i j k

v1 v2 v3w1 w2 w3

¯¯ = ¯ v2 v3

w2 w3

¯i−

¯v1 v3w1 w3

¯j+

¯v1 v2w1 w2

¯k.

Here the coefficients of i, j and k are two-by-two determinants:¯v2 v3w2 w3

¯= v2w3 − v3w2,¯

v1 v3w1 w3

¯= v1w3 − v3w1,¯

v1 v2w1 w2

¯= v1w2 − v2w1.

The two-by-two determinant that is the coefficient of i is obtained by deleting the first row and

the first column of the symbolic determinant. The two-by-two determinant that is the coefficient

of j is obtained by deleting the first row and the second column of the symbolic determinant.

The two-by-two determinant that is the coefficient of k is obtained by deleting the first row and

the third column of the symbolic determinant.

It can be shown that the cross product that is defined by the above expression has the properties

1-4.

Example 1 Let v = i− 3j+2k and w = 2i− j+ 4k.a) Determine v×w,b) Confirm that v×w is orthogonal to v and w.

c) Compute the area of the parallelogram spanned by v and w.

Solution

a)

v×w =

¯¯ i j k

1 −3 22 −1 4

¯¯

=

¯ −3 2−1 4

¯i−

¯1 22 4

¯j+

¯1 −32 −1

¯k

= −10i+ 5k.

v

wv x w

10

5

0

x

0 1 2 3

y

0

2

4

z

Figure 2


b)

(v×w) · v = (−10i+ 5k) · (i− 3j+2k) = 0,(v×w) ·w = (−10i+ 5k) · (2i− j+ 4k) = 0.

c) The area of the parallelogram spanned by v and w .is

||v×w|| =p102 + 52 =

√125.

¤

Example 2 Make use of the cross product to compute the area of the parallelogram that is

spanned by the vectors v = (2, 1) and w = (1, 2).

Solution

The cross product is defined for three dimensional vectors. Therefore, we consider the plane

to be embedded in R3, and consider the vectors v = (2, 1, 0) and w =(1, 2, 0) that span thesame parallelograms as the vectors v and w. Thus, we can compute the required area as the

magnitude of v× w.We have

v× w =

¯¯ i j k

2 1 01 2 0

¯¯

=

¯1 02 0

¯i−

¯2 01 0

¯j+

¯2 11 2

¯k = 3k.

Therefore, the area of the parallelogram that is spanned by the vectors v and w is 3. ¤

The Scalar Triple Product

The scalar triple product of the vectors u,v,w is

u · (v×w) =¯¯ u1 u2 u3v1 v2 v3w1 w2 w3

¯¯

The absolute value of the scalar triple product of u, v, w is the volume of the paral-

lelepiped that is spanned by u, v and w: Since v×w is perpendicular to the parallelogramspanned by v and w, the height of the parallelepiped is

||u|| cos (θ) ,

where θ is the angle between u and v×w. Thus, the volume of the parallelepiped is

(height)× (area of the base) = ||u|| |cos (θ)| × ||v ×w|| = |u · (v×w)| .


v

wuv x w

Θ

Figure 3

Example 3 Let

u = (1, 0, 4) , v = (1, 1, 1) and w = (−1, 1, 2) .Compute the volume of the parallelepiped spanned by u,v and w.

Solution

The volume of the parallelepiped spanned by u,v and w .is

u · (v×w) =¯¯ 1 0 41 1 1−1 1 2

¯¯ = 9

(confirm) .¤

Planes

Assume that the equation of a plane is given as

ax+ by + cz = d.

If P0 = (x0, y0, z0) is a given point on the plane and P = (x, y, z) is an arbitrary point on theplane, we have

ax+ by + cz = d and ax0 + by0 + cz0 = d,

so that

a (x− x0) + b (y − y0) + c (z − z0) = 0.Thus, if we set

N = ai+ bj+ ck,

we have

N ·³−−→P0P

´= 0.

We refer to a vector N such that N ·³−−→P0P

´= 0 for each P on the plane as a vector that is

orthogonal to the plane or a normal vector for the plane.


P0

N

Figure 4

Conversely, assume that a point P0 = (x0, y0, z0) is a given point on the plane and

N = ai+ bj+ ck

is orthogonal to the plane. If P = (x, y, z) is an arbitrary point on the plane, then

N ·³−−→P0P

´= 0.

Thus,

(ai+ bj+ ck) · ((x− x0) i+ (y − y0) j+ (z − z0)k) = 0.Therefore,

a (x− x0) + b (y − y0) + c (z − z0) = 0so that

ax+ by + cz = d,

where

d = ax0 + by0 + cz0.

Example 4 Assume that N = (1,−1, 2) is orthogonal to the plane Π and that (−2, 3, 2) is apoint on Π. Find an equation for the plane.

Solution

Set P0 = (−2, 3, 2) and let P = (x, y, z) be an arbitrary point on the plane. We have

N ·³−−→P0P

´= 0

so that

(i− j+ 2k) · ((x+ 2) i+ (y − 3) j+ (z − 2)k) = 0.Thus,

(x+ 2)− (y − 3) + 2 (z − 2) = 0,i.e.,

x− y + 2z = −1¤


A plane is also determined by specifying three points on the plane. Assume that P0 = (x0, y0, z0) , P1 =(x1, y1, z1) and P2 = (x2, y2, z2) are on the plane. Then

N =−−−→P0P1 ×−−−→P0P2

is orthogonal to the plane. We can determine an equation of the plane as before.

Example 5 Find an equation for the plane that contains the points P0 = (1, 2, 1), P1 =(−1, 1, 2) and P2 = (3,−2, 1).Solution

We have −−−→P0P1 = (−2,−1, 1) and −−−→P0P2 = (2,−4, 0) .

Therefore,

N =−−−→P0P1 ×−−−→P0P2 =

¯¯ i j k

−2 −1 12 −4 0

¯¯

=

¯ −1 1−4 0

¯i−

¯ −2 12 0

¯j+

¯ −2 −12 −4

¯k

= 4i+ 2j+ 10k.

A point P = (x, y, z) is on the plane iff

N ·³−−→P0P

´= 0,

i.e.,

(4i+ 2j+ 10k) · ((x− 1) i+ (y − 2) j+ (z − 1)k) = 0.Therefore,

4 (x− 1) + 2 (y − 2) + 10 (z − 1) = 0,so that

4x+ 2y + 10z = 18,

¤

Problems

In problems 1-4 determine v×w.

1.

v = (2, 1, 0) , w = (1, 3, 0)

2.

v = 3j+ 2k, w = −2j+ k

3.

v = (3, 1, 4) , w = (−2, 3, 1)4.

v = −2i+ j+ 4k, w = i− 2j+ 5k

In problems 5 and 6, Compute the area of the parallelogram spanned by v and w by making

use of the cross product.


5.

v = (−1, 3) , w = (2, 6)

6.

v = −i+ 3j− 2k, w = 4i− j− 6k

In problems 7 and 8, compute the volume of the parallelepiped spanned by u,v and w by making

use of the Scalar Triple Product.

7.

u = (2, 1, 3) , v = (3, 1, 0) , w = (1, 0, 4)

8.

u = 2i+ 2j+ 5k, v = 3i− j− 4k, w = i− jIn problems 9-12, assume that N is orthogonal to the plane Π and that P0 is a point on Π. Findan equation for the plane.

9.

N = (3, 2, 1) , P0 = (1, 3, 4)

10.

N = (1,−2, 4) , P0 = (2, 0, 5)

11.

N = −i+ j+ 2k, P0 = (2, 1,−3)

12

N = i− 4j− k, P0 = (3, 1, 2)

In problems 13 and 14, find an equation for the plane that contains the points P0, P1 and P2,

by making use of the cross product.

13.

P0 = (1, 2, 2) , P1 = (0, 2, 1) , P2 = (1,−3, 4)14.

P0 = (3,−2,−1) , P1 = (0, 4, 1) , P2 = (1, 3,−2)

Chapter 12

Functions of Several Variables

In this chapter we will discuss the differential calculus of functions of several variables. We will

study functions of a single variable that take values in two or three dimensions. The images

of such functions are parametrized curves that may model the trajectory of an object. We

will calculate their velocity and acceleration. Geometrically, we will calculate tangents

to curves and their curvature. Then we will take up the differential calculus real-valued

functions of several variables. We will study their rates of change in different directions

and their maxima and minima. In the process we will generalize the ideas of local linear

approximation and the differential from functions of a single variable to functions of several

variables.

12.1 Parametrized Curves, Tangent Vectors and Velocity

Parametrized Curves

Definition 1 Assume that σ is a function from an interval J to the plane R2. If

σ (t) = (x (t) , y (t)), then x (t) and y (t) are the component functions of σ. The letter t isthe parameter. The image of σ, i.e., the set of points C = (x (t) , y (t)) : t ∈ J is said tobe the curve that is parametrized by the function σ. We say that the function σ is a

parametric representation of the curve C.

We will use the notation σ : J → R2 to indicate that σ is a function from J into R2. Wecan identify the point σ (t) = (x (t) , y (t)) with its position vector, and regard σ as a vector-valued function whose values are two-dimensional vectors. We may use the notation σ (t) =x (t) i + y (t) j. It is useful to imagine that σ (t) is the position of a particle in motion at thefictitious time t. Figure 1 illustrates a parametrized curve and the "motion of the particle" is

indicated by arrows. In many applications, the parameter does refer to time.

35

36 CHAPTER 12. FUNCTIONS OF SEVERAL VARIABLES

x

y

Σt

Figure 1: A parametrized curve in the plane

Example 1 Let σ (t) = (cos (t) , sin (t)), where t ∈ [0, 2π].The curve that is parametrized by the function σ is the unit circle, since

cos2 (t) + sin2 (t) = 1.

The arrows in Figure 2 indicate the motion of a particle that is at σ (t) at ”time" t. ¤

1-1

1

-1

x

y

Figure 2: σ (t) = (cos (t) , sin (t))

We have to distinguish between a vector-valued function and the curve that is

parametrized by that function, as in the following example.

Example 2 Let σ2 (t) = (cos (2t) , sin (2t)), where t ∈ [0, 2π].Since

cos2 (2t) + sin2 (2t) = 1,

the function σ2 parametrizes the unit circle, just as the function σ of Example 1. But the

function σ2 6= σ. For example,

σ2 (π/4) = (cos (π/2) , sin (π/2)) = (0, 1) ,

whereas,

σ (π/4) = (cos (π/4) , sin (π/4)) =

µ1√2,1√2

¶.

If we imagine that the position of a particle is determined by σ2 (t), the particle revolves aroundthe origin twice on the time interval [0, 2π], whereas, a particle whose position is determined byσ revolves around the origin only once on the ”time" interval [0, 2π]. ¤

12.1. TANGENT VECTORS AND VELOCITY 37

Lines are basic geometric objects. Let P0 = (x0, y0) be a given point and let v = v1i+ v2j bea given vector. With reference to Figure 2, a line in the direction of v that passes through the

point P0 can be parametrized by

σ (t) =−−−→OP 0 + tv = x0i+ y0j+ t (v1i+ v2j)

= (x0 + tv1) i+ (y0 + tv2) j.

Thus, the component functions of σ are

x (t) = x0 + tv1 and y (t) = y0 + tv2

Here, the parameter t varies on the entire number line.

x

y

P0

O

Σttv

Figure 3: A line through P0 in the direction of v

Example 3 Determine a parametric representation of the line in the direction of the vector

v = i− 2j that passes through the point P0 = (3, 4).Solution

σ (t) =−−−→OP 0 + tv = 3i+ 4j+ t (i−2j)

= (3 + t) i+ (4− 2t) j.Thus, the component functions of σ are

x (t) = 3 + t and y (t) = 4− 2t.Figure 4 shows the line.

3 6x

4

8

12

y

P0

O

tv

Σt

Figure 4


We can eliminate the parameter t and obtain the relationship between x and y:

t = x− 3⇒ y = 4− 2 (x− 3) = −2x+ 10.

Note that the same line can be represented parametrically by other functions. Indeed, the

direction can be specified by any scalar multiple of v, and we can pick an arbitrary point on the

line. For example, if we set t = 1 in the above representation, Q0 = (4, 2) is a point on the line,and w = 2v = 2i− 4j is a scalar multiple of v. The function

σ (u) =−−−→OQ0 + uw = 4i+ 2j+ u (2i− 4j)

= (4 + 2u) i+ (2− 4u) j

is another parametric representation of the same line. ¤

Example 4 Imagine that a circle of radius r rolls along a line, and that P (θ) is a point on thecircle, as in Figure 5 ( θ = 0 when P is at the origin).

x

y

rΘ, r

ΘP

rΘ, 0x

y

r

Figure 5

We have

x (θ) = rθ − r sin (θ) = r (θ − sin (θ))

and

y (θ) = r − r cos (θ) = r (1− cos (θ)) ,

as you can confirm. The parameter θ can be any real number. Let’s set

σ (θ) = (x (θ) , y (θ)) = (r (θ − sin (θ)) , r (1− cos (θ))) ,−∞ < θ < +∞.

The curve that is parametrized by the function σ : R→ R2 is referred to as a cycloid. Figure6 shows the part of the cycloid that is parametrized by

σ (θ) = (2θ − 2 sin (θ) , 2− 2 cos (θ)) , where θ ∈ [0, 8π] .

¤


x

y

Figure 6

Similarly, we can consider parametrized curves in three dimensions:

Definition 2 Assume that σ is a function from an interval J to the three-dimensional

space R3. If σ (t) = (x (t) , y (t) , z (t)), then x (t), y (t) and z (t) are the component func-tions of σ. The letter t is the parameter. The image of σ, i.e., the set of points C =(x (t) , y (t) , z(t)) : t ∈ J is said to be the curve that is parametrized by the functionσ.

We will use the notation σ : J → R3 to indicate that σ is a function from J into R3. Wecan identify the point σ (t) = (x (t) , y (t) , z (t)) with its position vector, and regard σ as a

vector-valued function whose values are three-dimensional vectors. We may use the notation

σ (t) = x (t) i+ y (t) j+ z (t)k.

You can imagine that σ (t) is the position at time t of a particle in motion.

Example 5 Let σ (t) = (cos (t) , sin (t) , t), where t ∈ R. The curve that is parametrized by σis a helix.

Since

cos2 (t) + sin2 (t) = 1,

the curve that is parametrized by σ lies on the cylinder x2+ y2 = 1. Figure 7 shows part of thehelix.¤

-10

1.0

-5

0

5

z

10

0.51.0

y

0.0 0.5

x0.0-0.5

-0.5-1.0 -1.0

Figure 7


Tangent Vectors

Definition 3 The limit of σ : J → R2 at t0 is the vector w if

limt→t0

||σ (t)−w|| = 0

In this case we write

limt→t0

σ (t) = w.

Proposition 1 Assume that σ (t) = x (t) i+ y (t) j. We have

limt→t0

σ (t) = w = w1i+ w2j

if and only if limt→t0 x (t) = w1 and limt→t0 y(t) = w2. Thus,

limt→t0

(x (t) i+ y (t) j) =

µlimt→t0

x (t)

¶i+

µlimt→t0

y (t)

¶j.

Proof

Assume that

limt→t0

x (t) = w1 and limt→t0

y(t) = w2.

Since

||σ (t)−w||2 = (x (t)− w1)2 + (y (t)− w2)2 ,we have

limt→t0

||σ (t)−w||2 = limt→t0

(x (t)− w1)2 + limt→t0

(y (t)− w2)2 = 0.

Thus, limt→t0 σ (t) = w = w1i+ w2j.

Conversely, assume that limt→t0 σ (t) = w = w1i+ w2. Since

|x (t)− w1| ≤ ||σ (t)−w|| and |y (t)− w2| ≤ ||σ (t)−w|| ,

and limt→t0 ||σ (t)−w|| = 0, we have

limt→t0

|x (t)− w1| = 0 and limt→t0

|y (t)− w2| = 0.

Therefore,

limt→t0

x (t) = w1 and limt→t0

y(t) = w2.

¥

Definition 4 The derivative of σ : J → R2 at t is

lim∆t→0

σ (t+∆t)− σ (t)∆t

and will be denoted bydσ

dt,dσ

dt(t) ,

dσ (t)

dtor σ0 (t) .

If C is the curve that is parametrized by σ and σ0 (t) is not the zero vector, then σ0 (t) is avector that is tangent to C at σ (t).


x

y

Σt

Σt t Σt

Σt t

Figure 8

x

y

Σt

Figure 9

Proposition 2 If σ (t) = x (t) i+ y (t) j and x0 (t), y0 (t) exist, then

dσ

dt=dx

dti+

dy

dtj.

Proof

σ (t+∆t)− σ (t)∆t

=1

∆t(x (t+∆t) i+ y (t+∆t) j− x (t) i− y (t) j)

=x (t+∆t)− x (t)

∆ti+

y (t+∆t)− y (t)∆t

j.

Therefore,

dσ

dt= lim∆t→0

σ (t+∆t)− σ (t)∆t

= lim∆t→0

µx (t+∆t)− x (t)

∆t

¶i+ lim

∆t→0

µy (t+∆t)− y (t)

∆t

¶j

=dx

dti+

dy

dtj

¥

Example 6 Let σ (t) = cos (t) i+ sin(t)j. Determine σ0 (t).


Solution

σ0 (t) =µd

dtcos (t)

¶i+

µd

dtsin (t)

¶j = − sin (t) i+ cos (t) j.

Note that σ0 (t) is orthogonal to σ (t):

σ0 (t) · σ (t) = − sin (t) cos (t) + cos (t) sin (t) = 0.

¤

1 1x

1

1

y

Σt

Σ't

Figure 10

Definition 5 Let C be the curve that is parametrized by the function σ : J → R2. If σ0 (t) 6= 0,the unit tangent to C at σ (t) is

T (t) =σ0 (t)||σ0 (t)||

The curve C is said to be smooth at σ (t) if σ0 (t) 6= 0.

With reference to the curve of Example 6, σ0 (t) = − sin (t) i+ cos (t) j is of unit length, since

||σ0 (t)|| =qsin2 (t) + cos2 (t) = 1.

Therefore, T (t) = σ0 (t) = − sin (t) i+ cos (t) j.

Definition 6 If C is the curve that is parametrized by σ : J → R2 and σ0 (t0) 6= 0, the line

that is in the direction of σ0 (t0) and passes through σ (t0) is the tangent line to C at σ (t0) .

Note that a parametric representation of the tangent line is the function L : R→ R2, where

L (u) = σ (t0) + uσ0 (t0) .

Another parametric representation is

l(u) = σ (t0) + uT (t0) .

Example 7 Let σ (θ) = (2θ − 2 sin (θ) , 2− 2 cos (θ)), as in Example 4, and let C be the curvethat is parametrized by σ.


a) Determine T (θ). Indicate the values of θ such that T (θ) exists.b) Determine parametric representations of the tangent line to C at σ (π/2).

Solution

a)

σ0 (θ) = (2− 2 cos (θ)) i+ 2 sin (θ) jTherefore,

||σ0 (θ)|| =q(2− 2 cos (θ))2 + 4 sin2 (θ)

= 2

q1− 2 cos (θ) + cos2 (θ) + sin2 (θ)

= 2√2p1− cos (θ).

Thus, σ0 (θ) = 0 if cos (θ) = 1, i.e., if θ = 2nπ, n = 0,±1,±2,±3, . . .. If θ is not an integermultiple of 2π, the unit tangent to C at σ (θ) is

T (θ) =σ0 (θ)||σ0 (θ)|| =

1

2√2p1− cos (θ) ((2− 2 cos (θ)) i+ 2 sin (θ) j) .

Note that

σ (2nπ) = (2θ − 2 sin (θ) , 2− 2 cos (θ))|θ=2nπ = (4nπ, 0) .A tangent line to C does not exist at such points. Figure 6 is consistent with this observation.

The curve C appears to have a cusp at (4nπ, 0). Let’s consider the case θ = 2π. If the curve isthe graph of the equation y = y (x), by the chain rule,

dy

dx=

dy

dθdx

dθ

=2 sin (θ)

2 (1− cos (θ)) .

Therefore,

limx→4π−

dy

dx= lim

θ→2π−2 sin (θ)

2 (1− cos (θ)) .

We have

limθ→2π

sin (θ) = 0 and limθ→2π

(1− cos (θ)) = 0.By L’Hôpital’s rule,

limθ→2π−

2 sin (θ)

2 (1− cos (θ)) = limθ→2π

2 cos (θ)

2 sin (θ).

Since sin (θ) < 0 if θ < 2π and θ is close to 2π, and

limθ→2π

2 cos (θ) = 2 > 0, limθ→2π−

2 sin (θ) = 0,

we have

limx→4π−

dy

dx= lim

θ→2π−2 sin (θ)

2 (1− cos (θ)) = limθ→2π

2 cos (θ)

2 sin (θ)= −∞.

Similarly,

limx→4π+

dy

dx= +∞.

Therefore, the curve has a cusp at (4π, 0).

b)


We have

σ (π/2) = (2θ − 2 sin (θ) , 2− 2 cos (θ))|θ=π/2 = (π − 2, 2) ,and

T (π/2) =1

2√2p1− cos (θ) ((2− 2 cos (θ)) i+ 2 sin (θ) j)

¯¯π/2

=1

2√2(2i+ 2j) =

1√2(i+ j) .

A parametric representation of the line that is tangent to C at σ (π/2) is

L (u) = σ (π/2) + uσ0 (π/2)= (π − 2) i+ 2j+ u (2 (i+ j))= (π − 2 + 2u) i+ (2 + 2u) j

¤

The above definitions extend to curves in the three-dimensional space in the obvious manner:

if σ : J → R3, the limit of σ is the vector w if

limt→t0

||σ (t)−w|| = 0,

and we write

limt→t0

σ (t) = w.

If σ (t) = x (t) i+ y (t) j+ z (t)k, then

limt→t0

σ (t) = w1i+ w2j+ w3k

if and only if

limt→t0

x (t) = w1, limt→t0

y (t) = w2 and limt→t0

z (t) = w3.

Thus,

limt→t0

(x (t) i+ y (t) j+ z (t)k) =

µlimt→t0

x (t)

¶i+

µlimt→t0

y (t)

¶j+

µlimt→t0

z (t)

¶k.

The derivative of σ is

dσ

dt= lim∆t→0

σ (t+∆t)− σ (t)∆t

=dx

dti+

dy

dtj+

dz

dtk,

and the unit tangent is

T (t) =1

||σ0 (t)||σ0 (t) ,

provided that ||σ0 (t)|| 6= 0.

Example 8 Let σ (t) = (cos (t) , sin (t) , t) as in Example 5, and let C be the curve that is

parametrized by σ.


a) Determine T (t). Indicate the values of t such that T (t)exists.b) Determine a parametric representation of the tangent line to C at σ (π/3).Solution

a)

σ0 (t) = − sin (t) i+ cos (t) j+ k.Therefore,

||σ0 (t)|| =qsin2 (t) + cos2 (t) + 1 =

√2.

Thus, T (t) exists for each t ∈ R and

T (t) =1

||σ0 (t)||σ0 (t) =

1√2(− sin (t) i+ cos (t) j+ k) .

b) We have

σ (π/3) = cos (π/3) i+ sin (π/3) j+π

3k =

1

2i+

√3

2j+

π

3k,

and

σ0 (π/3) = − sin (π/3) i+ cos (π/3) j+ k = −√3

2i+

1

2j+ k.

A parametric representation of the line that is tangent to C at σ (π/6) is

L (u) = σ (π/3) + uσ0 (π/3)

=1

2i+

√3

2j+

π

3k+ u

Ã−√3

2i+

1

2j+ k

!

=

Ã1

2−√3

2u

!i+

Ã√3

2+1

2u

!j+

³π3+ u

´k.

¤

Velocity

Assume that σ is a function from an interval on the number line into the plane or three-

dimensional space, and that a particle in motion is at the point σ (t) at time t. Let ∆t denotea time increment. The average velocity of the particle over the time interval determined by t

and t+∆t isdisplacement

elapsed time=σ (t+∆t)− σ (t)

∆t.

We define the instantaneous velocity at the instant t as the limit of the average velocity as ∆tapproaches 0:

Definition 7 If σ : J → R2 or σ : J → R3 and a particle is at the point σ (t) at time t, thevelocity v (t) of the particle at the instant t is

v (t) =dσ

dt= lim∆t→0

σ (t+∆t)− σ (t)∆t

.

Note that v (t)is a vector in the direction of the tangent to C at σ (t) if v (t) 6= 0.

Example 9 Let σ (t) = (2t− 2 sin (t)) i + (2− 2 cos (t)) j, as in Example 7, and let C be the

curve that is parametrized by σ. Determine v (3π/4), v (π) and v (2π).


Solution

As in Example 7,

v(t) = σ0 (t) = (2− 2 cos (t)) i+ 2 sin (t) j.Therefore,

v (3π/4) =

µ2− 2 cos

µ3π

4

¶¶i+ 2 sin

µ3π

4

¶j =

Ã2− 2

Ã−√2

2

!!i+ 2

Ã√2

2

!j

=³2 +√2´i+√2j,

v (π) = (2− 2 cos (π)) i+ 2 sin (π) j = 4i,and

v (2π) = (2− 2 cos (2π)) i+ 2 sin (2π) j = 0.Note that the direction of the motion is not defined at t = 2π since v (2π) = 0 (and the curveC is not smooth at σ (2π)). ¤

Problems

In problems 1-4 assume that a curve C is parametrized by the function σ.

a) Determine σ0 (t) , σ0 (t0) and the unit tangent T (t0) to C at σ (t0).b) Determine the vector-valued function L that parametrizes the tangent line to C at σ (t0)(specify the direction of the line by the vector σ0 (t0)).

1.

σ (t) =¡t, sin2 (4t)

¢, −∞ < t < +∞, t0 = π/3.

2.

σ (t) = (2 cos (3t) , sin (t)) , 0 ≤ t ≤ 2π, t0 = π/4.

3.

σ (t) =

µ3t

t2 + 1,3t2

t2 + 1

¶, −∞ < t <∞, t0 = 1.

4.

σ (t) = (4 cos (3t) , 4 sin (3t) , 2t) , −∞ < t <∞, t0 = π/2.

In problems 5 and 6, the position of a particle at time t is σ (t). Determine the velocity functionv (t), the velocity at the instant t0, and the speed at t0. Express the result in the ij-notation.5

σ (t) =¡e−t sin (t) , e−t cos (t)

¢, t0 = π

6.

σ (t) = (t, arctan (t)) , t0 = 1

12.2 Acceleration and Curvature

Acceleration

Acceleration is the rate of change of velocity:

Definition 1 Let v(t) be the velocity of an object at time t. The acceleration a(t) of the

object at time t is the derivative of v at that instant:

a(t) = v0(t) =dv

dt.

12.2. ACCELERATION AND CURVATURE 47

If σ (t) is the position of the object at time t, then v (t) = σ0 (t). Therefore, a(t) is the secondderivative of σ:

a(t) = σ00 (t) =d2σ

dt2.

Example 1 Let σ (t) = (cos (2t) , sin (2t)) be the position of an object at time t. Determinethe acceleration of the object.

Solution

Note that σ (t) is the position vector of a point on unit circle, since

||σ (t)||2 = cos2 (2t) + sin2 (2t) = 1.We have

v (t) =dσ

dt= −2 sin (2t) i+ 2cos (2t) j,

so that

a (t) =dv

dt= −4 cos (2t) i− 4 sin(2t)j = −4σ (t) .

Since a (t) = −4σ (t), and σ (t) is in the radial direction, the acceleration vector points towardsthe origin. In particular, acceleration is not the 0 vector, even though the speed of the object

is constant:

||v (t)|| =q4 sin2 (2t) + 4 cos2 (2t) = 2.

There is no paradox, since acceleration is the rate of change of the velocity vector, and that

vector is not constant, even though its magnitude is constant. Figure 1 indicates the directions

of the velocity and acceleration vectors (the vectors have been scaled to fit the picture). ¤

1 1x

11

y

vt

at

Figure 1

Example 2 Let

σ (t) = (sin (2t) , cos (3t))

a) Determine a (t).b) Determine a (π/4) and a (3π/4).

Solution

a) The velocity at the instant t is

v (t) =dσ

dt= 2 cos (2t) i− 3 sin (3t) j,


so that acceleration at t is

a (t) =dv

dt= −4 sin (2t) i− 9 cos (3t) j.

b)

a(π/4) = −4 sin(π/2)i− 9 cos(3π/4)j = −4i+92

√2j,

a (3π/4) = −4 sin (3π/2) i− 9 cos (9π/4) j = 4i− 92

√2j

Figure 2 shows a (π/4) and a (3π/4) (the vectors have been scaled to fit the picture). ¤

1 1x

1

1

y

Figure 2

The Moving Frame

We can attach a special rectangular coordinate system to any point on a parametrized curve,

with one of the axes along the unit tangent to the curve at that point. It is useful to determine

the components of the acceleration vector with respect to this “moving frame ".

Let’s begin by stating a general fact. If the magnitude of a vector-valued function is a constant,

its derivative at any point is orthogonal to the vector at that point:

Proposition 1 If ||V (t)|| is constant then V0 (t) is orthogonal to V (t) .

Proof

Since ||V (t)||2 is a constant,d

dt||V (t)||2 = 0.

The product rule is applicable to the dot product:

d

dt(V (t) ·W (t)) =

µd

dtV (t)

¶·W (t) +V (t) ·

µd

dtW (t)

¶(exercise). Therefore,

0 =d

dt||V (t)||2 = d

dt(V (t) ·V (t)) = 2V0 (t) ·V (t) .

Thus, V0 (t)⊥V (t), as claimed.¥Assume that a curve C is parametrized by the function σ : J → Rn, where n = 2 or n = 3.Let T (t) be the unit tangent to the curve C at σ (t). As a corollary to Proposition 1, T0 (t) isorthogonal to T (t), since ||T (t)|| = 1 for each t ∈ J.


Definition 2 The principal unit normal is

N (t) =T0 (t)||T0 (t)|| .

Thus, N (t) is a unit vector that is perpendicular to the unit tangent T (t).

Example 3 Let σ (t) = (cos (2t) , sin (2t)), t ∈ [0, 2π], so that σ parametrizes the unit circle.

Determine T (t) and N (t).

Solution

We have

σ0 (t) = −2 sin (2t) i+ 2 cos (2t) j,so that

||σ0 (t)|| =q4 sin2 (2t) + 4 cos2 (2t) = 2.

Therefore,

T (t) =σ0 (t)||σ0 (t)|| =

1

2(−2 sin (2t) i+ 2 cos (2t) j) = − sin (2t) i+ cos (2t) j.

Thus,

T0 (t) = −2 cos (2t) i− 2 sin (2t) j.Therefore,

||T0 (t)|| = 2.Thus,

N (t) =T0 (t)||T0 (t)|| =

1

2(−2 cos (2t) i− 2 sin (2t) j) = − cos (2t) i+ sin (2t) j = −σ (t) .

Figure 3 shows T (π/8) and N (π/8) (the vectors have been scaled to fit the picture). bNotethat the unit normal points towards the origin. ¤

1 1x

11

y

Tt

Nt

Figure 3

Definition 3 Let σ : J ⊂ R → R3 and let C be the curve that is parametrized by σ. The

unit binormal to C at σ (t) is


B (t) = T (t)×N (t)The plane that is determined by T (t) and N (t) is referred to as the osculating plane forthe curve C at σ (t).

Thus, the unit binormal B (t) is orthogonal to the osculating plane at σ (t).

Tt

Nt

Bt

Figure 4: The moving frame

Example 4 Let σ (t) = (cos (2t) , sin (2t) , t), where t ∈ [0, 2π], and let C be the curve that is

parametrized by σ. Determine the unit binormal to C at σ (π/4).

Solution

We have

dσ

dt= −2 sin (2t) i+ 2 cos (2t) j+ k,

and ¯¯dσ

dt

¯¯=√5,

so that

T (t) =1√5

dσ

dt= − 2√

5sin (2t) i+

2√5cos (2t) j+

1√5k.

Therefore,

dT

dt= − 4√

5cos (2t) i− 4√

5sin (2t) j.

Thus, ¯¯dT

dt

¯¯=

4√5,

so that

N (t) =

√5

4

µ− 4√

5cos (2t) i− 4√

5sin (2t) j

¶= − cos (2t) i− sin (2t) j.


(Note that N (t) ⊥ T (t), as it should be). Therefore,

B (t) = T (t)×N (t) =¯¯ i j k

− 2√5sin (2t) 2√

5cos (2t) 1√

5

− cos (2t) − sin (2t) 0

¯¯

=

¯2√5cos (2t) 1√

5

− sin (2t) 0

¯i−

¯ − 2√5sin (2t) 1√

5

− cos (2t) 0

¯j

+

¯ − 2√5sin (2t) 2√

5cos (2t)

− cos (2t) − sin (2t)¯k

=1√5sin (2t) i− 1√

5cos (2t) j+

µ2√5sin2 (2t) +

2√5cos2 (2t)

¶k

=1√5sin (2t) i− 1√

5cos (2t) j+

2√5k

(Confirm that B (t) is orthogonal to T (t) and N (t)) .¤

Tt

Nt

Bt

1

0

1

x

1

0

1

y

0

1

2

3

z

Figure 5

Arc Length

Assume that C is a smooth curve that is parametrized by σ : [a, b] → R2, where σ (t) =(x(t), y(t)). As we discussed in Section 10.4, the arc length of C isZ b

a

sµdx

dt

¶2+

µdy

dt

¶2dt,

provided that the curve C is traversed exactly once by σ (t) as t varies from a to b. If σ (t) isthe position of an object at time t, then

v(t) =dσ

dt=dx

dti+

dy

dtj

is the velocity and

||v (t)|| = ||σ0 (t)|| =sµ

dx

dt

¶2+

µdy

dt

¶is the speed of the object at time t. Therefore,Z b

a

||v (t)|| dt =Z b

a

sµdx

dt

¶2+

µdy

dt

¶2dt


is the distance traveled by the object over the time interval [a, b]. We can state that thedistance that is traveled by the object during the time interval [a, b] is the integralof the speed of the object over the interval [a, b]. We can identify the arc length of thecurve C that is parametrized by σ with distance traveled, provided that C is traversed exactly

once by σ (t) as t varies from a to b.

Similarly, if σ (t) = (x (t) , y (t) , z (t)) parametrizes a curve C in R3 and C is traversed by σ (t)exactly once as t varies from a to b, the arc length of C is

Z b

a

¯¯dσ

dt

¯¯dt =

Z b

a

sµdx

dt

¶2+

µdy

dt

¶2+

µdz

dt

¶2dt.

If an object is at the point σ (t) at the instant t, the speed of the object at that instant is||v (t)|| = ||σ0 (t)||, so that the distance traveled by the object over the time interval [a, b] iscalculated by the same integral.

Example 5 Let C be parametrized by σ (t) = (cos (2t) , sin (2t) , t), where t ∈ [0, 2π]. Calculatethe arc length of C.

Solution

We havedσ

dt= −2 sin (2t) i+ 2 cos (2t) j+ k.

Therefore,

||σ0 (t)|| =q4 sin2 (2t) + 4 cos2 (2t) + 1 =

√5.

Thus, arc length of the path C isZ 2π

0

||σ0 (t)|| dt =Z 2π

0

√5dt =

√5 (2π) = 2

√5π.

¤

Definition 4 Given a smooth function σ:[a, b] → Rn, where n = 2 or n = 3, the arc lengthfunction corresponding to σ is

s (t) =

Z t

a

||σ0 (τ)|| dτ .

Assume that the curve C is traversed exactly once by σ (t) as t varies from a to b, and the arc

length of C is L. By the Fundamental Theorem of Calculus,

ds

dt= ||σ0 (t)|| > 0,

since σ is smooth. Therefore the arc length function is an increasing function from [a, b] to [0, L].and has an inverse. If we denote the inverse of the arc length function by t (s), the functionσ (t (s)), where s varies from 0 to L provides another parametrization of C, and will be referred

to as the parametrization C with respect to arc length.

Example 6 Let σ (t) = cos (2t) i + sin (2t) j, 0 ≤ t ≤ π. and let C be the unit circle that is

parametrized by σ. Determine the parametrization of C with respect to arc length.


Solution

We have

s (t) =

Z t

0

||σ0 (τ)|| dτ =Z t

0

||−2 sin (2τ) i+ 2 cos (2τ) j|| dτ

=

Z t

0

q4 sin2 (2τ) + 4 cos2 (2τ)dτ =

Z t

0

2dτ = 2t.

Since 0 ≤ t ≤ π, we have 0 ≤ s ≤ 2π. In this case,the arc length function is simply thelinear function s = 2t on the interval [0,π]. The inverse is t (s) = s/2, where s ∈ [0, 2π]. Theparametrization of C with respect to arc length is

σ (t (s)) = σ (s/2) = cos (s) i+ sin (s) j,

where 0 ≤ s ≤ 2π.¤In many cases, it is not feasible to obtain the arc length function and/or its inverse explicitly.

Nevertheless, the existence of the parametrization is a useful concept, as we will see in the

following subsections.

Curvature

Intuitively, the curvature of a curve is a measure of the rate at which it bends as a point

moves along that curve. The parametrization of the curve with respect to arc length provides a

way to quantify curvature that is independent of a particular parametrization: Curvature is an

"intrinsic" geometric property of a curve.

Definition 5 Assume that C is parametrized by σ. The curvature of C at σ (t) is ||dT/ds||Thus, curvature is the magnitude of the rate of change of the unit tangent with

respect to arc length. If we denote the curvature of C at σ (t) by κ (t), we have

κ (t) =

¯¯dT

ds

¯¯,

where dT/ds is evaluated at σ (t).

Proposition 2 The curvature is given by the expression

κ (t) =1

||σ0 (t)||¯¯dT

dt

¯¯Proof

Since s0 (t) > 0, the inverse t (s) exists. We have T = T (t (s)). The chain rule for functions ofa single variable leads to the expression

dT

ds=dt

ds

dT

dt,

as you can confirm as an exercise. Since t(s) denotes the inverse of the function defined by s(t),we have

dt

ds=

1

ds

dt

.


Therefore,dT

ds=dt

ds

dT

dt=

1

ds

dt

dT

dt=

1

||σ0 (t)||dT

dt.

Since the principal unit normal is

N(t) =

dT

dt¯¯dT

dt

¯¯ ,we have

dT

dt=

¯¯dT

dt

¯¯N(t).

Therefore,

κ (t) =

¯¯dT

ds

¯¯=

1

||σ0 (t)||dT

dt=

1

||σ0 (t)||¯¯dT

dt

¯¯||N(t)|| = 1

||σ0 (t)||¯¯dT

dt

¯¯,

since ||N(t)|| = 1.¥

Remark 1 Assume that C is a parametrized curve in the plane, and the unit tangent is ex-

pressed as

T (t) = cos (α (t)) i+ sin (α (t)) j,

where α (t) is the angle of inclination of T (t) with respect to the positive direction of the x-axis.Then

κ (t) =

¯dα

ds

¯.

Indeed,

dT

ds= − sin (α (t)) dα

dsi+ cos (α (t))

dα

dsj =

dα

ds(− sin (α (t)) i+ cos (α (t)) j) = dα

dsu (t) ,

where ||u (t)|| = 1.Therefore, ¯¯dT

ds

¯¯=

¯dα

ds

¯Thus, curvature is indeed a measure of the rate at which the tangent to the curve turns. ♦

Definition 6 Assume that the curve C is parametrized by the function σ. The radius of

curvature of C at σ (t) is 1/κ (t).

Example 7 Let σ (t) = (a cos (t) , a sin (t)), where a > 0, and let C be the curve that is

parametrized by σ. Determine the curvature and radius of curvature of C at σ (t) .

Solution

Note that C is the circle of radius a centered at the origin. We have

σ0 (t) = −a sin (t) i+ a cos(t)j,so that

||σ0 (t)|| = a,


and

T (t) =σ0 (t)||σ0 (t)|| =

1

a(−a sin (t) i+ a cos(2t)j) = − sin (t) i+ cos (t) j.

Therefore,

T0 (t) = − cos (t) i− sin (t) j.Thus

κ (t) =1

||σ0 (t)||¯¯dT

dt

¯¯=1

a||− cos (t) i− cos (t)|| = 1

a.

Therefore, the curvature of the circle at any point is the reciprocal of its radius. The radius of

curvature at any point is1

κ (t)=11

a

= a,

as it should be. ¤

Example 8 Assume that the curve C is parametrized by σ (t) = (cos (2t) , sin (2t) , t). Deter-mine the curvature and the radius of curvature of C at σ (t) .

Solution

As in Example 4

dσ

dt= −2 sin (2t) i+ 2 cos (2t) j+ k,

and ¯¯dσ

dt

¯¯=√5

so that

T (t) =1√5

dσ

dt= − 2√

5sin (2t) i+

2√5cos (2t) j+

1√5k.

Therefore, ¯¯dT

dt

¯¯=

4√5,

Thus,

κ (t) =1

||σ0 (t)||¯¯dT

dt

¯¯=

1√5

µ4√5

¶=4

5

and the radius of curvature is1

κ (t)=5

4

at any t. ¤

Tangential and Normal Components of Acceleration

Proposition 3 We have

a (t) =

µd

dt||v (t)||

¶T (t) + ||v (t)||

¯¯dT

dt

¯¯N (t) .

In particular, acceleration is in the osculating plane that is determined by T (t) and N (t).


Proof

a (t) =d

dtv (t) =

d

dt

µ||v (t)|| v (t)||v (t)||

¶=d

dt(||v (t)||T (t))

=

µd

dt||v (t)||

¶T (t) + ||v (t)||

µd

dtT (t)

¶.

Since

N (t) =

dT

dt¯¯dT

dt

¯¯ ,we have

dT

dt=

¯¯dT

dt

¯¯N (t) .

Therefore,

a (t) =

µd

dt||v (t)||

¶T (t) + ||v (t)||

µd

dtT (t)

¶=

µd

dt||v (t)||

¶T (t) + ||v (t)||

¯¯dT

dt

¯¯N (t) .

¥

Proposition 4

a (t) =

µd

dt||v (t)||

¶T (t) + κ (t) ||v (t)||2N (t) = d2s

dt2T (t) + κ (t)

µds

dt

¶2N (t)

and

κ (t) =||σ00 (t)× σ0 (t)||

||σ0 (t)||3 =||a (t)× v (t)||||v (t)||3

Remark 2 Note that the tangential component of acceleration is the "scalar acceleration"

s00 (t). The normal component is due to curvature, and it is curvature×(speed)2. The aboveformula for curvature is convenient for calculations. Also note that

||a (t)||2 =µd

dt||v (t)||

¶2+³κ (t) ||v (t)||2

´2,

since T (t) ⊥N (t) .♦The Proof of Proposition4

By Proposition 3,

a (t) =

µd

dt||v (t)||

¶T (t) + ||v (t)||

¯¯dT

dt

¯¯N (t) .

Since

κ (t) =1

||v (t)||¯¯dT

dt

¯¯,


we have ¯¯dT

dt

¯¯= κ (t) ||v (t)|| .

Therefore,

a (t) =

µd

dt||v (t)||

¶T (t) + ||v (t)||

¯¯dT

dt

¯¯N (t)

=

µd

dt||v (t)||

¶T (t) + ||v (t)||κ (t) ||v (t)||N (t)

=

µd

dt||v (t)||

¶T (t) + κ (t) ||v (t)||2N (t) .

Now,

a (t)× v (t) =µµ

d

dt||v (t)||

¶T (t) + κ (t) ||v (t)||2N (t)

¶× (||v (t)||T (t))

=

µd

dt||v (t)||

¶||v (t)||T (t)×T (t) + κ (t) ||v (t)||3N (t)×T (t)

= κ (t) ||v (t)||3N (t)×T (t) ,

since T (t)× T (t) = 0. Therefore,

||a (t)× v (t)|| = κ (t) ||v (t)||3 ||N (t)×T (t)|| = κ (t) ||v (t)||3 ,

since ||N (t)×T (t)|| = 1. Thus,

κ (t) =||a (t)× v (t)||||v (t)||3 =

||σ00 (t)× σ0 (t)||||σ0 (t)||3 ,

as claimed .¥

Example 9 Let σ (t) = (cos (2t) , sin (2t) , t), as in Example.8. Calculate κ (t) by making useof Proposition 4, and determine the tangential and normal components of acceleration.

Solution

We have

σ0 (t) = −2 sin (2t) i+ 2 cos (2t) j+ k,and

σ00 (t) = −4 cos (2t) i− 4 sin (2t) j.Therefore,

||σ0 (t)|| =√5.

We also have

σ00 (t)× σ0 (t) =

¯¯ i j k

−4 cos (2t) −4 sin (2t) 0−2 sin (2t) 2 cos (2t) 1

¯¯

=

¯ −4 sin (2t) 02 cos (2t) 1

¯i−

¯ −4 cos (2t)−2 sin (2t)

01

¯j+

¯ −4 cos (2t) −4 sin (2t)−2 sin (2t) 2 cos (2t)

¯k

= −4 sin (2t) i+ 4 cos (2t) j− 8k.


Therefore,

||σ00 (t)× σ0 (t)|| = √16 + 64 =√80.

Thus,

κ (t) =||σ00 (t)× σ0 (t)||

||σ0 (t)||3 =

√80

5√5=4√5

5√5=4

5.

We haveds

dt= ||v (t)|| = ||σ0 (t)|| = 5,

so that the tangential component of acceleration is 0:

d2s

dt2= 0.

The normal component of acceleration is

κ (t)

µds

dt

¶2=

µ4

5

¶(25) = 20.

¤

Proposition 5 If σ (t) = x (t) i+ y (t) j and C is the planar curve that is parametrized by σ,the curvature of C at σ (t) can be expressed as

κ (t) =|x0 (t) y00 (t)− y0 (t)x00 (t)|³(x0 (t))2 + (y0 (t))2

´3/2 .If C is the graph of y = f (x) ,

κ (x) =|f 0 (x)|³

1 + (f 0 (x))2´3/2

Proof

Since

σ0 (t) = x0 (t) i+ y0 (t) j,σ00 (t) = x00 (t) i+ y00 (t) j,

we have

||σ0 (t)|| =³(x0 (t))2 + (y0 (t))2

´1/2,

and

σ00 (t)× σ0 (t) =

¯¯ i j k

x0 (t) y0 (t) 0x00 (t) y00 (t) 0

¯¯

=

¯y0 (t) 0y00 (t) 0

¯i−

¯x0 (t) 0x00 (t) 0

¯i+

¯x0 (t) y0 (t)x00 (t) y00 (t)

¯k

= (x0 (t) y00 (t)− y0 (t)x00(t))k.


Therefore,

||σ00 (t)× σ0 (t)|| = |x0 (t) y00 (t)− y0 (t)x00(t)| .Thus,

κ (t) =||σ00 (t)× σ0 (t)||

||σ0 (t)||3 =|x0 (t) y00 (t)− y0 (t)x00(t)|µ³(x0 (t))2 + (y0 (t))2

´1/2¶3 = |x0 (t) y00 (t)− y0 (t)x00 (t)|³(x0 (t))2 + (y0 (t))2

´3/2 .In particular, if σ (x) = (x, f (x)),

κ (x) =|x0 (x) y00 (x)− y0 (x)x00 (x)|³(x0 (x))2 + (y0 (x))2

´3/2 =|f 00 (x)|³

1 + (f 0 (x))2´3/2 .

¥

Example 10 Let σ (θ) = (2θ − 2 sin (θ) , 2− 2 cos (θ)) and let C be the curve that is parame-

trized by σ. Determine the curvature of C at σ (π/4) and σ (π/2) .

Solution

We have

x (θ) = 2θ − 2 sin (θ) and y (θ) = 2− 2 cos (θ) .Therefore,

x0 (θ) = 2− 2 cos (θ) , x00 (θ) = 2 sin (θ) ,y0 (θ) = 2 sin (θ) , y00 (θ) = 2 cos (θ) .

Thus,

x0 (θ) y00 (θ)− y0 (θ)x00 (θ) = (2− 2 cos (θ)) (2 cos (θ))− (2 sin (θ)) (2 sin (θ))= 4 cos (θ)− 4 cos2 (θ)− 4 sin2 (θ)= 4 cos (θ)− 4,

and

(x0 (θ))2 + (y0 (θ))2 = (2− 2 cos (θ))2 + (2 sin (θ))2= 4− 4 cos (θ) + 4 cos2 (θ) + 4 sin2 (θ)= 8− 4 cos (θ) .

Thus,

κ (θ) =4 |cos (θ)− 1|8 (2− cos (θ))3/2

=4 (1− cos (θ))8 (2− cos (θ))3/2

Therefore,

κ (π/4) =4 (1− cos (π/4))8 (2− cos (π/4))3/2

=1−√2

2

2

Ã2−√2

2

!3/2 ∼= 9. 961 74× 10−2

and

κ (π/2) =4 (1− cos (π/2))8 (2− cos (π/2))3/2

=1

2¡23/2

¢ ∼= 0.176 777 .¤


Problems

In problems 1-4 the position of a particle at time t is σ (t). Determine the acceleration functiona (t) and the acceleration at the instant t0.

1.

σ (t) = (6 + 3 cos (4t) , 6 + 3 sin (4t)) , t0 = π/12

2.

σ (t) = (6t− 6 sin (2t) , 6t− 6 cos (2t)) , t0 = 03.

σ (t) = (t, arctan (t)) , t0 = 1

4.

σ (t) =¡tet, et

¢, t0 = 2

In problems In problems 5 and 6, assume that a curve C is parametrized by the function σ.

Determine the unit tangent T (t0), the principal unit normalN (t0) and the unit binormalB (t0),if applicable:

5.

σ (t) = (4 + 2 cos (t) , 3 + 2 sin (t)) , t0 = π/6

6.

σ (t) = (cos (3t) , sin (3t) , 4t) , t0 = π/2.

In problems 7 and 8, assume that a curve C is parametrized by the function σ.

a) Express the arc length function s (t) in terms of an integral,b) [C] Calculate s (t0) approximately with the help of the numerical integrator of your compu-tational utility:

7.

σ (t) = (2 cos (t) , 3 sin (t)) , t ≥ 0, t0 = π

8.

σ (t) =¡tet, et

¢, t ≥ 0, t0 = 1.

In problems 9 and 10 assume that a curve C is parametrized by the function σ. Calculate the

curvature at t0.

Hint: It is practical to use the formula

κ (t) =||σ00 (t)× σ0 (t)||

||σ0 (t)||3

9.

σ (t) = (3 cos (t) , 2 sin (t) , 0) , t ≥ 0, t0 = π/2

10.

σ (t) =¡t, et, 0

¢, t ≥ 0, t0 = 2.

In problems 11 and 12, σ (t) is the position of an object at time t. Determine the tangentialcomponent

d

dt||v (t)||

¯t=t0

and the normal component

κ (t0) ||v (t0)||2

of the acceleration of the object at t0.

12.3. REAL-VALUED FUNCTIONS OF SEVERAL VARIABLES 61

Hint: It is practical to use the formula

κ (t) =||σ00 (t)× σ0 (t)||

||σ0 (t)||3

11.

σ (t) = (cos (2t) , sin (2t) , 0) , t ≥ 0, t0 = π/6

12.

σ (t) =¡t, t2, 0

¢, t ≥ 0, t0 = 1.

12.3 Real-Valued Functions of Several Variables

In this section we will introduce scalar-valued (i.e., real-valued) functions of several independent

variables. The emphasis will be on functions of two variables. We will be able to visualize the

behavior of such functions since their graphs are surfaces in the three-dimensional space.

Real-Valued Functions of Two Variables

Definition 1 A real-valued (or scalar-valued) function f two independent variables x and

y is a rule that assigns a unique real number f (x, y) to each point (x, y) in a subset D of R2.The set D is the domain of f and the number f (x, y) is the value of f at (x, y). The rangeof f is the set of all possible values of f . The graph of f is the surface in R3 that consists ofpoints of the form (x, y, f (x, y)). If we set z = f (x, y), then z is the dependent variable of fand the graph of f is a subset of the Cartesian coordinate plane where the axes are labeled as

x, y and z.

We may refer to a real-valued function f with domain D by the notation f : D → R. As inthe case of a function of a single variable, we may refer to "the function f (x, y)" if the rulethat defines the function is defined by the single expression f (x, y). In such a case, it should beunderstood that the domain of f is its natural domain, i.e., the set of all (x, y) ∈ R2 such thatf (x, y) is defined.

Example 1 Let f (x, y) = x2+y2 for each (x, y) ∈ R2. The domain of f is the entire plane R2.The range of f is the set of all nonnegative numbers [0,+∞). The graph of f in the xyz-space isthe surface that consists of points (x, y, z) where (x, y) is an arbitrary point in R2 and x = x2+y2.Figure 1 shows the part of the graph of f in the viewing window [−3, 3]× [−3, 3]× [0, 18]. Eventhough such a picture depends on the viewing window, we may simply refer to "the graph of

f". ¤

0

5

10z

2

15

2

y

0

x

0-2

-2

Figure 1


Example 2 Let f (x, y) = x2 − y2 for each (x, y) ∈ R2. Figure 2 shows the graph of f . ¤

5

-10

0z

10

5

y

0

x0

-5 -5

Figure 2

Assume that f : D ⊂ R2 → R. A horizontal slice of the graph of f is the intersection of thegraph of f with a plane that is horizontal to the xy-plane. The projection of such a curve onto

the xy-plane is a level curve of f . Thus, a level curve of f is a curve in the xy-plane that is the

graph of an equation f (x, y) = c, where c is a constant. We may also consider vertical slices ofthe graph of f in order to have a better idea about the behavior of the function. Vertical slices

corresponding to x = c or y = c, where c is a constant, are especially useful for visualization.

Example 3 Let z = f (x, y) = x2+y2, as in 1. The level curve of f corresponding to z = c > 0is the graph of the equation

x2 + y2 = c.

This is a circle of radius√c centered at the origin. As c increases, the radius if the circle

increases. Figure 3 displays some level curves of f . The darker shading indicates smaller values

of f .

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

Figure 3

A vertical slice of the graph of f corresponding to the plane x = c projects onto the yz-planeas the graph of the equation

z = c2 + y2.

This is a parabola. Figure 4 displays some of these parabolas.

12.3. REAL-VALUED FUNCTIONS OF SEVERAL VARIABLES 63

-3 -2 -1 0 1 2 3

10

20

y

z

Figure 4

Similarly, a vertical slice of the graph of f corresponding to the plane y = c projects onto thexz-plane as the parabola that is the graph of the equation

z = x2 + c2.

¤

Example 4 Let z = f (x, y) = x2− y2, as in Example 2. The level curve of f corresponding toz = c is the graph of the equation

x2 − y2 = c.This is a hyperbola. Figure 5 displays some of these hyperbolas.

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

Figure 5

As in the previous example, the darker shading indicates that the function is decreasing, and

the lighter color indicates that the function is increasing. This is consistent with the saddle

shape of the graph as displayed in Figure 2

A vertical slice of the graph of f corresponding to the plane x = c projects onto the yz-planeas the graph of the equation

z = c2 − y2.This is a parabola. Figure 6 displays some of these parabolas.

-3 -2 -1 1 2 3

-10

10

y

z

Figure 6


Similarly, a vertical slice of the graph of f corresponding to the plane y = c projects onto thexz-plane as the parabola that is the graph of the equation

z = x2 − c2.

Figure 7 displays some of these parabolas.

-3 -2 -1 1 2 3

-10

10

x

z

Figure 7

¤

Real-Valued Functions of Three or More Variables

As in the case of two independent variables, a function f of three independent variables x, y and

z is a rule that assigns a real number f (x, y, z) to each (x, y, z) in a subset D of R3, referredto as the domain of f . If we set w = f (x, y, z), then the graph of f is the subset of the four-dimensional space R4 that consists of points of the form (x, y, z, f (x, y, z)), where (x, y, z) rangesover D. Since we cannot visualize the four-dimensional space, we cannot display the graph of

f as in the case of two independent variables. Nevertheless, we can consider the level surfaces

of f , and that is helpful in the understanding of the function: A level surface of f (x, y, z)consists of the points in R3 such that

f (x, y, z) = c,

where c is some constant.

Example 5 Let

f (x, y, z) = x2 − y2 − z2.Figure 8 and Figure 9 display the level surfaces of f such that f (x, y, z) = 1 and f (x, y, z) = −1,respectively. ¤

Figure 8

12.4. PARTIAL DERIVATIVES 65

Figure 9

In many applications, functions of more than three variables are encountered as well. Even

though we will discuss the ideas of differential and integral calculus mainly within the contexts

of functions of two or three variables, we will indicate possible extensions to more than three

variables.

Problems

[C] In problems 1-6, you will need to use your plotting device:

a) Plot the graph of the function f as the graph of the equation z = f (x, y) in an appropriateviewing window,

b) Plot some of the level curves of f .

1.

f (x, y) = (x− 2)2 + 2 (y − 1)2

2.

f (x, y) = 3 (x+ 1)2 + (y − 2)2

3.

f (x, y) = 4x2 − y2

4.

f (x, y) = (y − 2)2 − (x+ 1)2

5.

f (x, y) =¡x2 + y2

¢e−√x2+y2

6.

f (x, y) = sin (x) cos (y)

[C] In problems 7-10 you will need to use your plotting device. Plot some of the level surfacesof f (x, y, z).

7.

f (x, y, z) = x2 + y2 + z2

8.

f (x, y, z) = 4x2 + 9y2 + 16z2

9.

f (x, y, z) = x2 − y2 + z2

10.

f (x, y, z) = 4x2 − 9y2 − 16z2

12.4 Partial Derivatives

In this section we will discuss the rate of change of a function of several variables when one

the variables varies. This leads to the partial derivatives of the functions with respect to its

independent variables. We begin with the notion of the limit of a such a function at a point.


Limits and Continuity

As in the case of a real-valued function of a single variable, the limit of f (x, y) as (x, y) ap-proaches the point (x0, y0) is the number L if f (x, y) is as close to L as desired provided that(x, y) 6= (x0, y0) and (x, y) is sufficiently close to (x0, y0). In this case we write

lim(x,y)→(x0,y0)

f (x, y) = L.

Here is the precise definition:

Definition 1 Assume that f is a real-valued function of two variables and f (x, y) is definedat (x, y) if the distance of (x, y) from (x0, y0) is small enough, with the possible exception of(x0, y0) itself. The limit of f at (x0, y0) is the number L if, given any ε > 0 there exists δ > 0such that

(x, y) 6= (x0, y0) and ||(x− x0, y − y0)|| < δ

⇒|f (x, y)− L| < ε.

The definition has the following geometric interpretation: Given any ε > 0 there exists a disk Dδ

of radius δ centered at (x0, y0) such that for any (x, y) ∈ Dδ we have |f (x, y)− f (x0, y0)| < ε.

x

y

x0 , y0

x, y

Δ

Figure 1

As in the case of a function of a single variable, the real-valued function f (x, y) is said to becontinuous at (x0, y0) if

lim(x,y)→(x0,y0)

f (x, y) = f (x0, y0) .

The definitions of the limits and continuity of real-valued functions of more than two variables

are similar. For example,

lim(x,y,z)→(x0,y0,z0)

f (x, y, z) = L

if, given any ε > 0 there exists δ > 0 such that

(x, y, z) 6= (x0, y0, z0) and ||(x− x0.y − y0, z − z0)|| < δ

⇒|f (x, y, z)− L| < ε.

As a rule of the thumb, the functions that we will deal with will be continuous at

almost each point of their domain. At the end of this section you can find an example that

illustrates the rigorous proof of the continuity of a function of two variables. Such proofs are

best left to a course in advanced Calculus.


The Definition and Evaluation of Partial Derivatives

Let’s begin with functions of two variables.

Definition 2 The partial derivative of f with respect to x at (x, y) is

∂f

∂x(x, y) = lim

∆x→0f (x+∆x, y)− f (x, y)

∆x,

and the partial derivative of f with respect to y at (x, y) is

∂f

∂y(x, y) = lim

∆y→0f (x, y +∆y)− f (x, y)

∆y.

x

y

x, y x x, y

x, y y

Figure 2

We will also use the notations

fx, ∂xf , fy and ∂yf

Sincef (x+∆x, y)− f (x, y)

∆xis the average rate of change of f as the first independent variable varies from x to x+∆x,and

∂f

∂x(x, y) = lim

∆x→0f (x+∆x, y)− f (x, y)

∆x,

we can interpret fx (x, y) as the rate of change of f with respect to x at (x, y). Similarly,we can interpret fy (x, y) as the rate of change of f with respect to y at (x, y) .

Since y is kept fixed in the evaluation of fx (x, y), we can simply apply the rules for thedifferentiation of a function of a single variable to evaluate fx (x, y) by treating y asa constant. Similarly, we treat x as a constant and apply the familiar rules of differentiation

in order to evaluate fy (x, y).

Example 1 Let f (x, y) = 36− 4x2 − y2. Evaluate the partial derivatives ∂xf and ∂yf .

Solution

In order to evaluate fx, we treat y as a constant. Thus,

∂f

∂x=

∂

∂x

¡36− 4x2 − y2¢ = −4 ∂

∂x

¡x2¢− ∂

∂x

¡y2¢= −8x.

Similarly, we treat x as a constant to evaluate fy:

∂f

∂y=

∂

∂y

¡36− 4x2 − y2¢ = −2y.

¤


Example 2 Let f (x, y) = cos (x− y). Evaluate the partial derivatives ∂xf and ∂yf .

Solution

By the chain rule,

∂f

∂x(x, y) =

∂

∂xcos (x− y) =

Ãd

ducos (u)

¯u=x−y

!µ∂

∂x(x− y)

¶= − sin (x− y) (1) = − sin (x− y) ,

and

∂f

∂y(x, y) =

∂

∂ycos (x− y) =

Ãd

ducos (u)

¯u=x−y

!µ∂

∂y(x− y)

¶= − sin (x− y) (−1) = sin (x− y) .

¤The idea of partial differentiation extends to functions of more than two variables in a straight-

forward manner. In order to evaluate the partial derivative with respect to a specific variable, we

treat all the other variables as constants, and apply the rules for the differentiation of functions

of a single variable.

Example 3 Let

f (x, y, z) =px2 + y2 + z2,

so that f (x, y, z) is the distance of the point (x, y, z) from the origin. Determine ∂xf , ∂yf and

∂zf .

Solution

We have

∂f

∂x(x, y, z) =

∂

∂x

¡x2 + y2 + z2

¢1/2=

1

2

¡x2 + y2 + z2

¢−1/2µ ∂

∂x

¡x2 + y2 + z2

¢¶=

1

2

¡x2 + y2 + z2

¢−1/2(2x)

=xp

x2 + y2 + z2.

Similarly,∂f

∂y(x, y, z) =

ypx2 + y2 + z2

and∂f

∂z(x, y, z) =

zpx2 + y2 + z2

.

¤

The Geometric Interpretation of Partial Derivatives

We can visualize the graph of z = f (x, y) as a surface in the three-dimensional space withCartesian coordinates x, y and z. Let’s pick a point (x0, y0) in the xy-plane and consider theplane y = y0. This is a plane that is parallel to the xz-plane and intersects the graph of f

along a curve, as illustrated in Figure 3. We will refer to the curve as the x-coordinate curve

through the point (x0, y0, f (x0, y0)).


Figure 3

Since y is kept fixed at y0 and z = f (x, y0) on the x-coordinate curve through the point(x0, y0, f (x0, y0)), this curve can be parametrized by the function

σ1 (x) = (x, y0, f (x, y0)) .

The assignment x→ f (x, y0) defines a function of the single variable x whose derivative at x0 is∂xf (x0, y0), by the definition of the partial derivative. Therefore, we can determine a tangentvector to the x-coordinate curve through the point at (x0, y0, f (x0, y0)) as

dσ1

dx(x0) =

µ1, 0,

∂f

∂x(x0, y0)

¶= i+

∂f

∂x(x0, y0)k.

Similarly, the plane x = x0 is parallel to the yz-plane and intersects the graph of f along a curvethat can be parametrized by the function

σ2 (y) = (x0, y, f (x0, y)) .

We will refer to this curve as the y-coordinate curve that passes through (x0, y0, f (x0, y0)).A tangent vector to the y-coordinate curve through the point at (x0, y0, f (x0, y0)) canbe determined as

dσ2

dy(y0) =

µ0, 1,

∂f

∂y(x0, y0)

¶= j+

∂f

∂y(x0, y0)k.

Example 4 Let f (x, y) = 36−4x2−y2, as in Example 1. Determine the lines that are tangentat (1, 3, f (1, 3)) to the coordinate curves passing through that point.

Solution

As in Example 1∂f

∂x= −8x and ∂f

∂y= −2y.

Thus,∂f

∂x(1, 3) = −8x|x=1 = −8,

and∂f

∂y(1, 3) = −2y|y=3 = −6.

Therefore, a tangent vector to the x-coordinate curve passing through (1, 3, f (1, 3)) = (1, 3, 23)can be determined as µ

1, 0,∂f

∂x(1, 3)

¶= (1, 0,−8) = i− 8k.


The line that is tangent to this curve at (1, 3, f (1, 3)) can be parametrized by

L1 (u) = (1, 3, f (1, 3)) + u (1, 0,−8)= (1, 3, 23) + u (1, 0,−8) = (1 + u, 3, 23− 8u) ,

where u ∈ R.Similarly, the vector µ

0, 1,∂f

∂y(1, 3)

¶= (0, 1,−6) = j− 6k

is tangent to the y-coordinate curve passing through (1, 3, f (1, 3)) = (1, 3, 23). The line that istangent to this curve at (1, 3, f (1, 3)) can be parametrized by

L2 (u) = (1, 3, f (1, 3)) + u (0, 1,−6)= (1, 3, 23) + u (0, 1,−6) = (1, 3 + u, 23− 6u) .

Figure 4 illustrates the relevant coordinate curves and tangent lines. ¤

Figure 4

Example 5 Let f (x, y) = cos (x− y) as in Example 2. Determine the lines that are tangentat (π, 2π/3, f (π, 2π/3)) to the coordinate curves passing through that point.

Solution

As in Example 2,∂f

∂x(x, y) = − sin (x− y) ,

and∂f

∂y(x, y) = sin (x− y) .

Thus,

∂f

∂x(π, 2π/3) = − sin (x− y)|x=π, y=2π/3 = − sin

µπ − 2π

3

¶= − sin

³π3

´= −√3

2,

and∂f

∂y(π, 2π/3) = sin (x− y)|x=π, y=2π/3 = sin

µπ − 2π

3

¶= sin

³π3

´=

√3

2.

Therefore, the vector µ1, 0,

∂f

∂x(π, 2π/3)

¶=

Ã1, 0,−

√3

2

!


is tangent to the x-coordinate curve at (π, 2π/3, f (π, 2π/3)).We have

f (π, 2π/3) = cos

µπ − 2π

3

¶= cos

³π3

´=1

2.

Thus,the tangent line to the x-coordinate curve at (π, 2π/3, f (π, 2π/3)).can be parametrized by

L1 (u) =

µπ,2π

3, f (π, 2π/3)

¶+ u

Ã1, 0,−

√3

2

!=

µπ,2π

3,1

2

¶+ u

Ã1, 0,−

√3

2

!

=

Ãπ + u,

2π

3,1

2−√3

2u

!.

The vector µ0, 1,

∂f

∂y(π, 2π/3)

¶=

Ã0, 1,

√3

2

!is tangent to the y-coordinate curve at (π, 2π/3, 1/2).. The tangent line can be parameterizedby

L2 (u) =

µπ,2π

3,1

2

¶+ u

Ã0, 1,

√3

2

!=

Ãπ,2π

3+ u,

1

2+

√3

2u

!.

Figure 5 illustrates the relevant coordinate curves and tangent lines. ¤

x

z

y

Figure 5

Higher-Order Partial Derivatives

If f is a function of x and y, fx and fy are also functions of x and y. Therefore, we can

differentiate them with respect to x and y. Thus, we can compute

(fx)x =∂

∂x

µ∂f

∂x

¶.

This is a second-order partial derivative of f . We set

fxx (x, y) =∂2f

∂x2(x, y) = ∂x∂xf (x, y) =

∂

∂x

µ∂f

∂x

¶and refer to this second-order partial derivative as the second partial derivative of f with

respect to x.


Similarly, the second partial derivative of f with respect to y is

fyy (x, y) =∂2f

∂y2(x, y) = ∂y∂yf (x, y) =

∂

∂y

µ∂f

∂y

¶.

The mixed second-order partial derivatives are

fxy (x, y) =∂2f

∂y∂x(x, y) = ∂y∂xf (x, y) =

∂

∂y

µ∂f

∂x

¶,

and

fyx (x, y) =∂2f

∂x∂y(x, y) = ∂x∂yf (x, y) =

∂

∂x

µ∂f

∂y

¶.

Example 6 Let f (x, y) = e−x2−2y2 . Determine the second-order partial derivatives of f .

Solution

We have∂f

∂x(x, y) = −2xe−x2−2y2 and ∂f

∂y(x, y) = −4ye−x2−2y2 .

Therefore,

∂2f

∂x2(x, y) =

∂

∂x

³−2xe−x2−2y2

´= −2e−x2−2y2 − 2x

³−2xe−x2−2y2

´= −2e−x2−2y2 + 4x2e−x2−2y2 ,

∂2f

∂y2(x, y) =

∂

∂y

³−4ye−x2−2y2

´= −4e−x2−2y2 − 4y

³−4ye−x2−4y2

´= −4e−x2−2y2 + 16y2e−x2−2y2 ,

∂2f

∂y∂x(x, y) =

∂

∂y

³−2xe−x2−2y2

´= −2x

³−4ye−x2−2y2

´= 8xye−x

2−2y2 ,

and∂2f

∂x∂y(x, y) =

∂

∂x

³−4ye−x2−2y2

´= −4y

³−2xe−x2−2y2

´= 8xye−x

2−2y2 .

¤Note that

∂2f

∂x∂y(x, y) =

∂2f

∂y∂x(x, y)

in the above example. That is not an accident:

Theorem 1 If ∂∂f and ∂∂f are continuous in some open set D, then they are

equal on D.

The proof of this theorem is left to a course in advanced calculus.

The generalization of second-order partial derivatives to functions of more than two variables is

straightforward. Partial derivatives of order higher than two are also defined and calculated in

the obvious manner. For example,

∂3f

∂x2∂y(x, y) =

∂2

∂x2

µ∂f

∂y

¶(x, y) .


A Rigorous Proof of Continuity (Optional)

Example 7 If f (x, y) = x2 − y2 then f is continuous at (3, 2) .Solution

Let’s set x = 3 + h and y = 2 + k. Then

|f (3 + h, 2 + k)− f (3, 2)| =¯(3 + h)

2 − (2 + k)2 − 5¯

=¯h2 + 6h− k2 − 4k¯

≤ |h|2 + 6 |h|+ |k|2 + 4 |k|= |h| (|h|+ 6) + |k| (|k|+ 4) .

Let’s impose the restriction that dist ((x, y) , (3, 2)) < 1 so that |h| < 1 and |k| < 1. Then

|f (3 + h, 2 + k)− f (3, 2)| < 7 |h|+ 5 |k| < 7 (|h|+ |k|) .

Note that

(|h|+ |k|)2 ≤ (2max (|h| , |k|))2 = 4max³|h|2 , |k|2

´≤ 4

³|h|2 + |k|2

´so that

(|h|+ |k|) ≤ 2q|h|2 + |k|2 = 2dist ((x, y) , (3, 2)) .

Therefore,

|f (3 + h, 2 + k)− f (3, 2)| < 7 (|h|+ |k|) ≤ 14dist ((x, y) , (3, 2)) .Thus, if

dist ((x, y) , (3, 2)) < δ = min³1,

ε

14

´then

|f (x, y)− f (3, 2)| < ε.

Therefore f is continuous at (3, 2). ¤

Problems

In problems 1-14 compute the partial derivatives of the given function with respect to all of the

independent variables:

1.

f (x, y) = 4x2 + 9y2

2.

f (x, y) = 6x2 − 5y23.

f (x, y) =p2x2 + y2

4.

f (r, θ) = r2 cos (θ)

5.

f (x, y) = e−x2−y2

6.

f (x, y) = ln¡x2 + y2

¢

7.

f (x, y) = sin³p

x2 − y2´

8.

f (r, θ) = er2

tan (θ)

9.

f (x, y) = arctan

Ãyp

x2 + y2

!

10.

f (x, y) = arccos

Ãxp

x2 + y2

!


11.

f (x, y, z) =1p

x2 + y2 + z2

12.

f (x, y, z) = xyzex−y+2z

13.

f (x, y, z) = arcsin

Ã1p

x2 + y2 + z2

!

14.

f (ρ,ϕ, θ) = ρ cos (ϕ) sin (θ)

In problems 15 - 18, let C1 be the x-coordinate curve and let C2 be the y-coordinate curve that

passes through (x0, y0, f(x0, y0)).a) Determine tangent vectors to C1 and C2 at (x0, y0, f (x0, y0)),b) Determine parametric representations of the lines that are tangent toC1 and C2 at (x0, y0, f (x0, y0)).

15.

f (x, y) =px2 + y2, (x0, y0) = (3, 1) .

16.

f (x, y) = 10− x2 − y2, (x0, y0) = (2, 1) .

17.

f (x, y) = ex2+y2 , (x0, y0) = (1, 0) .

18.

f (x, y) = x2 − y2, (x0, y0) = (3, 2) .

In problems 19-24, compute the indicated higher-order partial derivatives:

19.

f (x, y) = 4x2 + 9y2,∂2f

∂x2,

∂2f

∂x∂y

20.

f (x, y) =1p

x2 + y2,∂2f

∂y2,

∂2f

∂y∂x

21.

f (x, y) = e−x2+y2 ,

∂2f

∂x2,

∂2f

∂y∂x

22.

f (x, y) = sin³p

x2 − y2´,∂2f

∂y2,

∂2f

∂x∂y

23.

f (x, y, z) = xyzex−y+2z,∂2f

∂x∂z,

∂2f

∂y∂z

24.

f (ρ,ϕ, θ) = ρ cos (ϕ) sin (θ) ,∂2f

∂ϕ∂θ,

∂2f

∂ρ∂θ

In problems 25 and 26, evaluate fxy and fyx and confirm that they are equal:

25.

f (x, y) = ln¡x2 + 4y2

¢ 26.

f (x, y) = x2yex2−2y2

12.5 Linear Approximations and the Differential

Local Linear Approximations

Assume that f is a function of a single variable that is differentiable at x0. The linear ap-

proximation to f based at x0 is the linear function

Lx0 (x) = f (x0) + f0 (x0) (x− x0) .

The graph of Lx0 is the tangent line to the graph of f at (x0, f (x0)). We saw that Lx0 (x)approximates f (x) very well if x is close to the basepoint x0. Indeed,

f (x0 +∆x) = Lx0 (x0 +∆x) +∆xQ (∆x) ,

12.5. LINEAR APPROXIMATIONS AND THE DIFFERENTIAL 75

where

lim∆x→0

Q (∆x) = 0.

Thus, the magnitude of the error in the approximation of f (x0 +∆x) by Lx0 (x0 +∆x) is muchsmaller than |∆x| if |∆x| is small.What is the appropriate generalization of the idea of local linear approximation to scalar func-

tions of more than one variable? Let’s begin with functions of two variables. Assume that

f (x, y) is defined for each (x, y) in some disk containing the point (x0, y0), and that the partialderivatives ∂xf (x0, y0) and ∂yf (x0, y0) exist. We would like to determine a linear function Lthat approximates f well near (x0, y0). It is natural to require that L (x0, y0) = f (x0, y0). Thus,we can express L as

L (x, y) = f (x0, y0) +A (x− x0) +B (x− x0) ,where A and B are constants. The graph of L is a plane that passes through the point

(x0, y0, f (x0, y0)). It is also natural to require that

∂L

∂x(x0, y0) =

∂f

∂x(x0, y0) and

∂L

∂y(x0, y0) =

∂f

∂y(x0, y0) .

We have∂L

∂x(x, y) = A and

∂L

∂y(x, y) = B

for each (x, y) ∈ R2. Therefore, we will set

A =∂f

∂x(x0, y0) and B =

∂f

∂y(x0, y0) .

Thus,

L (x, y) = f (x0, y0) +∂f

∂x(x0, y0) (x− x0) + ∂f

∂y(x0, y0) (y − y0) .

Definition 1 The linear approximation to f based at (x0, y0) is the linear function

L(x0,y0) (x, y) = f (x0, y0) +∂f

∂x(x0, y0) (x− x0) + ∂f

∂y(x0, y0) (y − y0) .

The graph of L(x0,y0) is the tangent plane to the graph of f at (x0, y0, f (x0, y0)).

Example 1 Let f (x, y) = 36 − 4x2 − y2. Determine the linear approximation to f based at(1, 3) .

Solution

We have∂f

∂x(x, y) = −8x and ∂f

∂y(x, y) = −2y.

Therefore,∂f

∂x(1, 3) = −8 and ∂f

∂y(1, 3) = −6,

and f (1, 3) = 23. Therefore, the linear approximation to f based at (1, 3) is

L(1,3) (x, y) = f (1, 3) +∂f

∂x(1, 3) (x− 1) + ∂f

∂y(1, 3) (y − 3)

= 23− 8 (x− 1)− 6 (y − 3) .


Figure 1 illustrates the graphs of f and L(1,3). Note that the graph of L(1,3) is consistent with

the intuitive idea of a tangent plane to the surface that is the graph of f . ¤

Figure 1

Remark 1 If we set

z = L(x0,y0) (x, y) = f (x0, y0) +∂f

∂x(x0, y0) (x− x0) + ∂f

∂y(x0, y0) (y − y0) ,

we can express the equation of the tangent plane to the graph of f at (x0, y0, f (x0, y0)) as

−∂f∂x(x0, y0) (x− x0)− ∂f

∂y(x0, y0) (y − y0) + (z − f (x0, y0)) = 0.

Thus, the vector

N(x0,y0) = −∂f

∂x(x0, y0) i− ∂f

∂y(x0, y0) j+ k

is orthogonal to the tangent plane. With reference to Example 1,

N(1.3) = −∂f∂x(1, 3) i− ∂f

∂y(1.3) j+ k =− (−8) i− (−6) j+ k =8i+ 6j+ k.

In Section 12.4 we saw that the vector

i+∂f

∂x(x0, y0)k

is tangential to the x-coordinate curve that passes through (x0, y0, f (x0, y0)) at that point.Similarly, the vector

j+∂f

∂y(x0, y0)k

is tangential to the y-coordinate curve that passes through (x0, y0, f (x0, y0)) at that point. Itis reasonable to declare that the plane that is tangent to the graph of f at (x0, y0, f (x0, y0)) isspanned by these vectors. A normal vector to that plane can be determined as the cross product

of these tangent vectors. Indeed,

N(x0,y0) =

¯¯ i j k

1 0 ∂xf (x0, y0)0 1 ∂yf (x0, y0)

¯¯ = −∂xf (x0, y0) i− ∂yf (x0, y0) j+ k

Thus, we are led to the same idea of a tangent plane geometrically and via the idea of a local

linear approximation. ♦


We say that a function f of a single variable is differentiable at x0 if f0 (x0) exists. Shall we

say that a function of two variables is differentiable at a point if its partial derivatives exist at

that point? In the case of a function of a single variable, we know that differentiability implies

continuity. It is natural to require that a function of two variables should be continuous at a

point if it is declared to be differentiable at that point. The following example shows that it is

not suitable to define differentiability in terms of the existence of partial derivatives:

Example 2 Let

f (x, y) =

⎧⎨⎩ x2y

x4 + y2if (x, y) 6= (0, 0) ,

0 if (x, y) = (0, 0)

Both partial derivatives of f exist at (0, 0):

∂f

∂x(0, 0) = lim

h→0f (h, 0)− f (0, 0)

h= limh→0

(0) = 0,

∂f

∂y(0, 0) = lim

h→0f (0, h)− f (0, 0)

h= limh→0

(0) = 0.

But the function is not continuous at (0, 0). Indeed,

limy→0

f (0, y) = 0, limx→0

f (x, 0) = 0

but

limx→0

f¡x, x2

¢= lim

x→0x4

2x4= limx→0

1

2=1

26= 0.

Figure 2 shows the graph of f . The picture gives an indication of the erratic behavior of the

function near the origin. ¤

-0.52

0.0z

1

0.5

y

0

2-1 1

x0

-1-2 -2

Figure 2

We will define the differentiability of a function of two variables by generalizing the approxima-

tion property of local linear approximations for a function of a single variable. Note that the

value of L(x0,y0) (x, y) at (x0 +∆x, y0 +∆y) is

L(x0,y0) (x0 +∆x, y0 +∆y) = f (x0, y0) +∂f

∂x(x0, y0) (x0 +∆x− x0) + ∂f

∂y(x0, y0) (y0 +∆y − y0)

= f (x0, y0) +∂f

∂x(x0, y0)∆x+

∂f

∂y(x0, y0)∆y.


Definition 2 Assume that f (x, y) is defined for each (x, y) in some open disk that contains(x0, y0) and that the partial derivatives of f at (x0, y0) exist. We say that f is differentiableat (x0, y0) if

f (x0 +∆x, y0 +∆x) = L(x0,y0) (x0 +∆x, y0 +∆y) + ||(∆x,∆y)||Q (∆x,∆y)= f (x0, y0) +

∂f

∂x(x0, y0)∆x+

∂f

∂y(x0, y0)∆y + ||(∆x,∆y)||Q (∆x,∆y) ,

where

lim∆x→0,∆y→0

Q (∆x,∆y) = 0.

Intuitively, we require that the magnitude of the error in the approximation of

f (x0 +∆x, y0 +∆x) by L(x0,y0) (x0 +∆x, y0 +∆y) to be much smaller than the distanceof (x0 +∆x, y0 +∆y) from the basepoint (x0, y0) near that point.

Example 3 Let f (x, y) = 36− 4x2 − y2, as in Example 1. Show that f is differentiable at theorigin.

Solution

We showed that

L(1,3) (x, y) = 23− 8 (x− 1)− 6 (y − 3) .Therefore

L (1 +∆x, 3 +∆y) = 23− 8 (1 +∆x− 1)− 6 (3 +∆y − 3) = 23− 8∆x− 6∆y.

Thus

f (1 +∆x, 3 +∆y)− L (1 +∆x, 3 +∆y)= 36− 4 (1 +∆x)2 − (3 +∆y)2 − (23− 8∆x− 6∆y)= 36− 4− 8∆x− 4 (∆x)2 − 9− 6∆y − (∆y)2 − 23 + 8∆x+ 6∆y= −4 (∆x)2 − (∆y)2 .

With reference to Definition 2, the error in the approximation is .

−4 (∆x)2 − (∆y)2 = ||(∆x,∆y)||Q (∆x,∆y) ,

so that

Q (∆x,∆y) =4 (∆x)2 + (∆y)2

||(∆x,∆y)||We need to show that lim∆x→0,∆y→0Q (∆x,∆y) = 0. Indeed,

Q (∆x,∆y) =4 (∆x)

2+ (∆y)

2

||(∆x,∆y)|| ≤4³(∆x)2 + (∆y)2

´||(∆x,∆y)|| = 4

||(∆x,∆y)||2||(∆x,∆y)|| = 4 ||(∆x,∆y)|| .

Therefore,


Q (∆x,∆y) = 0.

Thus, f is differentiable at (0, 0). ¤

Proposition 1 If f is differentiable at (x0, y0) then f is continuous at (x0, y0) .


Proof

Since f is differentiable at (x0, y0),

f (x0 +∆x, y0 +∆x) = f (x0, y0) +∂f

∂x(x0, y0)∆x+

∂f

∂y(x0, y0)∆y + ||(∆x,∆y)||Q (∆x,∆y) ,

where lim∆x→0,∆y→0Q (∆x,∆y) = 0. Therefore,

lim(∆x,∆y)→(0,0)

f (x0 +∆x, y0 +∆x) = f (x0, y0) +∂f

∂x(x0, y0)

µlim

(∆x,∆y)→(0,0)∆x

¶+

∂f

∂y(x0, y0)

µlim

(∆x,∆y)→(0,0)∆y

¶+ lim∆x→0,∆y→0

||(∆x,∆y)||Q (∆x,∆y)

= f (x0, y0) ,

Thus, f is continuous at (x0, y0) . ¥Proposition 1 shows that the function of Example 2 is not differentiable at (0, 0) since it is notcontinuous at (0, 0).

In general, it is not easy to show that a function is differentiable at a point in accordance with

Definition 2. On the other hand, the continuity of the partial derivatives ensures differentiability:

Theorem 1 Assume that the partial derivatives of f exist in some open disk that

contains (x0, y0), and that they are continuous at (x0, y0). Then f is differentiable at(x0, y0) .

The proof of Theorem 1is left to a course in advanced calculus.

Example 4 Let

f (x, y) =px2 + y2.

a) Show that f is differentiable at (3, 4) and determine the linear approximation to f based at(3, 4).b) Make use of the result of part a) in order to approximate f (3.1, 3.9). Compare the absoluteerror with the distance from the basepoint.

Solution

a) We have

∂f

∂x(x, y) =

∂

∂x

¡x2 + y2

¢1/2=1

2

¡x2 + y2

¢−1/2(2x) =

xpx2 + y2

.

Similarly,∂f

∂y(x, y) =

ypx2 + y2

.

Thus, the partial derivatives of f are continuous at any point other than the origin. By Theorem

1 f is differentiable at any point other than the origin. In particular, f is differentiable at (3, 4).We have

∂f

∂x(3, 4) =

3

5,

∂f

∂y(3, 4) =

4

5and f (3, 4) = 5.

Thus, the linear approximation to f based at (3, 4) is

L (x, y) = f (3, 4) +∂f

∂x(3, 4) (x− 3) + ∂f

∂y(3, 4) (y − 4)

= 5 +3

5(x− 3) + 4

5(y − 4) .


The graph of L is the tangent plane to the graph of f at (3, 4, f (3, 4)) = (3, 4, 5), as shown inFigure 3.

Figure 3

b) We have

f (3.1, 3.9) =

q(3.1)2 + (3.9)2 ∼= 4.981 97,

rounded to 6 significant digits. The corresponding value of the linear approximation is

L (3.1, 3.9) = 5 +3

5(0.1) +

4

5(−0.1) = 4. 98.

Therefore, the absolute error is approximately 1.97× 10−3. This number is much smaller thanthe distance of (3.1, 3.9) from the basepoint (3.4):q

(0.1)2 + (−0.1)2 =p2× 10−2 =

√2× 10−1 ∼= 0.14

¤

The Differential

Recall that the differential of a function of a single variable is a convenient way of keeping track

of local linear approximations as the basepoint varies. If f is differentiable at x,

f (x+∆x)− f (x) ∼= df (x,∆x) = df

dx(x)∆x,

and df (x,∆x) is the change along the tangent line (x, f (x)) corresponding to the increment∆x.Assume that f is a function of two variables that is differentiable at (x, y). For example, thepartial derivatives of f are continuous in some disk centered at (x, y). Let L be the linear

approximation to f based at (x, y), so that

L (u, v) = f (x, y) +∂f

∂x(x, y) (u− x) + ∂f

∂y(x, y) (v − y) .

Therefore,

L (x+∆x, y +∆y) = f (x, y) +∂f

∂x(x, y)∆x+

∂f

∂y(x, y)∆y.

Thus,

f (x+∆x, y +∆y) = f (x, y) +∂f

∂x(x, y)∆x+

∂f

∂y(x, y)∆y + ||(∆x,∆y)||Q (∆x,∆y) ,


where


Q (∆x,∆y) = 0.

Therefore,

f (x+∆x, y +∆y)− f (x, y) ∼= ∂f

∂x(x, y)∆x+

∂f

∂y(x, y)∆y,

and the magnitude of the error is much smaller than ||(∆x,∆y)|| if ||(∆x,∆y)|| is small. Theright-hand side defines the differential of f :

Definition 3 Assume that f is differentiable at (x, y). The differential of f is defined by theexpression

df (x, y,∆x,∆y) =∂f

∂x(x, y)∆x+

∂f

∂y(x, y)∆y.

Thus, the differential of a function of two variables is a function of four variables. For a given

(x, y), the function

(∆x,∆y)→ ∂f

∂x(x, y)∆x+

∂f

∂y(x, y)∆y

is a linear function. We have

f (x+∆x, y +∆y)− f (x, y) ∼= df (x, y,∆x,∆y)

if ||(∆x,∆y)|| is small.It is traditional to denote ∆x and ∆y as by dx and dy within the context of differentials. Thus,

df (x, y, dx, dy) =∂f

∂x(x, y) dx+

∂f

∂y(x, y) dy.

It is also traditional to be cryptic and simply write

df =∂f

∂x(x, y) dx+

∂f

∂y(x, y) dy

as the differential of the function f .

Example 5 Let f (x, y) =px2 + y2, as in Example 4.

a) Express the differential of f .

b) Make use of the differential in order to approximate f (1.9, 5.2) .

Solution

a) As in Example 4,

∂f

∂x(x, y) =

xpx2 + y2

and∂f

∂y(x, y) =

ypx2 + y2

.

Therefore,

df (x, y,∆x,∆y) =∂f

∂x(x, y)∆x+

∂f

∂y(x, y)∆y =

xpx2 + y2

∆x+yp

x2 + y2∆y.

In the traditional notation.

df =xp

x2 + y2dx+

ypx2 + y2

dy


b) It is reasonable to set x = 2 and y = 5 in order to approximate f (1.9, 5.2). Thus,

f (1.9, 5.2)− f (2, 5) ∼= df (2, 5,−0.1, 0.2) = ∂f

∂x(2, 5) (−0.1) + ∂f

∂y(2, 5) (0.2)

=2√29(−0.1) + 5√

29(0.2) ∼= 0.148 556.

Therefore,

f (1.9, 5.2) ∼= f(2, 5) + 0.148 556. =√29 + 0.148 556 ∼= 5. 533 72.

We have

f (1.9, 5.2) =

q(1.9)

2+ (5.2)

2 ∼= 5. 536 24,rounded to 6 significant digits. Thus, the absolute error in the approximation of f (1.9, 5.2) viathe differential is approximately 2.5× 10−3. This is much smaller than the distance of (1.9, 5.2)from the basepoint (2, 5) (which is approximately 0.2236). ¤

Functions of more than two variables

The generalization of the idea of local linear approximations and the related idea of the differ-

ential to functions of more than two variables is straightforward. For example, if f is a function

of three variables, the linear approximation to f based at (x0, y0, z0) is

L (x, y, z) = f (x0, y0, z0) +∂f

∂x(x0, y0, z0) (x− x0)

+∂f

∂y(x0, y0, z0) (y − y0) + ∂f

∂z(x0, y0, z0) (z − z0) .

We say that f is differentiable at (x0, y0, z0) if

f (x0 +∆x, y0 +∆x, z0 +∆z) = f (x0, y0, z0) +∂f

∂x(x0, y0, z0)∆x+

∂f

∂y(x0, y0, z0)∆y +

∂f

∂z(x0, y0, z0)∆z

+ ||(∆x,∆y,∆z)||Q (∆x,∆y,∆z) ,where

lim∆x→0,∆y→0,∆z→0

Q (∆x,∆y,∆z) = 0

Differentiability implies continuity and a sufficient condition for differentiability is the continuity

of the partial derivatives.

The differential of f is

df (x, y, z,∆x,∆y,∆z) =df

∂x(x, y, z)∆x+

∂f

∂y(x, y, z)∆y +

∂f

∂z(x, y, z)∆z.

We have

f (x+∆x, y +∆x, z +∆z)−f (x, y, z) = df (x, y, z,∆x,∆y,∆z)+||(∆x,∆y,∆z)||Q (∆x,∆y,∆z) ,where

lim∆x→0,∆y→0,∆z→0

Q (∆x,∆y,∆z) = 0

Cryptically,

df =df

∂xdx+

∂f

∂ydy +

∂f

∂zdz.

As in the case of two variables,

f (x+ dx, y + dy, z + dz)− f (x, y, z) ∼= dfif ||(dx, dy, dz)|| is small.


Example 6 Let f (x, y, z) =px2 + y2 − z2.

a) Determine the linear approximation to f based at (2, 2, 1). Make use of the result to approx-imate (2.1, 1.8, 0.9)b) Determine the differential of f . Make use of the result to approximate f (−3.1, 1.1, 1.9) .Solution

a) We have

∂f

∂x(x, y, z) =

xpx2 + y2 − z2 ,

∂f

∂y(x, y, z) =

ypx2 + y2 − z2 ,

∂f

∂z(x, y, z) = − zp

x2 + y2 − z2 .

Therefore,∂f

∂x(2, 2, 1) =

2√7,∂f

∂y(2, 2, 1) =

2√7and

∂f

∂z(2, 2, 1) = − 1√

7.

Thus, the linear approximation to f based at (2, 2, 1) is

L (x, y, z) = f (2, 2, 1) +∂f

∂x(2, 2, 1) (x− 2) + ∂f

∂y(2, 2, 1) (y − 2) + ∂f

∂z(2, 2, 1) (z − 1)

=√7 +

2√7(x− 2) + 2√

7(y − 2)− 1√

7(z − 1) .

Therefore,

f (2.1, 1.8, 0.9) ∼= L (2.1, 1.8, 0.9)=√7 +

2√7(0.1) +

2√7(−0.2)− 1√

7(−0.1) ∼= 2. 607 95

Note that

f (2.1, 1.8, 0.9) =

q(2.1)2 + (1.8)2 − (0.9)2 ∼= 2. 615 34.

Thus, the absolute error in the approximation is approximately 6.8 × 10−2. The distance of(2.1, 1.8, 0.9) from the basepoint (2, 2, 1) isq

(0.1)2+ (0.2)

2+ (0.1)

2 ∼= 0.245We see that the absolute error in the approximation is much smaller than the distance from the

basepoint.

b) In the traditional notation, the differential of f is

df =df

∂xdx+

∂f

∂ydy +

∂f

∂zdz

=xp

x2 + y2 − z2 dx+yp

x2 + y2 − z2 dy −zp

x2 + y2 − z2 dz.

It is reasonable to choose (−3, 1, 2) as the basepoint in order to approximate f (−3.1, 1.1, 1.9),Thus,

f (−3.1, 1.1, 1.9)− f (−3, 1, 1) ∼= df (−3, 1, 2,−0.1, 0.1,−0.1)= − 3√

6(−0.1) + 1√

6(0.1)− 2√

6(−0.1)

=0.6√6.


Therefore,

f (−3.1, 1.1, 1.9) ∼= f (−3, 1, 2) + 0.6√6=√6 +

0.6√6∼= 2. 694 44.

We have

f (−3.1, 1.1, 1.9) =q(−3.1)2 + (1.1)2 − (1.9)2 ∼= 2. 685 14

Thus, the absolute error in the approximation of f (−3.1, 1.1, 1.9) via the differential is approxi-mately 9.3× 10−3. This is much smaller than the distance of (−3.1, 1.1, 1.9) from the basepoint

(−3, 1, 2) (approximately 0.173). ¤

Problems

In problems 1-5,

a) Determine the linear approximation to f based at (x0, y0) or (x0, y0, z0),

b) Make use of the result of part a) in order to approximate f (x1, y1) or f (x1, y1, z1).

c) [C] Calculate the absolute error in the approximation by assuming that your calculatorcalculates the exact value.

1.

f (x, y) =¡x2 + y2

¢3/2, (x0, y0) = (3, 4) , (x1, y1) = (3.1, 3.9)

2.

f (x, y) = e−x sin (y) , (x0, y0) = (0,π/2) , (x1, y1) = (0.2,π/2 + 0.1)

3.

f (x, y) = arctan³yx

´, (x0, y0) =

³√3, 1´, (x1, y1) = (1.8, 0.8)

4.

f (x, y) = ex2+y2 , (x0, y0) = (1, 0) , (x1, y1) = (1.1,−0.2)

5.

f (x, y, z) =px2 + y2 + z2, (x0, y0, z0) = (1, 2, 3) , (x1, y1, z1) = (0.9, 2.2, 2.9)

In problems 6-8,

a) Determine the differential of f ,

b) Make use of the differential of f in order to approximate the indicated value of f :

c) [C] Calculate the absolute error in the approximation by assuming that your calculatorcalculates the exact value.

6.

f (x, y) =px2 + y2, f (12.1, 4.9)

7.

f (x, y, z) =√xyz, f (0.9, 2.9, 3.1)

8.

f (x, y) = arctan¡x2 + y2

¢, f (0.1,−1.2)

12.6. THE CHAIN RULE 85

12.6 The Chain Rule

In this section we will discuss various versions of the chain rule for functions of several variables.

Even though these rules are not as useful for the evaluation of derivatives as the chain rule for

functions of a single variable, they can be interpreted in ways that lead to useful general results,

as we will se;e within several contexts.

Let’s begin with the following version of the chain rule:

Proposition 1 Assume that f is a function of two variables, x and y are functions

of a single variable. If x and y are differentiable at t and f is differentiable at

(x (t) , y (t)) then

d

dtf (x (t) , y (t)) =

∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt

A Plausibility Argument:

Since f is differentiable at (x (t) , y (t)), if |∆t| is small,

f (x (t+∆t) , y (t+∆t))− f (x (t) , y (t)) ∼= ∂f

∂x(x (t) , y (t)) (x (t+∆t)− x (t))

+∂f

∂y(x (t) , y (t)) (y (t+∆t)− y (t))

∼= ∂f

∂x(x (t) , y (t))

dx

dt(t)∆t++

∂f

∂y(x (t) , y (t))

dy

dt(t)∆t.

Therefore,

f (x (t+∆t) , y (t+∆t))− f (x (t) , y (t))∆t

∼= ∂f

∂x(x (t) , y (t))

dx

dt(t) + +

∂f

∂y(x (t) , y (t))

dy

dt(t)

if |∆t| is small. Thus, it is plausible thatd

dtf (σ (t)) = lim

∆t→0f (x (t+∆t) , y (t+∆t))− f (x (t) , y (t))

∆t

=∂f

∂x(x (t) , y (t))

dx

dt(t) + +

∂f

∂y(x (t) , y (t))

dy

dt(t)

¥We can express the above version of the chain rule in more practical (albeit imprecise) ways

df

dt=

∂f

∂x

dx

dt+

∂f

∂y

dy

dt

If z = f (x, y),dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt

Example 1 Let f (x, y) = x2 − y2 and let x (t) = cos (t) , y (t) = sin (t).a) Determine

d

dtf (x (t) , y (t))

by applying the chain rule and directly.


b) Determined

dtf (x (t) , y (t))

¯t=π/3

.

Solution

a) We have x (t) = cos (t) and y (t) = sin (t). Thus.

∂f

∂x= 2x,

∂f

∂y= −2y, dx

dt= − sin (t) , dy

dt= cos (t) .

Therefore,

df

dt=

∂f

∂x

dx

dt+

∂f

∂y

dy

dt= −2x sin (t)− 2y cos (t) = −2 cos (t) sin (t)− 2 sin (t) cos (t)

= −4 cos (t) sin (t) .As for direct evaluation, we have f (cos (t) , sin (t)) = cos2 (t)− sin2 (t). Therefore,

df

dt= −2 cos (t) sin (t)− 2 sin (t) cos (t) = −4 cos (t) sin (t) .

b) By a),

df

dt(π/3) = −4 cos (π/3) sin (π/3) = −4

µ1

2

¶Ã√3

2

!= −4

√3.

¤Here is another version of the chain rule:

Proposition 2 Assume that f is a function of the single variable and x is a function

of two variables. If x is differentiable at (u, v) and f is differentiable at x (u, v) then

∂

∂uf (x (u, v)) =

df

dx(x (u, v))

∂x

∂u(u, v) and

∂

∂vf (x (u, v)) =

df

dx(x (u, v))

∂x

∂v(u, v)

Practical notation: If z = f (x),

∂z

∂u=dz

dx

∂x

∂uand

∂z

∂v=dz

dx

∂x

∂v

The Plausibility of Proposition 2

If |∆u| is small,

f (x (u+∆u, v))− f (x (u, v)) ∼= fµx (u, v) +

∂x

∂u(u, v)∆u

¶− f (x (u, v))

∼= f (x (u, v)) + df

dx(x (u, v))

∂x

∂u(u, v)∆u− f (x (u, v))

=df

dx(x (u, v))

∂x

∂u(u, v)∆u

Therefore,f (x (u+∆u, v))− f (x (u, v))

∆u∼= df

dx(x (u, v))

∂x

∂u(u, v)

if |∆u| is small. Thus, we should have∂

∂uf (x (u, v)) = lim

∆u→0f (x (u+∆u, v))− f (x (u, v))

∆u=df

dx(x (u, v))

∂x

∂u(u, v)

¥


Example 2 Let u = f (x− at), where f is a differentiable function of a single variable and ais a constant. Show that u is a solution of the wave equation

∂2u

∂t2= a2

∂u

∂x2.

Solution

Set w (x, t) = x− at, so that u = f (w (x, t)). By the chain rule,∂u

∂t=df

dw(w (x, t))

∂w

∂t=df

dw(w (x, t)) (−a) = −a df

dw(w (x, t)) .

Therefore,

∂2u

∂t2= −a ∂

∂t

µdf

dw(w (x, t))

¶= −a

µd2f

dw2(w (x, t))

∂w

∂t

¶= −a d

2f

dw2(w (x, t)) (−a) = a2 d

2f

dw2(x− at)

Similarly,∂u

∂x=df

dw(w (x, t))

∂w

∂x=df

dw(w (x, t)) (1) =

df

dw(w (x, t))

and

∂2u

∂x2=

∂

∂x

µdf

dw(w (x, t))

¶=d2f

dw2(w (x, t))

∂w

∂x

=d2f

dw2(w (x, t)) (1) =

d2f

dw2(x− at)

Therefore,∂2u

∂t2= a2

d2f

dw2(x− at) = a2 ∂

2u

∂x2

¤Here is still another version of the chain rule:

Proposition 3 Assume that f = f (x, y), x = x (u, v) and y = y (u, v). are differentiablefunctions. Then,

∂f

∂u(x (u, v) , y (u, v)) =

∂f

∂x(x (u, v) , y (u, v))

∂x

∂u(u, v) +

∂f

∂y(x (u, v) , y (u, v))

∂y

∂u(u, v) ,

∂f

∂v(x (u, v) , y (u, v)) =

∂f

∂x(x (u, v) , y (u, v))

∂x

∂v(u, v) +

∂f

∂y(x (u, v) , y (u, v))

∂y

∂v(u, v) .

Practical notation:

∂f

∂u=

∂f

∂x

∂x

∂u+

∂f

∂y

∂y

∂u,

∂f

∂v=

∂f

∂x

∂x

∂v+

∂f

∂y

∂y

∂v.

Practical hint to remember the rule: The differential of f is

df =∂f

∂xdx+

∂f

∂ydy.

In order to express ∂uf , replace dx by ∂x/∂u and dy by ∂y/∂u.


Example 3 Let r and θ be polar coordinates so that x = r cos (θ) and y = r sin (θ). Show that

∂f

∂r= cos (θ)

∂f

∂x+ sin (θ)

∂f

∂y

∂f

∂θ= −r sin (θ) ∂f

∂x+ r cos (θ)

∂f

∂y

Solution

By the chain rule, as in Proposition 3,

∂f

∂r=

∂f

∂x

∂x

∂r+

∂f

∂y

∂y

∂r=

∂f

∂x

µ∂

∂r(r cos (θ))

¶+

∂f

∂y

µ∂

∂r(r sin (θ))

¶=

∂f

∂xcos (θ) +

∂f

∂ysin (θ)

= cos (θ)∂f

∂x+ sin (θ)

∂f

∂y,

and

∂f

∂θ=

∂f

∂x

∂x

∂θ+

∂f

∂y

∂y

∂θ=

∂f

∂x

µ∂

∂θ(r cos (θ))

¶+

∂f

∂y

µ∂

∂θ(r sin (θ))

¶=

∂f

∂x(−r sin (θ)) + ∂f

∂y(r cos (θ))

= −r sin (θ) ∂f∂x

+ r cos (θ)∂f

∂y.

¤

Example 4 Show that

∂2f

∂x2+

∂2f

∂y2=

∂2f

∂r2+1

r

∂f

∂r+1

r2∂2f

∂θ2

Solution

∂2f

∂r2=

∂

∂r

µ∂f

∂r

¶=

∂

∂r

µcos (θ)

∂f

∂x+ sin (θ)

∂f

∂y

¶= cos (θ)

∂

∂r

µ∂f

∂x

¶+ sin (θ)

∂

∂r

µ∂f

∂y

¶= cos (θ)

µcos (θ)

∂

∂x

µ∂f

∂x

¶+ sin (θ)

∂

∂y

µ∂f

∂x

¶¶+ sin (θ)

µcos (θ)

∂

∂x

µ∂f

∂y

¶+ sin (θ)

∂

∂y

µ∂f

∂y

¶¶= cos2 (θ)

∂2f

∂x2+ 2 sin (θ) cos (θ)

∂2f

∂y∂x+ sin2 (θ)

∂2f

∂y2.


∂2f

∂θ2=

∂

∂θ

µ∂f

∂θ

¶=

∂

∂θ

µ−r sin (θ) ∂f

∂x+ r cos (θ)

∂f

∂y

¶= −r cos (θ) ∂f

∂x− r sin (θ) ∂

∂θ

µ∂f

∂x

¶− r sin (θ) ∂f

∂y+ r cos (θ)

∂

∂θ

µ∂f

∂y

¶= −r cos (θ) ∂f

∂x− r sin (θ)

µ−r sin (θ) ∂

∂x

µ∂f

∂x

¶+ r cos (θ)

∂

∂y

µ∂f

∂x

¶¶− r sin (θ) ∂f

∂y+ r cos (θ)

µ−r sin (θ) ∂

∂x

µ∂f

∂y

¶+ r cos (θ)

∂

∂y

µ∂f

∂y

¶¶= −r cos (θ) ∂f

∂x− r sin (θ) ∂f

∂y+ r2 sin2 (θ)

∂2f

∂x2+ r2 cos2 (θ)

∂2f

∂y2

− 2r2 sin (θ) cos (θ) ∂2f

∂y∂x

Therefore,

∂2f

∂r2+1

r

∂f

∂r+1

r2∂2f

∂θ2

= cos2 (θ)∂2f


∂2f

∂y∂x+ sin2 (θ)

∂2f

∂y2

+1

r

µcos (θ)

∂f

∂x+ sin (θ)

∂f

∂y

¶+1

r2

µ−r cos (θ) ∂f

∂x− r sin (θ) ∂f

∂y+ r2 sin2 (θ)

∂2f

∂x2+ r2 cos2 (θ)

∂2f

∂y2− 2r2 sin (θ) cos (θ) ∂2f

∂y∂x

¶= cos2 (θ)

∂2f


∂2f

∂y∂x+ sin2 (θ)

∂2f

∂y2

+1

rcos (θ)

∂f

∂x+1

rsin (θ)

∂f

∂y

− 1rcos (θ)

∂f

∂x− 1rsin (θ)

∂f

∂y+ sin2 (θ)

∂2f

∂x2+ cos2 (θ)

∂2f

∂y2− 2 sin (θ) cos (θ) ∂2f

∂y∂x

=∂2f

∂x2+

∂2f

∂y2

¤

In general if f is a function of x, y, z, . . . and

x = x (u, v, . . .) ,

y = y (u, v, . . .) ,

z = z (u, v, . . .) ,

etc., we have

∂f

∂u=

∂f

∂x

∂x

∂u+

∂f

∂y

∂y

∂u+ · · · ,

∂f

∂v=

∂f

∂x

∂x

∂v+

∂f

∂y

∂y

∂v+ · · · ,

etc..


Implicit Differentiation

Assume that y is defined implicitly as a function of x by the equation F (x, y) = 0. Thismeans that F (x, y (x)) = 0 for each x in an interval J . In Chapter 2 we discussed implicitdifferentiation for the computation of dy/dx. The chain rule for a function of two variables

enables us to describe the procedure for implicit differentiation in general terms. Indeed,

d

dxF (x, y (x)) = 0

for each x ∈ J . By the chain rule,

0 =d

dxF (x, y (x)) =

∂F (x, y)

∂x+

∂F (x, y)

∂y

dy

dx

Therefore,

dy

dx= −

∂F (x, y)

∂x∂F (x, y)

∂y

The expression makes sense if∂F (x, y)

∂y6= 0.

It can be shown that y can be expressed as a function of x in some open interval J that contains

the point x0 such that y (x0) = y0 if F (x0, y0) = 0, F (x, y) has continuous partial derivativesin a open disk containing (x0, y0) and we have

∂F (x, y)

∂y

¯(x,y)=(x0,y0)

6= 0

Under these conditions, we do have

dy

dx= −

∂F (x, y)

∂x∂F (x, y)

∂y

for each x in an open interval that contains x0. This is a version of the implicit function

theorem that is proven in a course on advanced calculus.

Example 5 Assume that y is defined as a function of x implicitly by the equation x3+y3 = 9xy.Figure 1 shows the graph of the equation.

-4 -2 2 4

-4

-2

2

4

x

y

Figure 1


If we set F (x, y) = x3 + y3 − 9xy, we have F (x, y (x)) = 0 for each x in some interval. As inthe previous discussion,

dy

dx= −

∂F (x, y)

∂x∂F (x, y)

∂y

= −3x2 − 9y

3y2 − 9x =9y − 3x23y2 − 9x.

We have F (2, 4) = 0, and

∂F (x, y)

∂y

¯(x,y)=(2,4)

= 3y2 − 9x¯(x,y)=(2,4)

= 30 6= 0.

Therefore, the equation F (x, y) = 0 defines y as a function of x such that y (2) = 4. We have

dy

dx

¯x=2

=9y − 3x23y2 − 9x

¯x=2,y=4

=4

5.

¤Under certain conditions, an equation of the form F (x, y, z) = 0 defines z implicitly as a functionof x and y. If z (x, y) is such a function that is differentiable, we can apply the chain rule inorder to express its partial derivatives. Indeed,

0 =∂

∂xF (x, y, z (x, y)) =

∂F

∂x+

∂F

∂z

∂z

∂x

and

0 =∂

∂yF (x, y, z (x, y)) =

∂F

∂y+

∂F

∂z

∂z

∂y,

where Fx, Fy and Fz are evaluated at (x, y, z (x, y)). Therefore,

∂z

∂x= −

∂F

∂x∂F

∂z

and∂z

∂y= −

∂F

∂y

∂F

∂z

,

provided that Fz 6= 0 at the relevant point (x, y, z (x, y)). A version of the implicit functiontheorem says that the above expressions make sense for each (x, y) in some open disk centeredat (x0, y0) if F (x0, y0, z0) = 0, Fz (x0, y0, z0) 6= 0 and z (x0, y0) = z0.Example 6 Assume that z (x, y) is defined implicitly by the equation

z2 − x2 − 2y2 − 3 = 0,and z (2, 1) = 3. Figure 2 shows the graph of the above equation.

Figure 2


a) Determine zx (x, y) and zy (x, y).b) Evaluate zx (2, 1) and zy (2, 1).

Solution

a) If we set F (x, y, z) = z2 − x2 − 2y2 − 3, we have F (x, y, z (x, y)) = 0. As in the previousdiscussion,

∂z

∂x= −

∂F

∂x∂F

∂z

=2x

2z= −x

z

and

∂z

∂y= −

∂F

∂y

∂F

∂z

=4y

2z=2y

z.

(Exercise: Obtain these expressions directly).

b)∂z

∂x(2, 1) = −2

3and

∂z

∂y(2, 1) =

2

3.

Note that

z =px2 + 2y2 + 3,

so that∂z

∂x=

1

2px2 + 2y2 + 3

(2x) =xp

x2 + 2y2 + 3=x

z

and∂z

∂y=

1

2px2 + 2y2 + 3

(4y) =2y

z

as before. ¤

Problems

In problems 1 and 2, computed

dtf (x (t) , y (t))

a) By making use of the chain rule,

b) Directly, from the expression for f (x (t) , y (t)) .

1.

f (x, y) =px2 + y2, x (t) = sin (t) , y (t) = 2 cos (t) .

2.

f (x, y) = e−x2+y2 , x (t) = 2t− 1, y (t) = t+ 1

In problems 3 and 4, compute

∂

∂uf (x (u, v)) and

∂

∂vf (x (u, v))


b) Directly, from the expression for f (x (u, v)):


3.

f (x) = ln (x) , x =1√

u2 + v2

4.

f (x) = sin¡x2¢, x = u− 4v

In problems 5 and 6, compute

∂

∂uf (x (u, v) , y (u, v)) and

∂

∂vf (x (u, v) , y (u, v))


b) Directly, from the expression for f (x (u, v) , y (u, v)):

5.

f (x, y) = arcsin

Ãyp

x2 + y2

!, x = u cos (v) , y = u sin (v)

(assume that −π/2 < v < π/2).6.

f (x, y) = arctan³yx

´, x = u+ 2v, y = u− 2v

7. Let z = f (x, y) and let r and θ be polar coordinates so that x = r cos (θ) and y = r sin (θ).Show that µ

∂z

∂x

¶2+

µ∂z

∂y

¶2=

µ∂z

∂r

¶2+1

r2

µ∂z

∂θ

¶28. Let z = f (x, y), x = et cos (θ) and y = et sin (θ). Show thatµ

∂z

∂x

¶2+

µ∂z

∂y

¶2= e−2t

"µ∂z

∂t

¶2+

µ∂z

∂θ

¶2#9.

a) Let u (x, t) = f (x+ at), where f is a differentiable function of a single variable and a is aconstant. Show that u is a solution of the wave equation

∂2u

∂t2= a2

∂2u

∂x2.

b) Determine u (x, 0) , u (x, 2) , u (4) if f (w) = sin (w) and a = π/4.c) [C] if f (w) = sin (w) plot u (x, 0) , u (x, 2) , u (x, 4), where −4π ≤ x ≤ 4π.10. Assume that z (x, y) is defined implicitly by the equation

z2 − x2 − y2 − 4 = 0,and z (1, 2) = −3.

5

xy

20

0 15-5

-5

-5

-3

0z

5


a) Determine zx (x, y) and zy (x, y) directly (i.e., without relying on a formula from the text).

b) Evaluate zx (1, 2) and zy (1, 2) and determine an equation for the plane that is tangent to thegraph of the given equation at (1, 2,−3).11. Assume that z (x, y) is defined implicitly by the equation

z2 + x2 − y2 − 9 = 0,and z (3, 6) = 6.

6

3x

z

6

y

a) Determine zx (x, y) and zy (x, y) directly (i.e., without relying on a formula from the text).

b) Evaluate zx (3, 6) and zy (3, 6). and determine an equation for the plane that is tangent tothe graph of the given equation at (3, 6, 6).

12.7 Directional Derivatives and the Gradient

Directional Derivatives

We will begin by considering functions of two variables. Assume that f has partial derivatives

in some open disk centered at (x, y). The partial derivative ∂xf (x, y) can be interpreted as therate of change of f in the direction of the standard basis vector i and the partial derivative

∂yf (x, y) can be interpreted as the rate of change of f in the direction of the standard basisvector j. We would like to arrive at a reasonable definition of the rate of f at (x, y) in thedirection of an arbitrary unit vector u = u1i+ u2j.

x

y

hu

P

Ph

Figure 1

Let h 6= 0. Let’s assume that −−→OPh = −−→OP + hu so thatPh = (x+ hu1, y + hu2) .

We define the average rate of change of f corresponding to the displacement−−→PPh as

f (Ph)− f (P )h

=f (x+ hu1, y + hu2)− f (x, y)

h

12.7. DIRECTIONAL DERIVATIVES AND THE GRADIENT 95

It is reasonable to define the rate of change of f at P in the direction of u as

limh→0

f (Ph)− f (P )h

.

Another name for this limit is ”directional derivative":

Definition 1 The directional derivative of f at (x, y) in the direction of the unit

vector u is

Duf (x, y) = limh→0

f (x+ hu1, y + hu2)− f (x, y)h

.

Note that Dif (x, y) is ∂xf (x, y) and Djf (x, y) is ∂yf (x, y). We can calculate the directionalderivative in an arbitrary direction in terms of ∂xf (x, y) and ∂yf (x, y):

Proposition 1 Assume that f is differentiable at (x, y) and that u = u1i+u2j is a unitvector. Then

Duf (x, y) =∂f

∂x(x, y)u1 +

∂f

∂y(x, y)u2.

Proof

Set

g (t) = f (x+ tu1, y + tu2) , t ∈ R.Then,

g0 (0) = limh→0

g (h)− g (0)h

= limh→0

f (x+ hu1, y + hu2)− f (x, y)h

= Duf (x, y)

Let’s set

v (x, t) = x+ tu1 and w (x, t) = y + tu2,

so that g (t) = f (v (x, t) , w (x, t)). By the chain rule,

d

dtg (t) =

d

dtf (x+ tu1, y + tu2)

=∂f

∂v(v,w)

dv

dt+

∂f

∂w(v, w)

dw

dt

=∂f

∂v(x+ tu1, y + tu2)

d

dt(x+ tu1) +

∂f

∂w(x+ tu1, y + tu2)

d

dt(y + tu2)

=∂f

∂v(x+ tu1, y + tu2)u1 +

∂f

∂w(x+ tu1, y + tu2)u2.

Therefore,

Duf (x, y) = g0 (0) =

∂f

∂v(x+ tu1, y + tu2)u1 +

∂f

∂w(x+ tu1, y + tu2)u2

¯t=0

=∂f

∂v(x, y)u1 +

∂f

∂w(x, y)u2

=∂f

∂x(x, y)u1 +

∂f

∂y(x, y)u2,

as claimed. ¥

Example 1 Let f (x, y) = 36 − 4x2 − y2 and let u be the unit vector in the direction of

v = −i+ 4j. Determine Duf (1, 3).


Solution

We have

∂f

∂x(x, y) =

∂

∂x

¡36− 4x2 − y2¢ = −8x and ∂f

∂y(x, y) =

∂

∂y

¡36− 4x2 − y2¢ = −2y.

Therefore,

∂f

∂x(1, 3) = −8 and ∂f

∂y(1, 3) = −6.

We have

||v|| =p1 + 42 =

√17.

Therefore,

u =v

||v|| =1√17(−i+ 4j) = − 1√

17i+

4√17j.

Thus,

Duf (1, 3) =∂f

∂x(1, 3)u1 +

∂f

∂y(1, 3)u2

= (−8)µ− 1√

17

¶+ (−6)

µ4√17

¶= − 16√

17.

¤

Remark 1 All directional derivatives may exist, but f may not be differentiable, as in the

following example:

Let

f (x, y) =

⎧⎨⎩ xy2

x2 + y4if (x, y) 6= (0, 0) ,

0 if (x, y) = (0, 0)

Figure 2 shows the graph of f .

2

x

0-0.5

2

y0

0.0z

-2-2

0.5

Figure 2

In Section 12.5 we saw that f is not continuous at (0, 0) so that it is not differentiable. Nev-ertheless, the directional derivatives of f exist in all directions at the origin. Indeed, let u be


an arbitrary unit vector, so that u = (cos (θ) , sin (θ)) for some θ. Let’s calculate Duf (0, 0),directly from the definition. We have

f (h cos (θ) , h sin (θ))

h=

h3 cos (θ) sin2 (θ)

h2 cos2 (θ) + h4 sin4 (θ)

h=

h3 cos (θ) sin2 (θ)

h3 cos2 (θ) + h5 sin4 (θ)

=cos (θ) sin2 (θ)

cos2 (θ) + h2 sin4 (θ).

Therefore,

Duf (0, 0) = limh→0

f (h cos (θ) , h sin (θ))

h=cos (θ) sin2 (θ)

cos2 (θ)=sin2 (θ)

cos (θ)

if cos(θ) 6= 0. The case cos(θ) = 0 corresponds to fy (0, 0) or −fy (0, 0). We have

fy (0, 0) = limh→0

f (0, h)− f (0, 0)h

= limh→0

0

h= 0.

Thus, the directional derivatives of f exist in all directions. ♦The gradient of a function is a useful notion within the context of directional derivatives, and

in many other contexts:

Definition 2 The gradient of f at (x, y) is the vector-valued function that assigns the vector

∇f (x, y) = ∂f

∂x(x, y) i+

∂f

∂y(x, y) j

to each (x, y) where the partial derivatives of f exist (read ∇f as "del f").We can express a directional derivative in terms of the gradient:

Proposition 2 The directional derivative of f at (x, y) in the direction of the unitvector u can be expressed as

Duf (x, y) =∇f (x, y) · uProof

By Proposition 1,

Duf (x, y) =∂f

∂x(x, y)u1 +

∂f

∂y(x, y)u2

By the definition of∇f (x, y), the right-hand side can be expressed as a dot product. Therefore,

Duf (x, y) =

µ∂f

∂x(x, y) i+

∂f

∂y(x, y) j

¶· (u1i+ u2j)

=∇f (x, y) · u¥Thus, the directional derivative of f at (x, y) in the direction of the unit vector u isthe component of ∇f (x, y) along u.

Example 2 Let f (x, y) = x2 + 4xy + y2.


a) Determine the gradient of f .

b) Calculate the directional derivative of f at (2, 1) in the directions of the vectors i + j andi− j.Solution

a) Figure 3 displays the graph of f .

0

4

50

z

100

24

y

0 2

x0-2

-2-4 -4

Figure 3

We have∂f

∂x(x, y) =

∂

∂x

¡x2 + 4xy + y2

¢= 2x+ 4y

and∂f

∂y(x, y) =

∂

∂y

¡x2 + 4xy + y2

¢= 4x+ 2y.

Therefore,

∇f (x, y) = ∂f

∂x(x, y) i+

∂f

∂y(x, y) j =(2x+ 4y) i+ (4x+ 2y) j.

Figure 4 displays ∇f (x, y) at certain points (x, y) as arrows located at these points.

-4 -3 -2 -1 1 2 3 4

-4

-3

-2

-1

1

2

3

4

x

y

Figure 4

b) A unit vector in the direction of i+ j is

u1 =1√2i+

1√2j,

and a unit vector in the direction of i− j is

u2 =1√2i− 1√

2j.

We have

∇f (2, 1) = (2x+ 4y) i+ (4x+ 2y) j|x=2,y=1 = 8i+ 10j.


Therefore,

Du1f (2, 1) =∇f (2, 1) · u1= (8i+ 10j) ·

µ1√2i+

1√2j

¶=

8√2+10√2=18√2= 9√2,

and

Du2f (2, 1) =∇f (2, 1) · u2= (8i+ 10j) ·

µ1√2i− 1√

2j

¶=

8√2− 10√

2= − 2√

2= −√2.

Thus, at (2, 1) the function increases in the direction of u1 and decreases in the direction ofu2. Figure 5 shows some level curves of f , the vectors u1 and u2, and a unit vector along

∇f (2, 1) . ¤

13

13

5

5

0

0

0

0

55

13

13

4 2 2 4x

4

2

1

2

4

y

2,1fu1

u2

Figure 5

Proposition 3 The maximum value of the directional derivatives of a function f at

a point (x, y) is ||∇f (x, y)||, the length of the gradient of f at (x, y), and it is attainedin the direction of ∇f (x, y). The minimum value of the directional derivatives of f

at (x, y) is − ||∇f (x, y)|| and it is attained the direction of −∇f (x, y) .Proof

Since Duf (x, y) =∇f (x, y) · u, and ||u|| = 1, we haveDuf (x, y) = ||∇f (x, y)|| ||u|| cos (θ) = ||∇f (x, y)|| cos (θ) ,

where θ is the angle between ∇f (x, y) and the unit vector u. Since cos (θ) ≤ 1,Duf (x, y) ≤ ||∇f (x, y)|| .

The maximum value is attained when cos (θ) = 1, i.e., θ = 0. This means that u is the unitvector in the direction of ∇f (x, y) .


Similarly,

Duf (x, y) = ||∇f (x, y)|| cos (θ)attains its minimum value when cos (θ) = −1, i.e., θ = π. This means that u = −∇f (x, y).The corresponding directional derivative is − ||∇f (x, y)|| .¥

Example 3 Let f (x, y) = x2 + 4xy + y2, as in Example 2. Determine the unit vector alongwhich the rate of increase of f at (2, 1) is maximized and and the unit vector along which therate of decrease of f at (2, 1) is maximized. Evaluate the corresponding rates of change of f .

Solution

In Example 2 we showed that ∇f (2, 1) = 8i+ 10j. Therefore,

||∇f (2, 1)|| = ||(8i+ 10j)|| = √64 + 100 = √164 = 2√41.

Thus, the value of f increases at the maximum rate of 2√41 in the direction of the unit vector

1

||∇f (2, 1)||∇f (2, 1) =1

2√41(8i+ 10j) =

4√41i+

5

41j.

The value of the function decreases at the maximum rate of 2√41 (i.e., the rate of change is −

2√41) in the direction of the unit vector

− 1

||∇f (2, 1)||∇f (2, 1) = −4√41i− 5

41j.

¤

The Chain Rule and the Gradient

A special case of the chain rule says that

d

dtf (x (t) , y (t)) =

∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt,

as In Proposition 1 of Section 12.6. The notion of the gradient enables us to provide a useful

geometric interpretation of the above expression. Let’s begin by expressing the chain rule as

follows:

Proposition 4 Assume that the curve C in the plane is parametrized by the function

σ where

σ (t) = (x(t), y(t)) .

Thend

dtf (σ (t)) =∇f (σ (t)) · dσ

dt

Proof

We have

d

dtf (σ (t)) =

d

dtf (x (t) , y (t)) ,

∇f (σ (t)) = ∂f

∂x(x (t) , y (t)) i+

∂f

∂y(x (t) , y (t)) ,


anddσ

dt=dx

dti+

dy

dtj.

Therefore,

∇f (σ (t)) · dσdt=

µ∂f

∂x(x (t) , y (t)) i+

∂f

∂y(x (t) , y (t))

¶·µdx

dti+

dy

dtj

¶=

∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt.

By the chain rule,

∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt=d

dtf (x (t) , y (t)) =

d

dtf (σ (t)) .

Thus,d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt,

as claimed. ¥

Remark 2 With the notation of Proposition 4, assuming that σ0 (t) 6= 0,

d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt=∇f (σ (t)) ·

dσ

dt(t)¯¯

dσ

dt(t)

¯¯ ¯¯dσdt(t)

¯¯

=

¯¯dσ

dt(t)

¯¯∇f (σ (t)) ·T (t) ,

where T (t) is the unit tangent to C at σ (t). The quantity∇f (σ (t))·T (t) is DT(t)f (σ (t)), thedirectional derivative of f at σ (t) in the direction of the unit tangent T (t), i.e., the componentof ∇f (σ (t)) along T (t). Thus,

d

dtf (σ (t)) =

¯¯dσ

dt(t)

¯¯DT(t)f (σ (t)) .

If σ (t) is the position of a particle at time t, ||σ0 (t)|| is its speed at t. In this case,d

dtf (σ (t)) = the speed of the object at t× component of ∇f (σ (t)) along T (t) .

♦

Proposition 5 Let C be the level curve of f that passes through the point (x0, y0).The gradient of f at (x0, y0) is perpendicular to C at (x0, y0), in the sense that ∇fis orthogonal to the tangent line to C at (x0, y0).

A Plausibility Argument for Proposition 5

We will assume that a segment of C near (x0, y0) can be parametrized by the function σ : J → R2such that σ (t0) = (x0, y0) and σ

0 (t0) 6= 0. By Proposition 4,d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt(t) .


Since C is a level curve of f , the value of f on C is a constant. Therefore,

d

dtf (σ (t)) = 0

for each t ∈ J . In particular,

0 =d

dtf (σ (t0)) =∇f (σ (t0)) · dσ

dt(t0) =∇f (x0, y0) · dσ

dt(t0) .

This implies that ∇f (x0, y0) is orthogonal to the vector σ0 (t0) that is tangent to C at σ (t0) =(x0, y0). ¥

Example 4

a) The point (1, 3) is on the curve C that is the graph of the equation x2 + 4xy + y2 = 22.Determine a vector that is orthogonal to C at (1, 3).b) Determine the line that is tangent to the C at (1, 3) .

Solution

a( We set f (x, y) = x2 + 4xy + y2, as in Example 2, so that C is a level curve of f . We have

∇f (1, 3) =∇f (1, 3) = (2x+ 4y) i+ (4x+ 2y) j|x=1,y=3 = 14i+ 10j.

Therefore, 14i+10j. is orthogonal to C at (1, 3). Figure 6 shows C and a vector along ∇f (1, 3).The picture is consistent with our claim.

1x

3

y

1,3

Figure 6

b) If P0 = (1, 3) and P = (x, y) is an arbitrary point on the line that is tangent to C at P0, wehave

∇f (1, 3) ·−−→P0P = 0.Therefore,

(14i+ 10j) · ((x− 1) i+ (y − 3)) = 0⇔

14 (x− 1) + 10 (y − 3) = 0.Figure 7 shows the tangent line.


1x

3

y

1,3

Figure 7

Functions of three or more variables

Our discussion of directional derivatives and the gradient extend to functions of more than two

variables in a straightforward fashion. Let’s consider functions of three variables to be specific.

Thus, assume that f is a scalar function of the variables x, y and z. If u = u1i+ u2j + u3kis a three-dimensional unit vector, the directional derivative of f at (x0, y0, z0) in the

direction of u is

Duf (x0, y0, z0) = limh→0

f (x0 + hu1, y0 + hu2, z0 + hu3)− f (x0, y0, z0)h

The gradient of f at (x, y, z) is

∇f (x, y, z) = ∂f

∂x(x, y, z) i+

∂f

∂y(x, y, z) j+

∂f

∂z(x, y, z)k.

If f is differentiable at (x0, y0, z0),

Duf (x0, y0, z0) =∂f

∂x(x0, y0, z0)u1 +

∂f

∂y(x0, y0, z0)u2 +

∂f

∂z(x0, y0, z0)u3

=∇f (x0, y0; z0) · uWe have

− ||∇f (x0, y0, z0)|| ≤ Duf (x0, y0, z0) ≤ ||∇f (x0, y0, z0)|| ,Thus, the rate of change of f at (x0, y0, z0) has the maximum value ||∇f (x0, y0, z0)|| in the direc-tion of∇f (x0, y0, z0). The function decreases at the maximum rate of ||∇f (x0, y0, z0)|| (i.e., therate of change is − ||∇f (x0, y0, z0)||) at the point (x0, y0, z0) in the direction of −∇f (x0, y0, z0).If σ : J → R3, and σ (t) = (x (t) , y (t) , z (t)), by the chain rule,

d

dtf (x (t) , y (t) , z (t))

=∂f

∂x(x (t) , y (t) , z (t))

dx

dt(t) +

∂f

∂y(x (t) , y (t) , z (t))

dy

dt(t) +

∂f

∂z(x (t) , y (t) , z (t))

dz

dt(t) .

so that.d

dtf (σ (t)) =∇f (σ (t)) · dσ

dt.


If S is a level surface of f = f (x, y, z) so that f has a constant value on S, and (x0, y0, z0)lies on S, then ∇f (x0, y0, z0) is orthogonal to S at (x0,y0, z0), in the sense that ∇f (x0, y0, z0)is orthogonal at (x0, y0, z0) to any curve on S that passes through (x0, y0, z0). Indeed, if such acurve C is parametrized by σ : J → R3 andσ (t0) = (x0, y0, z0) we have

d

dtf (σ (t)) = 0 for t ∈ J,

since f (σ (t)) has a constant value. Therefore,

∇f (σ (t)) · dσdt(t) =

d

dtf (σ (t)) = 0.

In particular,

∇f (x0, y0, z0) · dσdt(t0) =∇f (σ (t0)) · dσ

dt(t0) = 0,

so that ∇f (x0, y0, z0) is orthogonal to the vector σ0 (t0) that is tangential to the curve C. Itis reasonable to declare that the plane that is spanned by all such tangent vectors and passes

through (x0, y0, z0) is the tangent plane to the level surface S of f . Since ∇f (x0, y0z0)is normal to that plane, the equation of the tangent plane is

∇f (x0, y0z0) · ((x− x0) i+ (y − y0) j+ (z − z0)k) = 0,

i.e.,∂f

∂x(x0, y0, z0) (x− x0) + ∂f

∂y(x0, y0, z0) (y − y0) + ∂f

∂z(x0, y0, z0) (z − z0) = 0.

Example 5 Let f be the function such that

f(x, y, z) =x2

9+y2

4+ z2

a) Determine∇f (x, y, z) and the directional derivative of f at (2, 1, 1) in the direction of (1, 1, 1).

b) Determine the tangent plane to the surface

x2

9+y2

4+ z2 =

61

36

at (2, 1, 1).

Solution

a) We have

∇f (x, y, z) = ∂f

∂x(x, y, z) i+

∂f

∂y(x, y, z) j+

∂f

∂z(x, y, z) =

2x

9i+

y

2j+ 2zk.

Therefore,

∇f (2, 1, 1) = 4

9i+

1

2j+ 2k.

The unit vector u along (1, 1, 1) is

1√3i+

1√3j+

1√3k.


Therefore,

Duf (2, 1, 1) =∇f (2, 1, 1) · u=

µ4

9i+

1

2j+ 2k

¶·µ1√3i+

1√3j+

1√3k

¶=

4

9√3+

1

2√3+

2√3=53

54

√3.

b) We can express the equation of the required plane as

∇f (2, 1, 1) · ((x− 2) i+ (y − 1) j+ (z − 1)k) = 0,

i.e., µ4

9i+

1

2j+ 2k

¶· ((x− 2) i+ (y − 1) j+ (z − 1)k) = 0.

Thus,4

9(x− 2) + 1

2(y − 1) + 2 (z − 1) = 0.

Figure 8 illustrates the surface and the above tangent plane. ¤

Figure 8

Problems

In problems 1-6,

a) Compute the gradient of f

b) Compute the directional derivative of f at P in the direction of the vector v.

1.

f (x, y) = 4x2 + 9y2, P = (3, 4) , v = (−2, 1)2.

f (x, y) = x2 − 4x− 3y2 + 6y + 1, P = (0, 0) , v = (1,−1)3.

f (x, y) = ex2−y2 , P = (2, 1) , v = (−1, 3)

4.

f (x, y) = sin (x) cos (y) , P = (π/3,π/4) , v = (2,−3)


5.

f (x, y, z) = x2 − y2 + 2z2, P = (1,−1, 2) , v = (1,−1, 1)6.

f (x, y, z) =1p

x2 + y2 + z2, P = (2,−2, 1) , v = (3, 4, 1)

In problems 7 and 8,

a) Determine a vector v such that the rate at which f increases at P has its maximum value in

the direction of v, and the corresponding rate of increase of f,

b) Determine a vector w such that the rate at which f decreases at P has its maximum value

in the direction of w, and the corresponding rate of decrease of f.

7.

f (x, y) =1p

x2 + y2, P = (2, 3)

8.

f (x, y) = ln³p

x2 + y2´, P = (3, 4)


a) Computed

dtf (σ (t))

¯t=t0

by making use of the chain rule,

b) Compute the directional derivative of f in the tangential direction to the curve that is

parametrized by σ at the point σ (t0) .

9.

f (x, y) = ex2−y2 , σ (t) = (2 cos (t) , 2 sin (t)) , t0 = π/6

10

f (x, y) = arctan

Ãyp

x2 + y2

!, σ (t) =

µ1− t21 + t2

,2t

1 + t2

¶, t0 = 2.

In problems 11-13,

a) Find a vector that is orthogonal at the point P to the curve that is the graph of the given

equation,

b) Determine an equation of the line that is tangent to that curve at P :

11.

2x2 + 3y2 = 35, (2, 3)

12.

x2 − y2 = 4,³3,√5´

13.

e25−x2−y2 = 1, (3, 4)

In problems 14-17,

a) Find a vector that is orthogonal at the point P to the surface that is the graph of the given

equation

b) Determine an equation of the plane that is tangent to the given surface at P.

14.

z − x2 + y2 = 0, P = (4, 3, 7)15.

x2 − y2 + z2 = 1, P = (2, 2, 1)

12.8. LOCAL MAXIMA AND MINIMA 107

16.

x2 − y2 − z2 = 1, P = (3, 2, 2)

17.

x− sin (y) cos (z) = 0, P = (1,π/2, 0)

12.8 Local Maxima and Minima

The Definitions and Preliminary Examples

Definition 1 A function f has a local maximum at the point (x0, y0) if there exists an opendisk D containing (x0, y0) such that f (x, y) ≤ f (x0, y0) for each (x, y) ∈ D. A function f hasa local minimum at the point (x0, y0) if there exists an open disk D containing (x0, y0) suchthat f (x, y) ≤ f (x0, y0) for each (x, y) ∈ D.

Definition 2 The absolute maximum of a function f on the set D ⊂ R2 is f (x0, y0) iff (x, y) ≤ f (x0, y0) for each (x, y) ∈ D. The absolute minimum of a function f on the set

D ⊂ R2 is f (x0, y0) if f (x, y) ≥ f (x0, y0) for each (x, y) ∈ D.

Example 1 Let Q (x, y) = x2 + y2. Since Q(x, y) ≥ 0 for each (x, y) ∈ R2 and Q (0, 0) = 0, Qattains its absolute minimum value 0 on R2 at (0, 0). Of course, the only point at which Q hasa local minimum is also (0, 0). The function does not attain an absolute maximum on /R

2since

it attains values of arbitrarily large magnitude. For example,

limx→+∞Q (x, x) = lim

x→+∞ 2x2 = +∞.

¤

0

5

10z

15

2

y

20

x0

-2-2

Figure 1

Example 2 Let Q (x, y) = −x2 − y2. Q attains its maximum value 0 at (0, 0) .The functiondoes not attain an absolute minimum on R2 since it attains negative values of arbitrarily large

magnitude. For example,

limx→+∞Q (x, x) = lim

x→+∞¡−2x2¢ = −∞.

¤


-15

-10z

-5

2

0

2

y

0

x

0-2

-2

Figure 2

Example 3 Let Q (x, y) = x2−y2. The function does not have a local extremum or an absolutemaximum or minimum on R2. Indeed, the function attains positive and negative values ofarbitrarily large magnitude since

limx→+∞Q (x, 0) = lim

x→+∞x2 = +∞

and

limy→+∞Q (0, y) = lim

y→+∞−y2 = −∞.

-10

0z

2

10

2

y

0

x

0-2

-2

Fgiure 3

Figure 4 shows some level curves of Q. The darker color indicates a lower level on the graph of

Q. We have Q (x, 0) = x2 and Q (0, y) = −y2. Therefore, the restriction of Q to the x-axis has

a minimum at (0, 0) and the restriction of Q to the y-axis has a maximum at (0, 0). The pictureis consistent with these observations.¤

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

Figure 4


Definition 3 We say that (x0, y0) is a saddle point of Q if there exists a line that passes

through (x0, y0) on which Q has a local maximum at (x0, y0), and a similar line on which Q hasa local minimum at (x0, y0).

Thus, the function of Example 3 has a saddle point at (0, 0).

The Second Derivative Test for Local Extrema

If f is a differentiable function of a single variable and has a local extremum at x0, we must

have f 0 (x0) = 0. There is a similar condition for a function of two variables.

Proposition 1 If fx (x0, y0) and fy (x0, y0) exist and f has a local maximum or local

minimum at (x0, y0), then

∂f

∂x(x0, y0) = 0 and

∂f

∂y(x0, y0) = 0.

Proof

Set g (x) = f (x, y0). Thus, g is the restriction of f to the line y = y0. Since g is a function of asingle variable and has a local extremum at x0, we have

dg

dx(x0) =

d

dxf (x, y0)

¯x=x0

=∂f

∂x(x0, y0) = 0.

Similarly, if we set h (y) = f (x0, y), so that h is the restriction of f to the line x = x0, h has alocal extremum at y0. Therefore,

dh

dy(y0) =

d

dyf (x0, y)

¯y=y0

=∂f

∂y(x0, y0) = 0.

¥

Definition 4 We say that the point (x0, y0) is a stationary point of f if

∂f

∂x(x0, y0) = 0 and

∂f

∂y(x0, y0) = 0.

Thus, Proposition 1 asserts that a necessary condition for a f to have a local maximum

or minimum at (x0, y0) is that (x0, y0) is a stationary point of f , provided that f has

partial derivatives at (x0, y0). If we assume that f is differentiable at (x0, y0), the tangent planeto the graph of f at (x0, y0, f (x0, y0)) is the graph of the linear function

L (x, y) = f (x0, y0) + fx (x0, y0) (x− x0) + fy (x0, y0) (y − y0) = f (x0, y0) ,so that it is parallel to the xy-plane.

Figure 5


Remark 1 A function need not have a local extremum at a stationary point. For

example, if f (x, y) = x2 − y2, as in Example 3, we have∂f

∂x(x, y) = 2x and

∂f

∂y(x, y) = −2y,

so that the only stationary point of f is the origin (0, 0). We observed that f has a saddle pointat (0, 0). In particular, f does not have a local maximum or minimum at (0, 0) . ♦

Remark 2 A function can have a local extremum at a point even though its partial derivatives

do not exist at that point. For example, let

f(x, y) =px2 + y2.

We have∂f

∂x(x, y) =

1

2px2 + y2

(2x) =xp

x2 + y2

and∂f

∂y(x, y) =

ypx2 + y2

if (x, y) 6= (0, 0). You can confirm that the partial derivatives of f do not exist at (0, 0). Thefunction does have a local minimum at (0, 0), though. In fact,

f (x, y) =px2 + y2 ≥ 0 = f (0, 0)

for each (x, y) ∈ R2, so that the absolute minimum of f on R2. The graph of f is a cone, as

shown in Figure 6.♦

04

2

z

4

6

2

y

0 42

x

-2 0-2

-4 -4

Figure 6

Remark 3 ’Proposition 1 has a counterpart for a scalar function of any number of variables.

For example, if f is a function of x, y and z, has partial derivatives and a local extremum at

(x0, y0, z0), then∂f

∂x(x0, y0, z0) =

∂f

∂y(x0, y0, z0) =

∂f

∂z(x0, y0, z0) = 0.

♦

Example 4 Let

f (x, y) = 2x2 + 2xy + y2

Determine the stationary points of f . Make use of your graphing utility to plot the graph of f .

Does the picture suggest that f has a local maximum or minimum at each stationary point?


Solution

We have∂f

∂x(x, y) = 4x+ 2y and

∂f

∂y(x, y) = 2x+ 2y.

Therefore, (x, y) is a stationary point of f if and only if

4x+ 2y = 0,

2x+ 2y = 0.

The only solution is (0, 0). Thus, (0, 0) is the only stationary point of f .Figure 7 displays the graph of f . The picture suggests that f has a local (and absolute) minimum

at (0, 0). ¤

50

100

200

300

z

y

0-5

x

0 -55

Figure 7

There is a counterpart of the second derivative test for the local extrema of a function of a

single variable for functions of several variables. We will discuss the second derivative test only

functions of two variables since the discussion for more than two variables requires machinery

from linear algebra that you may not have at your disposal yet.

Theorem 1 (The Second Derivative Test for extrema) Assume that (x0, y0) is a station-ary point f , and that f has continuous second-order partial derivatives in some open set that

contains (x0, y0). Let

D (x0, y0) = fxx (x0, y0) fyy (x0, y0)− (fxy (x0, y0))2 .

1. If D (x0, y0) > 0 and fxx (x0, y0) > 0, f has a local minimum at (x0, y0) ,

2. If D (x0, y0) > 0 and fxx (x0, y0) < 0, f has a local maximum at (x0, y0) ,3. If D (x0, y0) < 0, (x0, y0) is a saddle point of f .

Remark 4 We will refer to

D (x, y) = fxx (x, y) fyy (x, y)− (fxy (x, y))2

as the discriminant of f at (x, y). The discriminant can be expressed as a determinant:

D (x, y) =

¯fxx (x, y) fxy (x, y)fyx (x, y) fyy (x, y)

¯= fxx (x, y) fyy (x, y)− (fxy (x, y))2 ,

since fxy (x, y) = fyx (x, y), under the assumption that the second-order partial derivatives of fare continuous.


The second derivative test does not provide information about the nature of the stationary point

(x0, y0) if the discriminant of f is 0 at (x0, y0). For example, let

f (x, y) = x4 + y4, g (x, y) = −x4 − y4 and h (x, y) = x4 − y4.Then (0, 0) is the only stationary point of each function. The discriminant vanishes at (0, 0) ineach case. The function f attains its minimum value 0 at (0, 0), g attains its maximum value 0

at (0, 0) and h has a saddle point at (0, 0), as you can confirm easily. ♦

Example 5 Let

f (x, y) = 2x2 + 2xy + y2,

as in Example 4. Determine the nature of the stationary point of f .

Solution

As in Example 4,∂f

∂x(x, y) = 4x+ 2y and

∂Q

∂y(x, y) = 2x+ 2y,

and (0, 0) is the only stationary point of QWe have

fxx (x, y) = 4, fyy (x, y) = 2 and fxy (x, y) = 2

at each (x, y) ∈ R2. Therefore,

D (0, 0) = fxx (0, 0) fyy (0, 0)− (fxy (0, 0))2 = 4 (2)− 22 = 4 > 0.Since fxx (0, 0) = 4 > 0, f has a local minium at (0, 0), as we inferred in Example 4 from the

graph of f . Figure 8 shows some level curves, and the darker color indicates a lower level on the

graph of f . The picture is consistent with our assertion that f has a local minimum at (0, 0). ¤

-4 -2 0 2 4

-4

-2

0

2

4

Figure 8

Remark 5 The function of Example 5 is a quadratic function. The nature of the stationary

point (x0, y0) of a quadratic function can be determined by completing the squares in terms ofthe powers of X = x− x0 and Y = y − y0 (we translate the origin to (x0, y0)). For example, if

f (x, y) = 2x2 + 2xy + y2,

as in Example 5, the only stationary point is (0, 0). We have

f (x, y) = 2¡x2 + xy

¢+ y2

= 2³x+

y

2

´2− y

2

2+ y2 = 2

³x+

y

2

´2+y2

2.


Therefore,

f (x, y) ≥ 0 = f (0, 0)for each (x, y) ∈ R2. This shows that f attains its absolute minimum at (0, 0). ♦

Example 6 Let f (x, y) = x2. Determine the nature of the stationary points of f .

Solution

Since∂f

∂x(x, y) = 2x and

∂f

∂y(x, y) = 0,

(x, y) is a stationary point of f if x = 0 and y is an arbitrary real number. Thus, any point(0, y) on the y-axis is a stationary point of f . Since

∂2f

∂x2(x, y) = 2, .

∂2f

∂y∂x(x, y) = 0 and

∂2f

∂x2(x, y) = 0,

the discriminant of f is

D (x, y) =

µ∂2f

∂x2(x, y)

¶µ∂2f

∂x2(x, y)

¶−µ

∂2f

∂y∂x(x, y)

¶2= 0.

This is another case where the second derivative test does not give any information about the

nature of the stationary points of the function.

Note that

f (x, y) = x2 ≥ 0and f (0, y) = 0 for each y ∈ R, so that f attains its absolute minimum on each point of the

y-axis. Figure 9 shows the graph of f which is a cylinder that has the same cross section (a

parabola) on any plane that is parallel to the xz-plane. ¤

04

5

z

10

15

24

y

0 2

x

0-2-2

-4 -4

Figure 9

Example 7 Let

f (x, y) = x2 + 4xy + y2

Determine the nature of the stationary points of f .

Solution

We have∂f

∂x(x, y) = 2x+ 4y and

∂f

∂y(x, y) = 4x+ 2y,


Therefore (x, y) is a stationary point of f if and only if

2x+ 4y = 0,

4x+ 2y = 0.

The only solution is (0, 0). Thus, (0, 0) is the only stationary point of f .We have

fxx (x, y) = 2, fxy (x, y) = 4 and fyy (x, y) = 2,

so that

D (x, y) = (fxx (x, y)) (fyy (x, y))− (fxy (x, y))2 = (2) (2)− 16 = −12 < 0Therefore, f has a saddle point (0, 0). Figure 10 shows the graph of f and Figure 11 shows somelevel curves of f . The pictures are consistent with the fact that f has a saddle point at (0, 0).¤

4

0

z

50

2

100

y

0 42

-2

x

0-2

-4 -4

Figure 10

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

Figure 11

Remark 6 With reference to Example 7, we can determine directions in which f (x, y) increasesor decreases by completing the squares:

f (x, y) = x2 + 4xy + y2 = (x+ 2y)2 − 4y2 + y2= (x+ 2y)

2 − 3y2.Thus,

f (x, 0) = x2,

so that the restriction of f to the x-axis has its minimum at (0, 0). We also have

f (−2y, y) = −3y2,so that the restriction of to the line x = −2y has its maximum value at (0, 0). ♦


Example 8 Let

f (x, y) =1

3x3 +

1

3y3 − xy.

Determine the nature of the stationary points of f/

Solution

Figure 12 shows the graph of f . The picture does not give a clear indication about the stationary

points of f . In any case, our analysis will clarify matters.

-10-2

-5

z

0

-1

5

x

02

11

y

0-1

2 -2

Figure 12

We have

fx (x, y) = x2 − y and fy (x, y) = y2 − x

so that (x, y) is a stationary point of f if and only if

x2 − y = 0,y2 − x = 0.

From the first equation, y = x2. Therefore, the second equation leads to the following:

x4 − x = 0⇒ x¡x3 − 1¢ = 0⇒ x = 0 or x = 1.

If x = 0, then y = 0, Therefore, (0, 0) is a stationary point. If x = 1, then y = 1. Therefore,(1, 1) is a stationary point.

We have

fxx (x, y) = 2x, fxy (x, y) = −1 and fyy (x, y) = 2y.

Therefore

D (x, y) = (2x) (2y)− (−1)2 = 4xy − 1.

Thus,

D (0, 0) = −1 < 0

Therefore, f has a saddle point at (0, 0). Figure 13 shows some level curves of f near (0, 0). Thepicture is consistent with our assertion that (0, 0) is a saddle point of f . Clearly, f increases insome directions and decreases in some directions.


-0.4 -0.2 0 0.2 0.4

-0.4

-0.2

0

0.2

0.4

Figure 13

At (1, 1),

fxx (1, 1) = 2, fxy (1, 1) = −1 and fyy (1, 1) = 2.

Therefore,

D = (2) (2)− 1 = 3 > 0.

We have fxx (1, 1) = 2 > 0. Therefore, f has a local minimum at (1, 1). The level curves inFigure 14 are consistent with our assertion.¤

-1 -0.5 0 0.5 1 1.5 2-1

-0.5

0

0.5

1

1.5

2

Figure 14

Example 9 Let

f (x, y) =¡y2 − x2¢ e−x2−y2 .

Determine the nature of the critical points of f .

Solution

Figure 15 shows the graph of f . The picture indicates that f has several local extrema and at

least one saddle point. Our analysis will clarify the picture.


Figure 15

Then,

fx (x, y) =∂

∂x

³¡y2 − x2¢ e−x2−y2´

= −2xe−x2−y2 − ¡y2 − x2¢ 2xe−x2−y2= −2xe−x2−y2 ¡1 + ¡y2 − x2¢¢

and

fy (x, y) =∂

∂y

³¡y2 − x2¢ e−x2−y2´

= 2ye−x2−y2 − 2y ¡y2 − x2¢ e−x2−y2

= 2ye−x2−y2 ¡1− ¡y2 − x2¢¢ .

Therefore,

−2x ¡1 + ¡y2 − x2¢¢ = 0 and 2y ¡1− ¡y2 − x2¢¢ = 0If x = 0, then

2y¡1− y2¢ = 0⇔ y = 0 or y = ±1

Therefore, (0, 0), (0,−1) and (0, 1) are stationary points.If y2 − x2 = −1, then 4y = 0 ⇒ y = 0, so that x = ±1. Therefore, (−1, 0) and (1, 0) arestationary points.

We have

∂2f (x, y)

∂x2= −2e−x2−y2 + 10x2e−x2−y2 − 2e−x2−y2y2 + 4x2e−x2−y2y2 − 4x4e−x2−y2 ,

∂2f (x, y)

∂y2= 2e−x

2−y2 − 10e−x2−y2y2 + 2x2e−x2−y2 + 4e−x2−y2y4 − 4x2e−x2−y2y2,∂2f (x, y)

∂y∂x= −4 ¡−y2 + x2¢ yxe−x2−y2

The discriminant of f is

D (x, y) =

µ∂2f (x, y)

∂x2

¶µ∂2f (x, y)

∂y2

¶−µ∂2f (x, y)

∂y∂x

¶2Table 1 displays the values the second-order partial derivatives and the discriminant of f at the

stationary points, and the nature of the stationary points. Figure 16 shows some level curves of

f . The picture is consistent with our conclusions.


(x, y) fxx fyy fxy D (x, y)(0, 0) −2 2 0 −4 saddle point

(0, 1) −4e−1 −4e−1 0 16e−2 maximum

(0,−1) −4e−1 −4e−1 0 16e−2 maximum

(1, 0) 4e−1 4e−1 0 16e−2 minimum

(−1, 0) 4e−1 4e−1 0 16e−2 minimum

Table 1

-2 -1 0 1 2

-2

-1

0

1

2

Figure 16

A Plausibility Argument for the Second Derivative Test

Proposition 2 Let Q (x, y) be a quadratic function. Let (x0, y0) be an arbitrary point. Then,

Q (x, y) = Q (x0, y0) +∂Q

∂x(x0, y0) (x− x0) + ∂Q

∂y(x0, y0) (y − y0)

+1

2

∂2Q

∂x2(x− x0)2 + ∂2Q

∂x∂y(x− x0) (y − y0) + 1

2

∂2Q

∂y2(y − y0)2

Proof

We can express Q (x, y) as

Q (x, y) = A (x− x0)2 + 2B (x− x0) (y − y0) + C (y − y0)2 +D (x− x0) +E (y − y0) + F.Then.

∂Q

∂x= 2A (x− x0) + 2B (y − y0) +D,

∂Q

∂y= 2B (x− x0) + 2C (y − y0) +E,

∂2Q

∂x2= 2A,

∂2Q

∂y2= 2C,

∂2Q

∂y∂x= 2B.

Therefore,

∂Q

∂x(x0, y0) = D,

∂Q

∂y(x0, y0) = E,

∂2Q

∂x2(x0, y0) = 2A,

∂2Q

∂y2(x0, y0) = 2C,

∂2Q

∂y∂x(x0, y0) = 2B


¥As a result of Proposition 2, the quadratic function that has the same value as f at (x0, y0) andhas the same first and second partial derivatives at (x0, y0) is uniquely determined:

Definition 5 Assume that f has continuous second-order partial derivatives in some open set

that contains (x0, y0). The quadratic approximation to f based at (x0, y0) is

Q (x, y) = f (x0, y0) +∂f

∂x(x0, y0) (x− x0) + ∂f

∂y(x0, y0) (y − y0)+

+1

2

∂2f

∂x2(x0, y0) (x− x0)2 + ∂2f

∂y∂x(x0, y0) (x− x0) (y − y0) + 1

2

∂2f

∂y2(x0, y0) (y − y0)2

We have Q (x, y) ∼= f (x, y) if (x, y) is close to (x0, y0) and the magnitude of the error is muchsmaller than ||(x− x0, y − y0)||2 if ||(x− x0, y − y0)|| is small. In fact, the errorR (x− x0, y − y0)is such that

lim(x,y)→(x0,y0)

R (x− x0, y − y0)||(x− x0, y − y0)||2

= 0.

We leave the proof of this fact to a course in advanced calculus. Note that the linear approx-

imation to f based at (x0, y0) is

L (x, y) = f (x0, y0) +∂f

∂x(x0, y0) (x− x0) + ∂f

∂y(x0, y0) (y − y0) ,

so that

Q (x, y) = L (x, y)

+1

2

∂2f

∂x2(x0, y0) (x− x0)2 + ∂2f

∂y∂x(x0, y0) (x− x0) (y − y0) + 1

2

∂2f

∂y2(x0, y0) (y − y0)2 .

If (x0, y0) is a stationary point of f , we have

∂f

∂x(x0, y0) = 0 and

∂f

∂y(x0, y0) = 0,

so that the quadratic approximation to f based at (x0, y0) is

Q (x, y) = f (x0, y0) +1

2

∂2f

∂x2(x0, y0) (x− x0)2

+∂2f

∂y∂x(x0, y0) (x− x0) (y − y0) + 1

2

∂2f

∂y2(x0, y0) (y − y0)2

Since Q (x, y) approximates f (x, y) very well near (x0, y0), it is reasonable to expect that thenature of the stationary point (x0, y0) of f will be revealed by analyzing the behavior of thequadratic function Q (x, y).We can set X = x− x0, Y = y − y0 and

Q (X,Y ) = Q (X + x0, Y + y0)− f (x0, y0)= +

1

2

∂2f

∂x2(x0, y0)X

2 +∂2f

∂y∂x(x0, y0)XY +

1

2

∂2f

∂y2(x0, y0)Y

2,

so that (0, 0) is the stationary point of Q and Q (0, 0) = 0. We might as well assume that(x0, y0) = (0, 0) and Q (0, 0) = 0 and simplify the notation. Thus, we assume that

Q (x, y) =1

2Qxxx

2 +Qxyxy +1

2Qyyy

2.


Note the the second-order partial derivatives and the discriminant of a quadratic function are

constants.

1. Let’s assume that D = QxxQyy − (Qxy)2 > 0 and Qxx > 0. We complete the square:

Q (x, y) =1

2Qxx

µx2 +

2Qxyxy

Qxx

¶+1

2Qyyy

2

=1

2Qxx

µx+

Qxy

Qxxy

¶2− 12

Q2xy

Qxxy2 +

1

2Qyyy

2

=1

2Qxx

µx+

Qxy

Qxxy

¶2+1

2

QxxQyy −Q2xyQxx

y2.

Therefore, Q (x, y) ≥ 0 for each (x, y) and Q (x, y) = 0 iff (x, y) = (0, 0). Thus, Q attains its

minimum value 0 at (0, 0) (there is no maximum).

2. Let’s assume that D = QxxQyy − (Qxy)2 > 0 and Qxx < 0. Since

Q (x, y) =1

2Qxx

µx+

Qxy

Qxxy

¶2+1

2

QxxQyy −Q2xyQxx

y2,

and Qxx < 0 andQxxQyy −Q2xy

Qxx< 0,

we have

Q (x, y) ≤ 0for each (x, y) and Q (x, y) = 0 iff (x, y) = (0, 0). Thus, Q attains its maximum value 0 at (0, 0)(there is minimum).

3. Assume that D = QxxQyy − (Qxy)2 < 0. If Qxx 6= 0,

Q (x, y) =1

2Qxx

µx+

Qxy

Qxxy

¶2+1

2

QxxQyy −Q2xyQxx

y2

If Qxx > 0,

Q (x, 0) =1

2Qxxx

2 > 0 if x 6= 0.On the other hand, if

x+Qxy

Qxxy = 0,

then

Q (x, y) =1

2

QxxQyy −Q2xyQxx

y2 < 0

if y 6= 0. Therefore, Q has a saddle point at (0, 0) .If Qxx < 0, then

Q (x, 0) =1

2Qxxx

2 < 0 if x 6= 0,whereas if

x+Qxy

Qxxy = 0

then

Q (x, y) =1

2

QxxQyy −Q2xyQxx

y2 > 0

12.9. ABSOLUTE EXTREMA AND LAGRANGE MULTIPLIERS 121

if y 6= 0. Therefore Q has a saddle point at (0, 0) .There is a similar argument which shows that f has a saddle point at (0, 0) if Qyy 6= 0.If we have both Qxx and Qyy equal to 0, then

Q (x, y) = Qxyxy,

and

D = −Q2xy < 0⇒ Qxy 6= 0,so that

Q (x, x) = Qxyx2 and Q (x,−x) = −Qxyx2.

Again, we have saddle a saddle point at (0, 0) (there are many lines along which Q increases aswe move away from (0, 0) and many lines along which Q decreases.

Problems

In problems 1 - 8 determine the nature of the critical points of f (maximum minimum, or saddle

point):

1.

f (x, y) = x2 − y2 + xy + x− y2.

f (x, y) = x2 + y2 − xy + y3.

f (x, y) = x2 + y2 + 2xy − 104.

f (x, y) = x2 + y2 + 3xy − 3y5.

f (x, y) = x2 − 3xy + 5x− 2y + 6y2 + 86.

f (x, y) = 3x− x3 − 3xy2

7.

f (x, y) = −x4 − y4 − 4xy + 1

16.

8.

f (x, y) = x3 − 12xy + 8y3

12.9 Absolute Extrema and Lagrange Multipliers

Absolute Extrema

As we saw in Section 13.6, a function need not have an absolute maximum or minimum on a

given set. For example, if f (x, y) = x2 + y2 then f does not have an absolute maximum on R2

since f attains arbitrarily large values on R2. Indeed,

limx→+∞ f (x, x) = lim

x→+∞ 2x2 = +∞.

The function does have an absolute minimum:

f(x, y) = x2 + y2 ≥ 0 = f(0, 0).


0

5

10z

15

2

y

20

x

0-2

-2

Figure 1: f(x, y) = x2 + y2

On the other hand, if we set g (x) = −f (x) = −x2 − y2, then g does not have an absoluteminimum, since g attains negative values of arbitrarily large magnitude. The function has an

absolute maximum on R2 since

g (x, y) = −x2 − y2 ≤ 0 = g (0, 0) .

-15

-10z

-5

2

0

2

y

0

x

0-2

-2

Figure 2: g(x, y) = −x2 − y2

In the above cases, the underlying set (R2) is unbounded. A function need not attain an absoluteextremum on a bounded set either. For example, if f (x, y) = x2 + y2 and D is the open unit

disk©(x, y) : x2 + y2 < 1

ª, f does not attain an absolute maximum on D since

f(x, y) = x2 + y2 < 1

for each (x, y) ∈ D and f has values that are arbitrarily close to 1. Indeed,

limx2+y2=r2

r→1f (x, y) = lim

r→1r2 = 1.

Thus, the only candidate for the absolute maximum of f on D is 1, but f does not attain the

value 1 in D.

Just as in the case of functions of a single variable, a continuous function of several variables

attains its absolute extrema on a closed and bounded set. Let’s state the theorem for a function

of two variables:

Theorem 1 Assume that D ⊂ R2 is closed and bounded, and f is a scalar func-

tion that is continuous on D. Then f attains its absolute maximum and absolute

minimum on D.

We leave the proof of Theorem 1 to a course in advanced calculus.

Just as in the case of functions of a single variable, we can implement the following strategy in

order to find the absolute extrema of a function f that is continuous on a closed and bounded

set D ⊂ R2 :


1. Determine the stationary points of f in the interior of D and the points in the

interior of D where f does not have partial derivatives. Calculate the corresponding

values of f .

2. Determine the maximum and minimum values of f on the boundary of D.

3. Determine the absolute extrema of f on D by comparing the values that have

been computed in steps 1 and 2.

Example 1 Let f (x, y) = x2 + y2. Determine the absolute maximum and minimum of f on

the square D = [−2, 2]× [−2, 2] .Solution

The only stationary point of f is (0, 0) and f (0, 0) = 0.The restriction of f to the side L1 = (x,−2) : −2 ≤ x ≤ 2 of D is

f (x,−2) = x2 + 4.

Therefore, the maximum on f on L1 is f (2,−2) = 8 and the minimum of f on L1 is f (0,−2) = 4.Since f (x, 2) = x2 + 4 = f (x,−2), the maximum on f on L2 = (x, 2) : −2 ≤ x ≤ 2 is also 8and the minimum of f on L2 is 4.Similarly, the maximum of f on the other sides of D is 8 and the minimum of f on the other

sides of D is 4.

Thus, the absolute maximum of f on D is 8 and the absolute minimum of f is 0. ¤

Lagrange Multipliers

There is a useful necessary condition for a function to have a local extremum on a curve:

Theorem 2 (Lagrange Multipliers) Assume that C is a smooth curve in R2 that is

the graph of the equation g (x, y) = c, where c is a constant. If the restriction of thedifferentiable function f to C has a local extremum at the point (x0, y0) then thereexists a constant λ such that

∇f (x0, y0) = λ∇g (x0, y0) ,

i.e.,

∂f

∂x(x0, y0) = λ

∂g

∂x(x0, y0) ,

∂f

∂y(x0, y0) = λ

∂g

∂y(x0, y0)

Proof

Since the curve C is assumed to be smooth, we can assume that a segment of C that contains

(x0, y0) in its interior can be parametrized by a function σ : J → R2 such that σ (t0) = (x0, y0),where t0 is in the interior of the interval J . Thus, f σ has a local extremum at t0, so that

df (σ (t))

dt

¯t=t0

= 0.

By the chain rule,

df (σ (t))

dt

¯t=t0

=∇f (σ (t0)) · dσdt(t0) =∇f (x0, y0) · dσ

dt(t0) = 0


Therefore, ∇f (x0, y0) is orthogonal to the vector σ0 (t0) that is tangent to C at (x0, y0). SinceC is the graph of g (x, y) = c, we know that ∇g (x0, y0) is also orthogonal to C at (x0, y0).Therefore, ∇f (x0, y0) is parallel to ∇g (x0, y0). Thus, there is a scalar λ such that

∇f (x0, y0) = λ∇g (x0, y0) .

In terms of the components of the vectors,

∂f

∂x(x0, y0) = λ

∂g

∂x(x0, y0) ,

∂f

∂y(x0, y0) = λ

∂g

∂y(x0, y0)

¥

Remark 1 The fact that ∇f (x0, y0) is parallel to ∇g (x0, y0) implies that the level curve of fthat passes through (x0, y0) and the curve g (x, y) = 0 have a common tangent at (x0, y0).♦

With reference to the statement of Theorem 2, the number λ is called a Lagrange multiplier.

Note that we have to solve the three equations

∂f

∂x(x, y) = λ

∂g

∂x(x, y) .

∂f

∂y(x, y) = λ

∂g

∂y(x, y) ,

g (x, y) = c,

in the three unknowns x, y and λ.

Example 2 Determine the extrema of f (x, y) = y2 − x2 on the unit circle x2 + y2 = 1.

Solution

We set g (x, y) = x2 + y2, so that the constraint is g (x, y) = 1. We have

∇f (x, y) = (−2x, 2y) and ∇g (x, y) = (2x, 2y) .Therefore, ∇f (x, y) = λ∇g (x, y) iff

−2x = 2λx2y = 2λy

i.e.,

(1 + λ)x = 0

(1− λ) y = 0

and

x2 + y2 = 1

From the first equation, x = 0 or λ = −1. If x = 0, then y = ±1. Therefore, the candidates are(0,−1) and (0, 1).If λ = −1, then y = 0, so that x = ±1. Therefore, the candidates are (−1, 0) and (0, 1). Wehave

f (0,±1) = 1and f (±1, 0) = −1


Therefore, the minimum of f on the unit circle is −1 and the maximum is 1.

Figure 3 shows some level curves of f and the unit circle. Note that the level curves of f that

pass through the points determined with the help of Theorem 2.are tangent to the "constraint

curve" x2 + y2 = 1 at those points (Remark 1. ¤

2 1 0 1 2

2

1

0

1

2

0, 1

0, 1

1, 01, 0

Figure 3

Remark 2 We could have obtained the result directly:

On the circle,

f (x, y) = y2 − x2 = ¡1− x2¢− x2 = 1− 2x2,We have

d

dx

¡1− 2x2¢ = −4x = 0⇒ x = 0,

so that we obtain the points (0,±1). We must also take into account the endpoints ±1, so thatthe corresponding points are (−1, 0) and (1, 0). ♦

Example 3 Let f (x, y) = y2 − x2. Determine the maximum and minimum values of f inside

the unit disk

G =©(x, y) : x2 + y2 ≤ 1ª

Solution

In Example 2 we determined the maximum and minimum values of f on the boundary:

f (±1, 0) = −1 and f (0,±1) = 1

The critical point in the interior is (0, 0). We have f (0, 0) = 0..Therefore, the maximum value

of f in G is 1 and the minimum value is −1. ¤The theorem on Lagrange multipliers has counterparts for functions of any number of variables.

For example, assume that the surface S is the graph of the equation g (x, y, z) = c, where c

is a constant. Thus, S is a level surface of g. Also assume that f attains a local extremum

at P0 = (x0, y0, z0). We know that ∇g (P0) is orthogonal at P0 to any curve on S that passesthrough P0. Assume that C is a curve on S such that σ (t0) = P0. The function f (σ (t)) attainsa local extremum at t0. Therefore,

0 =df (σ (t))

dt

¯t=t0

=∇f (σ (t0)) · dσdt(t0)


Therefore, ∇f (P0) is also orthogonal to such a curve at P0. Thus, ∇f (P0) is parallel to∇g (P0). Therefore there exists a number λ such that

∇f (P0) = λ∇g (P0) .Therefore,

∂f

∂x(x0, y0, z0) = λ

∂g

∂x(x0, y0, z0)

∂f

∂y(x0, y0, z0) = λ

∂g

∂y(x0, y0, z0)

∂f

∂z(x0, y0, z0) = λ

∂g

∂z(x0, y0, z0)

and

g (x0, y0, z0) = c

Example 4 Let f (x, y, z) = x + y + z. Determine the maximum of f on the unit sphere

x2 + y2 + z2 = 1.

Solution

If we set g (x, y, z) = x2 + y2 + z2, the constraint equation is

g (x, y, z) = 1.

Therefore, we need to have ∇f (x, y, z) = λ∇g (x, y, z) for some λ. Since∇f (x, y, z) = (1, 1, 1) and ∇g (x, y, z) = (2x, 2y, 2z) ,

we obtain the equations

1 = 2λx

1 = 2λy

1 = 2λz,

and

x2 + y2 + z2 = 1.

Since 2λx = 1 we have λ 6= 0. Thus, the first three equations lead to the following equalities:

x = y = z =1

2λ

We substitute these equalities in the last equation:

3

4λ2= 1⇒ λ = ±

√3

2.

If λ =√3/2, we have

x = y = z =1

2λ=

1√3,

if λ = −√3/2, we havex = y = z =

1

2λ= − 1√

3


We have

f

µ1√3

¶=

3√3=√3,

and

f

µ− 1√

3

¶= −√3.

Therefore, the maximum value of f on the unit sphere is√3 and the minimum value is −√3.Note

that

∇f = (1, 1, 1)and

∇g (x, y, z) = (2x, 2y, 2z) ,so that

∇gµ1√3,1√3,1√3

¶=

2√3(1, 1, 1) .

Therefore

∇f =√3

2∇g

µ1√3,1√3,1√3

¶Thus, the graph of f (a plane) is "tangential" to the graph of the equation g (x, y, z) = 1 (asphere) at the point was determined. Figure 4 is consistent with this observation. ¤

Figure 4

Maximizing Output under Budgetary Constraints

Assume that the total value of a certain product produced by a firm depends on the capital

and the cost of labor. Thus, if we denote the value of the product by z, the capital by x and

the cost of labor by y, there exists a function f such that z = f (x, y). The firm would like to

maximize f subject to a constraint g (x, y) = qx + py = B, where p, q and B are constants.

Thus, the necessary conditions provided by the technique of Lagrange multipliers are

∂f

∂x= λq,

∂f

∂y= λp and py + qx = B.

Let X = qx and Y = py, so that X is the dollar value of capital and Y is the dollar value of

labor. Then,∂f

∂X=1

q

∂f

∂x= λ and

∂f

∂Y=1

p

∂f

∂y= λ.


Thus,∂f

∂X=

∂f

∂Y= λ

Thus, at the optimum production level, the marginal change in output per dollar’s worth of

additional capital investment is equal to the marginal change in output per dollar’s worth of

additional labor, and λ is the common value. Thus, at the optimum production, the exchange

of a dollar’s worth of capital for a dollar’s worth of labor does not change the output.

Another interpretation: Consider the optimal value as a function of the budget B. Set

x (B) and y (B) denote the optimal capital and labor, respectively corresponding to g (x, y) =qx+ py = B. We have

d

dBf (x (B) , y (B)) =

∂f

∂x

dx

dB+

∂f

∂y

dy

dB

We have

fx = λgx and fy = λgy

and (x, y) = (x (B) , y (B)). Therefore,

d

dBf (x (B) , y (B)) =

∂f

∂x

dx

dB+

∂f

∂y

dy

dB

= λ

µ∂g

∂x

dx

dB+

∂g

∂y

dy

dB

¶= λ

g (x (B) , y (B))

dB

Since g (x (B) , y (B)) = B, we have

g (x (B) , y (B))

dB= 1.

Therefore,d

dBf (x (B) , y (B)) = λ

Thus, the value of λ is the rate of the maximum value of f with respect to the budgetary

allocation.

Example 5 Assume that

z = f (x, y) = Axαy1−α,

where A > 0, α > 0 and α < 1. This is a Cobb-Douglas production function. The price ofcapital is q, the price of labor is p, and we have a budgetary constraint g (x, y) = qx+py =B, where B is a positive constant. We would like to maximize the value of the product subject

to the budgetary constraint.

Solution

We apply the technique of Lagrange multipliers. We need to have ∇f = λ∇g. Thus,αAxα−1y1−α = λq,

(1− α)Axαy−α = λp,

qx+ py = B.

Eliminating λ from the first two equations,

αAxα−1y1−α

q=(1− α)Axαy−α

p,


so that

αpy = (1− α) qx.

Therefore,

qx+1− α

αqx = B ⇒ 1

αqx = B ⇒ x =

αB

q,

and

py = B − qx = B − αB = (1− α)B ⇒ y =1− α

pB.

Therefore, the candidate for (x, y) in order to maximize the value of the output isµαB

q,1− α

pB

¶the corresponding value of λ is

λ =αAxα−1y1−α

q=

αA³αBq

´α−1 ³1−αp B

´1−αq

=αAαα−1Bα−1 (1− α)

1−αB1−α

qα−1pα−1q

=Aαα (1− α)

1−α

qαpα−1

The corresponding value of f is

f (x (B) , y (B)) = Axαy1−α = AµαB

q

¶αµ1− α

pB

¶1−α=ABαα (1− α)

1−α

qαpα−1

We do have

λ (B) =d

dBf (x (B) , y (B))

For example let A = 1, B = 1, α = 1/3. Then,

z = f (x, y) = x1/3y2/3.

subject to the constraint qx+ py = 1. Let q = 1 and p = 2. Then, x+ 2y = 1 is the constraint.The optimal values are

x =αB

q=

13

1=1

3and y =

23

2=1

3.

We have

z = x1/3y2/3 =1

31/3

µ1

3

¶2/3=1

3

Figure 5 shows the relevant level curve of f and the graph of the constraint x+ 2y = 1. ¤


0.0 0.5 1.0 1.5 2.0

0.0

0.5

1.0

1.5

2.0

13, 13

Figure 5

Problems

In problems 1-6, make use of Lagrange multipliers to determine the maximum and minimum

values of f on the set D:

1.

f (x, y) = x+ y; D = (x, y) : x2 + y2 = 4.2.

f (x, y) = 3x− 4y, D = (x, y) : x2 + y2 = 93.

f (x, y) = xy, D = (x, y) : 4x2 + y2 = 84.

f (x, y) = x+ y2, D = (x, y) : 2x2 + y2 = 15.

f (x, y) = x2 + 2y2, D = (x, y) : x2 + y2 ≤ 46.

f (x, y) = xy, D = (x, y) : x2 + 2y2 ≤ 1

Chapter 13

Multiple Integrals

In this chapter we will discuss the integrals of real-valued functions of several variables. Inte-

gration of functions of two variables involves double integrals. In certain cases it is useful

to make use of polar coordinates. Integration of functions of three variables involves triple

integrals. In certain cases it is useful to use cylindrical or spherical coordinates. Double

and triple integrals are needed for the calculation of volumes and mass. They are also essential

for the understanding of the major facts of vector analysis that will be taken up in the next

chapter.

13.1 Double Integrals over Rectangles

Let D be the rectangle [a, b]× [c, d] so that

D =©(x, y) ∈ R2 : a ≤ x ≤ b and c ≤ y ≤ dª .

Assume that f is a real-valued function that is continuous on D and f (x, y) ≥ 0 for each(x, y) ∈ D. We would like to determine the volume of the region G in the xyz-space that is

between the graph of z = f (x, y) and D.

z

d b

y xac

Figure 1: The integral of a positive-valued function is a volume

As in the case of the area problem in connection with single-variable functions, we will begin

with approximations. Let

a = x0 < x1 < · · · < xj−1 < xj < · · · < xm−1 < xm = b,

and

c = y0 < y1 < · · · < yk−1 < yk < · · · < yn−1 < yn = d,

131

132 CHAPTER 13. MULTIPLE INTEGRALS

be partitions of [a, b] and [c, d], respectively. Let

∆xj = xj − xj−1 and ∆yk = yk − yk−1denote the lengths of the corresponding intervals. The grid that consists of the lines x = xj ,

j = 0, 1, 2, . . . ,m and y = yk, k = 1, 2, . . . , n, forms a collection of rectangles

Rjk = [xj−1, xj ]× [yk−1, yk]that covers the rectangle D.

a b

c

d

x j1 x j

yk1

ykRjk

x

y

Figure 2: A rectangle is covered by smaller rectangles

We denote the area of Rjk by ∆Ajk, so that

∆Ajk = ∆xj ∆yk,

We sample a point¡x∗j , y

∗k

¢from the rectangle Rjk and approximate the volume of the part of

the region G between the graph of f and Rjk by the volume of the rectangular box with base

Rjk and height f¡x∗j , y

∗k

¢, namely,

f¡x∗j , y

∗k

¢∆Ajk = f

¡x∗j , y

∗k

¢∆xj ∆yk

z

y

x

Figure 3: The volume over Rjk is approximated by the volume of a box

The volume of G is approximated by the Riemann sum

nXk=1

mXj=1

f¡x∗j , y

∗k

¢∆Ajk =

nXk=1

mXj=1

f¡x∗j , y

∗k

¢∆xj ∆yk

13.1. DOUBLE INTEGRALS OVER RECTANGLES 133

Under the assumption that f is continuous on D, there is number that is approximated by such

a sum with desired accuracy provided that the rectangles Rjkare small enough. We will identify

the volume of G with that number and denote it by the symbolZ ZD

f (x, y) dA =

Z ZD

f (x, y) dxdy.

Thus, ¯¯ nXk=1

mXj=1

f¡x∗k, y

∗j

¢∆Ajk −

Z ZD

f (x, y) dA

¯¯

is as small as we please provided that the maximum of ∆Ajk is small enough. The sums andthe approximation property make sense even if f is of variable sign over D:

Definition 1 Let f be defined on the rectangle D = [a, b]× [c, d]. We say that f is integrableon D and its double integral on D isZ Z

D

f (x, y) dA

if ¯¯ nXk=1

mXj=1

f¡x∗k, y

∗j

¢∆Ajk −

Z ZD

f (x, y) dA

¯¯

as small as required provided that the maximum of ∆Ajk is small enough.

Even though we motivated the definition of the double integral as the volume of the region

under the graph of a positive-valued function, the definition makes sense when the integrand

has variable sign. Just as in the case of the integral of a function of a single variable, we can

interpret a double integral as "signed volume": If f is negative-valued on D, the double

integral of f on D is (−1)× (volume of the region between the graph of f and D). We may

refer to a double integral simply as an integral.

y

x

z

Figure 4: The integral of a negative-valued function is signed volume

Remark It is known that the (double) integralZ ZD

f (x, y) dA

exists if f is continuous on D. We leave the proof of this fact to a course in advanced

calculus. ♦


The following theorem enables us to reduce the calculation of a double integral to the calculation

of integrals in a single variable:

Theorem 1 (Fubini’s Theorem) Let f be continuous on D = [a, b]× [c, d]. We haveZ ZD

f (x, y) dA =

Z x=b

x=a

ÃZ y=d

y=c

f (x, y) dy

!dx =

Z y=d

y=c

ÃZ x=b

x=a

f (x, y) dx

!dy.

The proof of Fubini’s Theorem is left to a course in advanced calculus. Intuitively, the theorem

corresponds to rearranging the double sum

nXk=1

mXj=1

f¡x∗j , y

∗k

¢∆Ajk =

nXk=1

mXj=1

f¡x∗j , y

∗k

¢∆xj ∆yk

asmXj=1

ÃnXk=1

f¡x∗j , y

∗k

¢∆yk

!∆xj or

nXk=1

⎛⎝ mXj=1

f¡x∗j , y

∗k

¢∆xj

⎞⎠ ∆yk.

You can also make a connection with the method of slices that was used for the computation

certain volumes. If f is positive-valued on D, the integral

A (x) =

Z y=d

y=c

f (x, y) dy

is the area of a cross section of the region between the graph of f and the rectangle D. The

volume of that region can be calculated asZ x=b

x=a

A (x) dx =

Z x=b

x=a

ÃZ y=d

y=c

f (x, y) dy

!dx.

z

A(x)

y xx

Figure 5: Fubini’s Theorem is related to the method of slices

We can reverse the roles of x and y and obtain the equalityZ ZD

f (x, y) dA =

Z y=d

y=c

ÃZ x=b

x=a

f (x, y) dx

!dy

Remark 1 We may use the notation Z ZD

f (x, y) dxdy


to denote Z ZD

f (x, y) dA.

Thus, dxdy does not mean that we integrate with respect to x first, unless we indicate the limits

for x and y.

If there is no danger of confusion, we may dispense with the parentheses and setZ x=b

x=a

Z y=d

y=c

f (x, y) dydx =

Z x=b

x=a

ÃZ y=d

y=c

f (x, y) dy

!dx,

and Z y=d

y=c

Z x=b

x=a

f (x, y) dxdy =

Z y=d

y=c

ÃZ x=b

x=a

f (x, y) dx

!dy.

♦

Example 1 Let f (x, y) = x2 + y2.Determine the volume of the region bounded by the graphof z = f (x, y), the xy-plane, and the planes x = ±2, y = ±2.Solution

02

5

z

2

y

0

x

0

-2 -2

Figure 6

The volume is Z ZD

¡x2 + y2

¢dxdy,

where D is the rectangle [−2, 2]× [−2, 2]. Therefore,Z ZD

¡x2 + y2

¢dxdy =

Z y=2

y=−2

µZ x=2

x=−2

¡x2 + y2

¢dx

¶dy

=

Z y=2

y=−2

Ãx3

3+ y2x

¯x=2x=−2

!dy

=

Z y=2

y=−2

µµ8

3+ 2y2

¶−µ−83− 2y2

¶¶dy

=

Z y=2

y=−2

µ4y2 +

16

3

¶dy

=4

3y3 +

16

3y

¯2y=−2

=128

3.

¤


Example 2 Determine Z ZD

x sin (xy) dxdy,

where D is the rectangle [−π,π]× [0, 1].

Solution

z

yx

Figure 7

We have Z ZD

x sin (xy) dxdy =

Z y=1

y=0

µZ x=π

x=−πx sin (xy) dx

¶dy.

The inner integral requires integration by parts. Let’s reverse the order of integration:Z ZD

x sin (xy) dxdy =

Z x=π

x=−π

µZ y=1

y=0

x sin (xy) dy

¶dx..

The inner integral is Z y=1

y=0

x sin (xy) dy = − cos (xy)|10 = − cos (x) + 1.

Therefore, Z ZD

x sin (xy) dxdy =

Z x=π

x=−π(− cos (x) + 1) dx = − sin (x) + x|π−π = 2π.

¤

Remark 2 If D is a rectangular region in R2 so that D = [a, b]× [c, d], an integral of the formZ ZD

f (x) g (y) dxdy

can be expressed as a product of single-variable integrals:

Z ZD

f (x) g (y) dxdy =

ÃZ b

a

f (x) dx

!ÃZ d

c

g (y) dy

!.


Indeed, Z ZD

f (x) g (y) dxdy =

Z x=b

x=a

ÃZ y=d

y=c

f (x) g (y) dy

!dx

=

Z x=b

x=a

f (x)

ÃZ y=d

y=c

g (y) dy

!dx

=

ÃZ y=d

y=c

g (y) dy

!Z x=b

x=a

f (x) dx

=

ÃZ b

a

f (x) dx

!ÃZ d

c

g (y) dy

!

♦

Example 3 Evaluate Z ZD

sin (x) cos (y) dxdy,

where D is the rectangle [−π/2, 0]× [0,π/2].Solution

z

x

y

Figure 8

Z ZD

sin (x) cos (y) dxdy =

Z π/2

y=0

Z 0

x=−π/2sin (x) cos (y) dxdy

=

ÃZ 0

x=−π/2sin (x) dx

!ÃZ π/2

y=0

cos (y) dxdy

!³− cos (x)|x=0x=−π/2

´³sin (y)|π/2y=0

´= (−1) (1) = −1.

Note that the integral is negative, consistent with the fact that the integrand is negative on D.

¤

Problems

In problems 1-4, calculate the given iterated integral:


1. Z y=2

y=1

µZ x=3

x=0

¡x2 + 3xy2

¢dx

¶dy

2. Z x=1

x=0

ÃZ y=π/3

y=π/4

x sin (y) dy

!dx

3. Z y=4

y=2

µZ x=3

x==1

exyydx

¶dy

4. Z x=2

x=0

µZ y=3

y=1

xyp1 + y2dy

¶dx

In problems 5-8, calculate the double integral:

5. Z ZD

xy2

x2 + 1dA where D = (x, y) : 0 ≤ x ≤ 1, −2 ≤ y ≤ 2

6. Z ZD

xyex2ydA where D = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2

7. Z ZD

x2

1 + y2dA where D = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

8. Z ZD

y cos (x+ y) dA, D = (x, y) : 0 ≤ x ≤ π

3, 0 ≤ y ≤ π

2

13.2 Double Integrals over Non-Rectangular Regions

In this section we will discuss double integrals on regions that are not rectangular. Let D be

a bounded region in the Cartesian coordinate plane that has a boundary consisting of smooth

segments. As in the case of a rectangle, let’s begin with a function f that is continuous and

nonnegative on D. In order to approximate the volume of the region G in the xyz-space that is

between the graph of z = f (x, y) and D we form a mesh in the xy-plane that consists of vertical

and horizontal lines. In this case D is not necessarily the union of rectangles. Let’s consider

only those rectangles that have nonempty intersection with D and label them as

Rjk = [xj−1, xj ]× [yk−1, yk] , where j = 0, 1, 2, . . . ,m and k = 0, 1, 2, . . . , n.

We set

∆xj = xj − xj−1 and ∆yk = yk − yk−1.We can approximate the volume of G by a sum

nXk=1

mXj=1

f¡x∗j , y

∗k

¢∆Ajk =

nXk=1

mXj=1

f¡x∗j , y

∗k

¢∆xj ∆yk,

where¡x∗j , y

∗k

¢ ∈ Rjk..The double integral of f on D is the numberZ ZD

f (x, y) dA =

Z ZD

f (x, y) dxdy

13.2. DOUBLE INTEGRALS OVER NON-RECTANGULAR REGIONS 139

that can be approximated by such sums as accurately as desired provided that the maximum

of ∆Ajk is small enough. Such a number exists for any continuous function f on D even if the

sign of f may vary.

There are counterparts of the Fubini Theorem that allowed us to calculated a double integral

on a rectangle as an iterated integral. Assume that the region D in the xy-plane is bounded by

the lines x = a, x = b and the graphs of functions f (x) and g (x), as in Figure 1. We will referto such a region as an x-simple region, or as a region of type 1.

Figure 1

Assume that F is continuous on D. We can evaluate the double integral as an iterated integral:Z ZD

F (x, y) dA =

Z x=b

x=a

ÃZ y=g(x)

y=f(x)

F (x, y) dy

!dx.

We will leave the proof to a course in advanced calculus. The statement should be plausible, as

in the case of rectangular regions: As x varies from a to b,.the region is swept by rectangles of

"infinitesimal" thickness dx that extend from y = g (x) to y = f (x) .The roles of x and y may be reversed: A region D is a y-simple region, or a region of type

2, if it is bounded by the lines y = a, y = b and the graphs of x = f (y) and x = g (y), as inFigure 2. We have Z Z

D

F (x, y) dA =

Z y=b

y=a

ÃZ x=g(y)

x=f(y)

F (x, y) dx

!dxy.

x

y

y b

y a

x fy x gy

Figure 2


p1 + x2dxdy,

where D is the triangle that is bounded by y = x, x = 1 and the x-axis.


Solution

Figure 3 shows the triangle D. The integral is the volume of the region between the graph of

z =√1 + x2 and the triangle D in the xy-plane, as shown in Figure 4.

0 x1

0

1

02

2y

1

z

2

Figure 4

The region D is both x-simple and y-simple. We will treat it as an x-simple region. Thus,

Z ZD

p1 + x2dxdy =

Z 1

x=0

µZ x

y=0

p1 + x2dy

¶dx =

Z 1

x=0

p1 + x2

µZ x

y=0

1dy

¶dx

=

Z 1

x=0

p1 + x2 (y|x0) dx

=

Z x=1

x=0

p1 + x2xdx.

We set u = 1 + x2, so that du = 2xdx. Therefore,Z x=1

x=0

p1 + x2xdx =

1

2

Z 2

u=1

√udu =

1

2

Ãu3/2

3/2

¯2u=1

!=23/2 − 13

.

¤


x sin (y) dA,

where D is the region bounded by y = x, y = x2, x = 0 and x = 1.

Solution

Figure 5 shows the region D.

1x

1

y

Figure 5


This is a region of type 1 and of type 2. We will treat it as a region of type I, and integrate

with respect to y first:

Z ZD

x sin (y) dA =

Z x=1

x=0

µZ y=x

y=x2x sin (y) dy

¶dx =

Z x=1

x=0

x

µZ y=x

y=x2sin (y) dy

¶dx

=

Z x=1

x=0

x³− cos (y)|y=xy=x2

´dx

=

Z x=1

x=0

x¡− cos (x) + cos ¡x2¢¢ dx

= −Z 1

0

x cos (x) dx+

Z 1

0

x cos¡x2¢dx.

The first integral requires integration by parts: If we set u = x and dv = cos (x), then

du = dx and v =

Zcos (x) dx = sin (x) .

Thus, Zx cos (x) dx =

Zudv = uv −

Zvdu

= x sin (x)−Zsin (x) dx

= x sin (x) + cos (x) .

Therefore, Z 1

0

x cos (x) dx = sin(1) + cos (1)− cos (0) = sin(1) + cos (1)− 1.

As for the second integral, we set u = x2 so that du = 2xdx. Thus,Zx cos

¡x2¢dx =

1

2

Zcos (u) du =

1

2sin¡x2¢.

Therefore, Z 1

0

x cos¡x2¢dx =

1

2sin (1) .

Thus, Z ZD

x sin (y) dA = −Z 1

0

x cos (x) dx+

Z 1

0

x cos¡x2¢dx

= − sin (1)− cos (1) + 1 + 12sin (1)

= 1− cos (1)− 12sin (1) .

¤


yexdA

where D is the region bounded by x = y2/4, x = 1, y = 1 and y = 2.


Solution

Figure 6 shows the region D.

14

1 2x

1

2

y

Figure 6

This is a region of type 1 and of type 2. We will treat it as a region of type 2, and integrate

with respect to x first:

Z ZD

yex2

dA =

Z y=2

y=1

ÃZ x=1

x=y2/4

yexdx

!dy =

Z y=2

y=1

y

ÃZ x=1

x=y2/4

exdx

!dy

=

Z y=2

y=1

y³ex|1y2/4

´dy

=

Z y=2

y=1

y³e− ey2/4

´dy

=e

2y2 − 2ey2/4

¯21

= 2e− 2e− e2+ 2e1/4

= −e2+ 2e1/4.

¤

In some cases, changing the order of integration enables us to evaluate an integral that is

intractable in its original form, as in the following example.

Example 4 Evaluate the iterated integralZ 1

x=0

µZ y=2

y=2x

e−y2

dy

¶dx.

Solution

We cannot evaluate the given iterated integral in terms of familiar antiderivatives. The first

thing to do is to sketch the region of integration:


1 2x

1

2

y

y = 2x

Figure 7

We will reverse the order of integration:Z 1

x=0

µZ 2

y=2x

e−y2

dy

¶dx =

Z y=2

y=0

ÃZ x=y/2

x=0

e−y2

dx

!dy =

Z y=2

y=0

e−y2

ÃZ x=y/2

x=0

dx

!dy

=

Z y=2

y=0

e−y2³x|y/20

´dy

=

Z y=2

y=0

e−y2 y

2dy

= −14e−y

2

¯20

= −14e−4 +

1

4.

¤

Problems

In problems 1 and 2 evaluate the given iterated integral:

1. Z y=4

y=0

Z x=√y

x=0

xy2dxdy

2. Z θ=π/2

θ=0

Z r=cos(θ)

r=0

esin(θ)drdθ

In problems 3-6, sketch the region D and evaluate the given double integral:

3. Z ZD

4x3ydA,

where D is the region bounded by the graphs of y = x2 and y = 2x.

4. Z ZD

y√xdA,

where D is the region bounded by the graphs of y = x2, y = 0, x = 0 and x = 4.

5. Z ZD

y2exydA,

where D is the region bounded by the graphs of y = x, y = 1 and x = 0.

6. Z ZD

y2 sin¡x2¢dA,


where D is the region bounded by the graphs of y = −x1/3, y = x1/3, x = 0 and x = √π.In problems 7-9 determine the volume of the region D in R3:

7. D is bounded by the paraboloid z = x2+y2+1, the xy-plane, the surfaces y = x2 and y = 1.

8. D is bounded by the graphs of z = ex, z = −ey, x = 0, x = 1, y = 0 and y = 1.9. D is the tetrahedron bounded by the plane 3x+ 2y + z = 6 and the coordinate planes.

In problems 10-12,

a) Sketch the region of integration,

b) Evaluate the integral by reversing the order of integration:

10. Z y=1

y=0

Z 3

x=3y

ex2

dxdy

11. Z y=√π/2

y=0

Z x=√π/2

x=y

cos¡x2¢dxdy.

12. Z y=1

y=0

Z π/2

x=arcsin(y)

cos (x)p1 + cos2 (x)dxdy.

13.3 Double Integrals in Polar Coordinates

In some cases it is convenient to evaluate a double integral by converting it to an integral in

polar coordinates.

Assume that D is the set of points in the plane with polar coordinates (r, θ) such that α ≤ θ ≤ β

and a ≤ r ≤ b,and we would like to approximateZ ZD

f (x, y) dA

by a sum that will be constructed as follows: Let

α = θ0 < θ1 < θ2 < · · · < θj−1 < θj < · · · < θm = β,

be a partition of the interval [α,β], and let

a = r0 < r1 < r2 < · · · < rk−1 < rk < · · · < rn = b

be a partition of [a, b]. We set

∆θj = θj − θj−1 and ∆rk = rk − rk−1.

The rays θ = θj and the arcs r = rk form a grid that covers D, as illustrated in Figure 1.

13.3. DOUBLE INTEGRALS IN POLAR COORDINATES 145

x

y

r

r r1

r r2

Θ Α

rΘ

Θ Β

Figure 1

The area of the "polar rectangle that consists of the points with polar coordinates (r, θ), whererk−1 < r < rk and θj−1 < θ < θj is

1

2r2k∆θj −

1

2r2k−1∆θj =

1

2

¡r2k − r2k−1

¢∆θj =

1

2(rk + rk−1) (rk − rk−1)∆θj

= rk∆rk∆θj ,

where

rk =1

2(rk + rk−1) .

Therefore, we can approximate the part ofZ ZD

f (x, y) dA

that corresponds to that polar rectangle by

f³rk cos

³θj

´, r sin

³θj

´´rk∆rk∆θj,

where

θj =1

2(θj + θj−1)

and the total integral by the sum

mXj=1

nXk=1

f³rk cos

³θj

´, r sin

³θj

´´rk∆rk∆θj .

This is a Riemann sum for the integralZ θ=β

θ=α

Z r=r2

r=r1

f (r cos (θ) , r sin (θ)) rdrdθ.

Thus, it is reasonable to conjecture thatZ ZD

f (x, y) dA =

Z θ=β

θ=α

Z r=r2

r=r1


This is indeed true. As a matter of fact the following more general fact valid:


Theorem 1 Assume that D is the set of points in the plane with polar coordinates (r, θ) suchthat α ≤ θ ≤ β and

r1 (θ) ≤ r ≤ r2 (θ) ,where the functions r1 (θ) and r2 (θ) are continuous. If f is continuous on D thenZ Z

D

f (x, y) dA =

Z θ=β

θ=α

Z r2(θ)

r=r1(θ)


x

y

r r1Θ

r r2Θ

Θ Θ1

Θ Θ2

Figure 2

We leave the proof of Theorem 1 to a course in advanced calculus. We can view rdrdθ as the

area of a "polar rectangle" with sides rdθ and dr.

x

y

drrdΘ

Figure 3

The conversion of a double integral from Cartesian to polar coordinates is useful if the region

of integration and the integrand can be expressed conveniently in polar coordinates, as in the

following examples.

Example 1 Let D be the annular region between the circles of radius π/2 and π centered at

the origin. Evaluate Z ZD

sin³p

x2 + y2´

px2 + y2

dxdy,

by expressing the integral in polar coordinates.


x

y

Π2 Π

Figure 4

Solution

Z ZD

sin³p

x2 + y2´

px2 + y2

dxdy =

Z θ=2π

θ=0

Z r=π

r=π/2

sin (r)

rrdrdθ

=

Z θ=2π

θ=0

Z r=π

r=π/2

sin (r) drdθ

=

Z θ=2π

θ=0

³− cos (r)|ππ/2

´dθ

=

Z θ=2π

θ=0

(− cos (π) + cos (π/2)) dθ

=

Z θ=2π

θ=0

dθ = 2π.

Remark 1 ifD is the set of points in the plane with polar coordinates (r, θ) such that α ≤ θ ≤ β

and 0 ≤ r ≤ r (θ), the area of D is

1

2

Z θ=β

θ=α

r2 (θ) dθ.

Indeed,

Area of D =

Z ZD

dxdy =

Z θ=β

θ=α

Z r(θ)

r=0

rdrdθ =

Z θ=β

θ=α

1

2r2 (θ) dθ.

♦

Example 2 Compute the area inside one loop of the curve C that has the equation

r = cos (3θ)

in polar coordinates.

Solution

Figure 6 shows the graph of r = cos (3θ) in the Cartesian θr-plane and the curve C (in the

xy-plane).


Π 2 ΠΘ

1

1r

Π6 Π3 Π2

0.5 10.45x

0.5

0.5

y

Figure 6

Note that

cos (3θ) = 0 if 3θ =π

2⇔ θ =

π

6.

The part of the curve in the first quadrant is traced when θ increases from 0 to π/6. By

symmetry, the area inside one loop is

2

Ã1

2

Z θ=π/6

θ=0

r2 (θ) dθ

!=

Z π/6

0

cos2 (3θ) dθ =1

3

Z u=π/2

u=0

cos2 (u) du

=1

3

Ãu

2+cos (u) sin (u)

2

¯π/20

!=1

3

³π4

´=

π

12.

¤

Example 3 Show that Z ∞0

e−x2

dx =1

2

√π.

Solution

We have shown that the improper integralZ ∞0

e−x2

dx


converges. Since e−x2

defines an even function,Z ∞0

e−x2

dx =1

2

Z ∞−∞

e−x2

dx.

We have Z ∞−∞

e−x2

dx = limA→∞

Z A

−Ae−x

2

dx.

Now, ÃZ A

−Ae−x

2

dx

!2=

ÃZ A

−Ae−x

2

dx

!ÃZ A

−Ae−x

2

dx

!

=

ÃZ A

−Ae−x

2

dx

!ÃZ A

−Ae−y

2

dy

!

since the variable of integration is a dummy variable. Thus,ÃZ A

−Ae−x

2

dx

!2=

Z y=A

y=−A

Z x=A

x=−Ae−x

2

e−y2

dxdy =

Z ZRA

e−x2

e−y2

dxdy

=

Z ZRA

e−x2−y2dxdy

where RA is the square [−A,A]× [−A,A]. Therefore,µZ ∞−∞

e−x2

dx

¶2= limA→∞

ÃZ A

−Ae−x

2

dx

!2= lim

A→∞

Z ZRA

e−x2−y2dxdy.

We can evaluate the limit as

limR→∞

Z ZDR

e−x2−y2dxdy,

where DR is the disk of radius R centered at the origin. We transform the integral to polar

coordinates:Z ZDR

e−x2−y2dxdy =

Z θ=2π

θ=0

Z r=R

r=0

e−r2

rdrdθ

=

ÃZ θ=2π

θ=0

dθ

!ÃZ r=R

r=0

e−r2

rdr

!

= 2π

Ã−12e−r

2

¯R0

!= 2π

µ−12e−R

2

+1

2

¶= −πe−R2

+ π.

Therefore, µZ ∞−∞

e−x2

dx

¶2= lim

R→∞

Z ZDR

e−x2−y2dxdy

= limR→∞

³−πe−R2 + π

´= π,

so that Z ∞−∞

e−x2

dx =√π.


Thus, Z ∞0

e−x2

dx =1

2

Z ∞−∞

e−x2

dx =1

2

√π,

as claimed. ¤

Example 4 Determine the volume of the region between the paraboloid z = 4 − x2 − y2 andthe xy-plane.

Solution

Figure 7

The given paraboloid intersects the xy-plane along the circle

4− x2 − y2 = 0⇔ x2 + y2 = 4.

This a circle of radius 2 centered at the origin. The required volume isZ Zx2+y2≤4

¡4− x2 − y2¢ dxdy = Z 2π

θ=0

Z r=2

r=0

¡4− r2¢ rdrdθ = 2π

Z r=2

r=0

¡4r − r3¢ dr

= 2π

Ã2r2 − 1

4r4¯20

!= 2π (4) = 8π.

Problems

In problems 1-5,

a) Sketch the region D,

b) Evaluate the given double integral by transforming the integral to an integral in polar coor-

dinates:

1. Z ZD

xydA,

where

D = (x, y) : x2 + y2 ≤ 16 and x ≥ 0, y ≥ 0(i.e., D is the part of the disk of radius 4 centered at the origin that is in the first quadrant).

2. Z ZD

sin¡x2 + y2

¢dA,


where

D = (x, y) : x2 + y2 ≤ 4 and y ≥ 0(i.e., D is the part of the disk of radius 2 centered at the origin that is in the upper half-plane).

3. Z ZD

p9− x2 − y2dA,

where

D = (x, y) : x2 + y2 ≤ 9 and x ≥ 0(i.e., D is the part of the disk of radius 3 centered at the origin that is in the right half-plane).

4. Z ZD

e−x2−y2dA,

where

D = (x, y) : 1 ≤ x2 + y2 ≤ 4 and y ≥ 0(i.e., D is the annular region in the upper half-plane between the circles of radius 1 and 2

centered at the origin).

5. Z ZD

arctan³yx

´dA,

where

D = (x, y) : 1 ≤ x2 + y2 ≤ 4 and 0 ≤ y ≤ x.In problems 6 and 7, sketch the region D and determine the area of D by using a double integral

in polar coordinates:

6. D is the region inside one loop of the graph of r = sin (2θ) .

7. D is the region inside the cardioid r = 1 + cos (θ) and ou tside the circle r = 3cos (θ).

In problems 8 and 9, make use of polar coordinates to compute the volume of the region D in

R3:8. D is between the cone z =

px2 + y2 and the cylinder x2 + y2 = 4, and above the xy-plane,

as in the picture:

3

2

z1

022

y x

1 1

00

-1 -1

-2-2

9. D is between the sphere x2 + y2 + z2 = 16 and the cylinder x2 + y2 = 4, as in the picture:

4

-44

2

z 0

-2y

4

2

x-2

-4 -4



a) Sketch the region that is relevant to the given iterated integral,

b) Make use of polar coordinates to evaluate the integral:

10. Z x=3

x=−3

Z y=√9−x2

y=0

sin¡x2 + y2

¢dydx.

11. Z y=1

y=0

Z x=√2−y2

x=y

xdxdy

13.4 Applications of Double Integrals

In this section we will discuss some physical applications of double integrals and some appli-

cations to probability. In physical applications you may assume that the units are mks-units

(meter-kilogram-second).

Density and Mass

Assume that a thin plate is modeled as a region D in the xy-plane. If the plate has area A and

constant density ρ (per unit area) its mass is

m = ρA.

How about the mass of such a plate if its density is not constant but is a function ρ (x, y)? Areasonable approach is to cover D with a mesh of rectangles, as in the general definition of a

double integral on D. Assuming that ρ (x, y) is continuous, we can assume that density has theconstant value ρ

¡x∗j , y

∗k

¢on the rectangle Rjk = [xj−1, xj ] × [yk−1, yk] (we are using the same

notation as in the definition of a double integral). The mass of the portion of the plate on Rjkis approximated by

ρ¡x∗j , y

∗k

¢∆Ajk = ρ

¡x∗j , y

∗k

¢∆xj∆yk,

and the total mass is approximated by the sum

nXk=1

nXj=1

ρ¡x∗j , y

∗k

¢∆Ajk =

nXk=1

nXj=1

ρ¡x∗j , y

∗k

¢∆xj∆yk,

where the sum is over those rectangles Rjk that have nonempty intersection with the region D.

Since such Riemann sums approximate the double integral

m (D) =

Z ZD

ρ (x, y) dA

as accurately as required provided that the mesh is sufficiently fine, we will calculate the mass

m (D) of the plate as the above double integral.

Example 1 Assume that a thin plate of mass densitypx2 + y2 (kg/m2) is in the shape of a

disk of radius 2 meters centered at the origin. Calculate the mass of the plate.

Solution

The mass is

m (D) =

Z ZD2

px2 + y2dxdy

13.4. APPLICATIONS OF DOUBLE INTEGRALS 153

where D denotes the disk of radius 2 meters centered at the origin. We transform to polar

coordinates:

m (D) =

Z ZD

px2 + y2dxdy =

Z θ=2π

θ=0

Z 2

r=0

r2drdθ

= 2π

Ã1

3r3¯20

!= 2π

µ8

3

¶=16π

3∼= 16.755 2 kg.

¤

Moments and Center of Mass

The moment of a point mass m at (x, y) with respect to the x-axis is defined as my and itsmoment with respect to the y-axis is defined as mx. If the region D represents a thin plate that

has mass density ρ (x, y), as in our discussion of mass, we arrive at the definitions of momentsas follows: As before, we will cover D with a mesh of rectangles and approximate the mass

of the part of the plate corresponding to the small rectangle Rjk = [xj−1, xj ] × [yk−1, yk] byρ¡x∗j , y

∗k

¢∆xj∆yk, where

¡x∗j , y

∗k

¢is a point inside that rectangle. Let’s assume that this mass

is concentrated at a point¡x∗j , y

∗k

¢, so that its moment with respect to the x-axis is

y∗kρ¡x∗j , y

∗k

¢∆xj∆yk

and its moment with respect to the y-axis is

x∗jρ¡x∗j , y

∗k

¢∆xj∆yk.

The sum of the moments with respect to the x-axis is

nXk=1

nXj=1

y∗kρ¡x∗j , y

∗k

¢∆xj∆yk =

nXk=1

nXj=1

y∗kρ¡x∗j , y

∗k

¢∆Ajk,

and the sum of the moments with respect to the y-axis is

nXk=1

nXj=1

x∗jρ¡x∗j , y

∗k

¢∆xj∆yk =

nXk=1

nXj=1

x∗jρ¡x∗j , y

∗k

¢∆Ajk.

These are Riemann sums that approximate the integralsZ ZD

yρ (x, y) dxdy and

Z ZD

xρ (x, y) dxdy,

respectively. We will define the moment of the plate with respect to the x-axis as

Mx (D) =

Z ZD

yρ (x, y) dxdy =

Z ZD

yρ (x, y) dA,

and the moment of the plate with respect to the y-axis as

My (D) =

Z ZD

xρ (x, y) dxdy =

Z ZD

xρ (x, y) dA.

The center of mass of plate with mass m (D) is the point (x, y), where

x =My (D)

m (D)and y =

Mx (D)

m (D).

Physically, if we assume that the plate is supported only at the center of mass it would remain

horizontal.


Example 2 Assume that a plate is represented as a half-disk

D =©(x, y) : x2 + y2 ≤ 9, x ≥ 0ª ,

and has density

ρ (x, y) = x2 + y2.

Calculate the center of mass of the plate.

Solution

The mass of the plate is

m (D) =

Z ZD

¡x2 + y2

¢dxdy.

Let’s transform to polar coordinates:

m (D) =

Z ZD

¡x2 + y2

¢dxdy. =

Z θ=π/2

θ=−π/2

Z r=3

r=0

r3drdθ

= π

Ã1

4r4¯30

!=34π

4=81π

4∼= 63. 617 3.

The moment of the plate with respect to the y-axis is

My (D) =

Z ZD

xρ (x, y) dxdy =

Z ZD

x¡x2 + y2

¢dxdy.

We will transform to polar coordinates again:

My (D) =

Z ZD

x¡x2 + y2

¢dxdy =

Z θ=π/2

θ=−π/2

Z r=3

r=0

r cos (θ) r3drdθ

=

Z θ=π/2

θ=−π/2sin (θ)

µZ r=3

r=0

r4dr

¶dθ

=

ÃZ θ=π/2

θ=−π/2sin (θ) dθ

!µZ r=3

r=0

r4dr

¶

=³sin|π/2−π/2

´Ã r55

¯30

!

= (2)

µ35

5

¶=486

5

Therefore, the x-coordinate of the center of mass is

x =My (D)

m (D)=

486

581π

4

=24

5π∼= 1. 527 89

The moment of the plate with respect to the x-axis is

Mx (D) =

Z ZD

yρ (x, y) dxdy =

Z ZD

y¡x2 + y2

¢dxdy.

13.4. APPLICATIONS OF DOUBLE INTEGRALS 155

We will transform to polar coordinates:

Mx (D) =

Z ZD

y¡x2 + y2

¢dxdy =

Z θ=π/2

θ=−π/2

Z r=3

r=0

r sin (θ) r3drdθ

=

Z θ=π/2

θ=−π/2sin (θ)

µZ r=3

r=0

r4dr

¶dθ

=

ÃZ θ=π/2

θ=−π/2sin (θ) dθ

!µZ r=3

r=0

r4dr

¶

=³− cos (θ)|π/2−π/2

´Ã r55

¯30

!= 0

Therefore, the y-cordinate of the center of mass is 0. The center of mass is the point (24/5π, 0) ∼=(1. 527 89, 0). ¤

Probability

We discussed the probability density function f that corresponds to a single continuous random

variable X: The probability that X has values between a and b is

P (a ≤ X ≤ b) =Z b

a

f (x) dx.

If we have two random variables X and Y that have the joint density function f (x, y), theprobability that X has values between a and b, Y has values between c and d is

P (a ≤ X ≤ b, c ≤ Y ≤ d) =Z x=b

x=a

Z y=d

y=c

f (x, y) dxdy.

More generally, if D is a region in the xy-plane, the probability that the pair (X,Y ) has valuesin D is

P ((X,Y ) ∈ D) =Z Z

D

f (x, y) dxdy.

We have f (x, y) ≥ 0 for all (x, y) andZ ZR2f (x, y) dxdy = 1

since the probability that (X,Y ) attains some value is 1.The random variables are said to be independent random variables if

f (x, y) = f1 (x) f2 (y) .

Example 3 Assume that the independent random variables X and Y have distribution func-

tions

f1 (x) =

½0 if x < 0,

18e−x/8 if x ≤ 0 , and f2 (y) =

½0 if x < 0,

14e−y/4 if x ≤ 0 ,

respectively. Determine the probability that X + Y < 16.

Solution

Since X and Y are independent, their joint density function is

f (x, y) = f1 (x) f2 (y) =

½132e−x/8e−y/4 if x ≥ 0 and y ≥ 0,0 otherwise.


Therefore, the probability that X + Y < 16 isZ ZD

1

32e−x/8e−y/4dxdy,

where D is the triangular region in the first quadrant below the line x+ y = 16. We haveZ ZD

1

32e−x/8e−y/4dxdy =

Z x=16

x=0

Z 16−x

y=0

1

32e−x/8e−y/4dydx

=1

32

Z x=16

x=0

e−x/8µZ 16−x

y=0

e−y/4dy¶dx

=1

32

Z x=16

x=0

e−x/8µ−4e−y/4

¯16−x0

¶dx

=1

8

Z x=16

x=0

e−x/8³1− e−(4−x/4)

´dx

= 1 +1

e4− 2

e2∼= 0.747 645.

¤Recall that a random variable is normally distributed if it has a density function of the form

1

σ√2πe−(x−μ)

2/(2σ2),

where μ is the mean and σ is the standard deviation.

Example 4 Assume thatX and Y are independent normally distributed random variables with

means 5 and 6, and standard deviations 0.2 and 0.1, respectively. Calculate the probability that4.5 < X < 5.5 and 5.5 < Y < 6.5.

Solution

The density functions of X and Y are

f1 (x) =1

0.2√2πe−(x−5)

2/0.08 and f2 (y) =1

0.1√2πe−(y−6)

2/0.02,

respectively. Therefore,

P (4.5 < X < 5.5 and 5.5 < Y < 6.5)

=

Z y=6.5

y=5.5

Z x=5.5

x=4.5

f1 (x) f2 (y) dxdy

=

µZ y=6.5

y=5.5

f2 (y) dy

¶µZ x=5.5

x=4.5

f1 (x) dx

¶=

µZ y=6.5

y=5.5

1

0.1√2πe−(y−6)

2/0.02dy

¶µZ x=5.5

x=4.5

1

0.2√2πe−(x−5)

2/0.08dx

¶∼= (0.999 999 ) (0.987 581 ) = 0.987 58

Problems

In problems 1 and 2 determine the mass and center of mass of a thin plate that occupies the

region D and has mass density ρ (x, y):

13.5. TRIPLE INTEGRALS 157

1.

D = (x, y) : 0 ≤ x ≤ 4, 0 ≤ y ≤ 2 , ρ (x, y) = 1 + x+ y2. D is the half-disk

D =©(x, y) : x2 + y2 ≤ 9, y ≥ 0ª

and

ρ (x, y) = 81−px2 + y2

Assume that a thin plate occupies the region D in the xy-plane and has mass density ρ (x, y).The moment of inertia of the plate about the x-axis is

Ix =

Z ZD

y2ρ (x, y) dxdy

and the moment of inertia of the plate about the y-axis is

Iy =

Z ZD

x2ρ (x, y) dxdy.

The moment of inertia of the plate about the origin (the polar moment of inertia) is

I0 = Iy + Ix =

Z ZD

¡x2 + y2

¢ρ (x, y) dxdy.

In problems 3 and 4 calculate the Ix, Iy and I0 of a thin plate that occupies the region D and

has mass density ρ (x, y):

3. D is the interior of the circle that is centered at the origin and has radius r0. The density

has tha constant value ρ0.

4. D is the half-disk

D =©(x, y) : x2 + y2 ≤ 9, y ≥ 0ª

and

ρ (x, y) = 4−px2 + y2

5. Assume that X and Y are independent random variables that have distribution functions

f1 (x) =

½0 if x < 0,

16e−x/6 if x ≤ 0 , and f2 (y) =

½0 if x < 0,

12e−y/2 if x ≤ 0 ,

respectively. Determine the probability that X + Y < 8.

6 [C] Assume that X and Y are independent normally distributed random variables with means

3 and 5, and standard deviations 0.2 and 0.1, respectively. Make use of your computationalutility in order to calculate the probability that 2.5 < X < 3.5 and 4.5 < Y < 5.5 approximately.

13.5 Triple Integrals

As in the case of double integrals, let’s begin by considering a case that leads to a triple integral.

Assume that D is a rectangular box [a, b]× [c, d]× [e, f ] in R3, and f (x, y, z) is the continuousmass density (say, in kilograms per cubic meter) of some material that occupies D. We will

approximate the mass of the material by partitioning D via small rectangular boxes. Let

a = x0 < x1 < · · · < xj−1 < xj < · · · < xm−1 < xm = b,c = y0 < y1 < · · · < yk−1 < yk < · · · < yn−1 < yn = d,e = z0 < z1 < · · · < zl−1 < zl < · · · < zp−1 < zp = f


be partitions of the intervals, [a, b], [c, d] and [e, f ], respectively. We set

∆xj = xj − xj−1, ∆yk = yk − yk−1 and ∆zl = zl − zl−1.

The volume of the box

Rjkl = [xj − xj−1]× [yk − yk−1]× [zl − zl−1]

is

∆Vjkl = ∆xj∆yk∆zl.

We can select a point ¡x∗j , y

∗k, z∗p

¢ ∈ [xj − xj−1]× [yk − yk−1]× [zl − zl−1]arbitrarily in Rjkl and approximate the mass of the material in Rjkl by

f¡x∗j , y

∗k, z∗p

¢∆Vjkl = f

¡x∗j , y

∗k, z∗p

¢∆xj∆yk∆zl.

The total mass of the material is approximated by "a Riemann sum"

mXj=1

nXk=1

pXl=1

f¡x∗j , y

∗k, z∗p

¢∆Vjkl =

mXj=1

nXk=1

pXl=1

f¡x∗j , y

∗k, z∗p

¢∆xj∆yk∆zl.

It is known that such Riemann sums approximate a number that depends only on f and D

with desired accuracy provided that f is continuous and the maximum of the ∆Vjkl’s is smallenough. That number is called the triple integral of f on D and denoted asZ Z Z

D

f (x, y, z) dV =

Z Z ZD

f (x, y, z) dxdydz.

Thus ¯¯Z Z Z

D

f (x, y, z) dV −mXj=1

nXk=1

pXl=1

f¡x∗j , y

∗k, z∗p

¢∆Vjkl

¯¯

=

¯¯Z Z Z

D

f (x, y, z) dxdydz −mXj=1

nXk=1

pXl=1

f¡x∗j , y

∗k, z∗p

¢∆xj∆yk∆zl

¯¯

is as small as desired if the maximum of ∆xj∆yk∆zl is sufficiently small. The triple integralcorresponds to the mass of the material that occupies D if f is the density of the material. The

triple integral is defined for an arbitrary function that is continuous on D. We will see other

applications of triple integrals in Chapter 14.

As in the case of double integral, we can evaluate the triple integral as an iterated integral in

any order. For example,Z Z ZD

f (x, y, z) dV =

Z x=b

x=a

Z y=d

y=c

Z z=f

z=e

f (x, y, z) dzdydx.

Example 1 Evaluate Z Z ZD

f (x, y, z) dV,

where f (x, y, z) = yz + xz + xy and D is the rectangular box [0, 1]× [0, 2]× [0, 3].


Solution Z Z ZD

f (x, y, z) dV =

Z 3

z=0

Z 2

y=0

Z x=1

x=0

(yz + xz + xy) dxdydz.

We have Z x=1

x=0

(yz + xz + xy) dx = yzx+1

2x2z +

1

2x2y

¯1x=0

= yz +1

2z +

1

2y.

Therefore, Z 2

y=0

Z x=1

x=0

(yz + xz + xy) dxdy =

Z 2

y=0

µyz +

1

2z +

1

2y

¶dy

=1

2y2z +

1

2zy +

1

4y2¯2y=0

=1

2(4) z +

1

2(2) z +

1

4

¡22¢

= 2z + z + 1 = 3z + 1.

Thus, Z 3

z=0

Z 2

y=0

Z x=1

x=0

(yz + xz + xy) dxdydz =

Z 3

z=0

(3z + 1) dz

=3

2z2 + z

¯3z=0

=3

2(9) + 3 =

33

2.

¤As in the case of double integrals, the concept of the triple integral is generalized to regions D

in R3 other than rectangular boxes by considering Riemann sums

mXj=1

nXk=1

pXl=1

f¡x∗j , y

∗k, z∗p

¢∆xj∆yk∆zl

where the sum is over those rectangular boxes with nonempty intersection with the given region.

The triple integral of f on D is the numberZ Z ZD

f (x, y, z) dV =

Z Z ZD

f (x, y, z) dxdydz

that can be approximated by such sums with desired accuracy provided that the maximum of

∆xj∆yk∆zl’s is small enough. A triple integral can be evaluated as an iterated integral wherebythe given integral is reduced to a double integral. For example, Let D be a region in R3 thatconsists of points (x, y, z) where (x, y) varies in some region D2 in the xy-plane, and

g1 (x, y) ≤ z ≤ g2 (x, y) for each (x, y) ∈ D2.

If f is continuous on DZ Z ZD

f (x, y, z) dV =

Z ZD2

ÃZ z=g2(x,y)

z=g1(x,y)

f (x, y, z) dz

!dxdy


The roles of x, y and z may be interchanged

Note that Z Z ZD

dV

is simply the volume of D.


x2dV,

whereD is the tetrahedron bounded by the coordinate planes and the part of the plane x+y+z =1 in the first octant.

Figure 1

Solution

The region D is of type I. The projection of D onto the xy-plane is the triangle ∆ that is

bounded by x+ y = 1 and the coordinate axes.

Figure 2

For each (x, y) ∈ ∆, we have (x, y, z) ∈ D iff 0 ≤ z ≤ 1− x− y. Therefore,Z Z ZD

x2dV =

Z Z∆

µZ z=1−x−y

z=0

x2dz

¶dxdy =

Z 1

x=0

Z y=1−x

y=0

µZ z=1−x−y

z=0

x2dz

¶dydx.

We have Z z=1−x−y

z=0

x2dz = x2z¯z=1−x−yz=0

= x2 (1− x− y) = x2 − x3 − x2y.


Therefore, Z y=1−x

y=0

µZ z=1−x−y

z=0

x2dz

¶dy =

Z y=1−x

y=0

¡x2 − x3 − x2y¢ dy

= x2y − x3y − 12x2y2

¯y=1−xy=0

= x2 (1− x)− x3 (1− x)− 12x2 (1− x)2

=1

2x2 − x3 + 1

2x4

Thus. Z 1

x=0

Z y=1−x

y=0

µZ z=1−x−y

z=0

x2dz

¶dydx. =

Z x=1

x=0

µ1

2x2 − x3 + 1

2x4¶dx

=1

6x3 − 1

4x4 +

1

10x5¯10

=1

6− 14+1

10=1

60

¤


x2y2dV,

where D is the region that is bounded by the surfaces z = 1−x2 and z = x2−1, and the planesy = −2 and y = 2.

Figure 3

Solution

The surfaces z = 1 − x2 and z = x2 − 1 intersect along the lines that are determined by thesolution of the system

z = 1− x2,z = −1 + x2.

Thus,

0 = 2− 2x2 ⇒ x2 = 1⇒ x = ±1.


Therefore, z = 1−1 = 0. The lines of intersection are the lines x = ±1 in the xy-plane. Since Dis between the planes y = −2 and y = 2, the projection of D onto the xy-plane is the rectangle

[−1, 1]× [−2, 2]. The region D is type I. For any (x, y) ∈ [−1, 1]× [−2, 2], the point (x, y, z) ∈ Diff −1 + x2 ≤ z ≤ 1− x2. We can express the triple integral as an iterated integral:

Z Z ZD

x2y2dV =

Z x=1

x=−1

Z y=2

y=−2

ÃZ z=1−x2

z=−1+x2x2y2dz

!dydx.

We haveZ z=1−x2

z=−1+x2x2y2dz = x2y2

Z z=1−x2

z=−1+x2dz = x2y2

³z|1−x2−1+x2

´= x2y2

¡1− x2 + 1− x2¢ = 2x2y2 ¡1− x2¢ .

Therefore,Z Z ZD

x2y2dV =

Z x=1

x=−1

Z y=2

y=−22x2y2

¡1− x2¢ dydx = 2µZ x=1

x=−1

¡x2 − x4¢ dx¶µZ y=2

y=−2y2dy

¶= 2

Ã1

3x3 − 1

5x5¯1−1

!Ã1

3y3¯2−2

!

= 2

µ1

3− 15+1

3− 15

¶µ16

3

¶= 2

µ4

15

¶µ16

3

¶=128

45

¤


xdV,

where D is the region in the first octant that is bounded by the surface z = x2 + y2 and theplane z = 4.

Figure 4

Solution


The projection of D onto the xy-plane is part of the disk x2 + y2 ≤ 4 of radius 2 in the firstquadrant. If D2 denotes that quarter disk, and (x, y) ∈ D2, then (x, y, z) ∈ D iff

x2 + y2 ≤ z ≤ 4.

Therefore, Z Z ZD

xdV =

Z ZD2

x

µZ 4

z=x2+y2dz

¶dA,

The inner integral is Z 4

z=x2+y2dz = z|4x2+y2 = 4−

¡x2 + y2

¢.

Therefore, Z Z ZD

xdV =

Z ZD2

x¡4− ¡x2 + y2¢¢ dA.

Let’s convert to polar coordinates:Z ZD2

x¡4− ¡x2 + y2¢¢ dA = Z θ=π/2

θ=0

Z r=2

r=0

r cos (θ)¡4− r2¢ rdrdθ

=

ÃZ θ=π/2

θ=0

cos (θ) dθ

!µZ 2

r=0

¡4r2 − r4¢ dr¶

=³sin (θ)|π/20

´Ã 43r3 − 1

5r5¯20

!=4

323 − 1

525 =

64

15

¤

Example 5 Assume that D is the region in R3 that is bounded by the xy-plane, the cylinderx2 + y2 = 4 and the graph of z = x2 + 4, as shown in the picture:

Evaluate Z Z ZD

1

1 + x2 + y2dV.

Solution


The projection of the intersection of x2 + y2 = 4 and z = x2 + 4 onto the xy-plane is the circlex2 + y2 = 4. This is the circle of radius 2 centered at the origin. Set

D2 =©(x, y) : x2 + y2 ≤ 4ª .

We have Z Z ZD

dV =

Z ZD2

ÃZ z=x2+4

y=0

1

1 + x2 + y2dz

!dxdy

=

Z ZD2

1

1 + x2 + y2¡x2 + 4

¢dxdy.

Let’s use polar coordinates in the xy-plane, so that

x = r cos (θ) , y = r sin (θ) .

Then,Z ZD2

x2 + 4

1 + x2 + y2dxdy =

Z θ=2π

θ=0

Z 2

r=0

4 + r2 cos2 (θ)

1 + r2rdrdθ

=

Z θ=2π

θ=0

Z 2

r=0

4

1 + r2rdrdθ +

Z θ=2π

θ=0

Z 2

r=0

r3 cos2 (θ)

1 + r2drdθ.

We have Z θ=2π

θ=0

Z 2

r=0

4

1 + r2rdrdθ = 2π

Z 2

r=0

4

1 + r2rdr

= 8π

Ã1

2ln¡1 + r2

¢¯20

!= 4π ln (5) .

As for the second integral,Z θ=2π

θ=0

Z 2

r=0

r3 cos2 (θ)

1 + r2drdθ =

ÃZ θ=2π

θ=0

cos2 (θ) dθ

!µZ 2

r=0

r3

1 + r2dr

¶

=

ÃZ θ=2π

θ=0

µ1 + cos (2θ)

2

¶dθ

!µZ 2

r=0

r3

1 + r2dr

¶

=

Ã1

2θ +

1

4sin (2θ)

¯2π0

!Z 2

r=0

r3

1 + r2dr

= π

Z 2

r=0

r3

1 + r2dr.

Finally,r3

1 + r2= r − r

r2 + 1,

by division, so that

π

Z 2

r=0

r3

1 + r2dr = π

Ã1

2r2 − 1

2ln¡r2 + 1

¢¯20

!

= π

µ2− 1

2ln (5)

¶.


Therefore, Z ZD2

x2 + 4

1 + x2 + z2dxdz = 4π ln (5) + 2π − π

2ln (5)

= 2π +7π

2ln (5) .

¤

Problems

1. Find the volume of the region bounded by the surfaces

z = x2 + y2 and z = 18− x2 − y2.

04

10

2

z

20

30

y

04

2-2

x0

-2-4 -4

2. Find the volume of the region bounded by the surfaces

x2 + y2 = 4, x+ y + z = 1 and z = −10.

-10

0

2

z

10

y

0

-2 2

x0

-2

In problems 3-10 evaluate the given integral (Note that the symbol dxdydz stands for dV in

Cartesian coordinates and does not specify the order of integration):

3. Z Z ZD

x2dxdydz,

where

D = [0, 1]× [0, 1]× [0, 1] .4. Z Z Z

D

e−xyydxdydz,

where

D = [0, 1]× [0, 1]× [0, 1]


5. Z Z ZD

zex+ydxdydz,

where

D = [0, 1]× [0, 1]× [0, 2]6. Z Z Z

D

x2 cos (z) dxdydz,

where D is the region bounded by the planes z = 0, z = π/2, y = 0, y = 1, x = 0 andx+ y = 1.

0.01.0

0.5

1.0

z

1.5

y

0.5

1.0

x0.5

0.0 0.0

7. Z Z ZD

zdxdydz,

where D is the region bounded by the planes x = 0, y = 0, z = 0, z = 1, and the cylinderx2 + y2 = 1 with x ≥ 0 and y ≥ 0.

0

1

z

y

1

1

x0 0

8. Z Z ZD

x2dxdydz,

where D is the solid tetrahedron with vertices (0, 0, 0) , (1, 0, 0) , (0, 1, 0) , (0, 0, 1).

1

1

z

x

1

y

9. Z Z ZD

xydxdxydz,

13.6. TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 167

where D is bounded by the cylinders y = x2, x = y2, the xy-plane and the plane z = x+ y.

z

2

1

1

y

x

10. Z Z ZD

x2dydxdz,

where D is bounded by the cylinder x2 + y2 = 9, the plane y + z = 5 and the plane z = 1.

x

-2

0

5

z

0

2-2 y

02

13.6 Triple Integrals in Cylindrical and Spherical Coordi-

nates

Cylindrical Coordinates

We can extend polar coordinates to the three dimensional space R3 as follows: If the Cartesiancoordinates of a point P ∈ R3 are x, y and z, then r and θ are polar coordinates for the point

(x, y) in the plane. Thus,x = r cos (θ) and y = r sin (θ) ,

so that

r =px2 + y2, cos (θ) =

xpx2 + y2

and sin (θ) =yp

x2 + y2.

The polar angle is determined up to an additive constant that is an integer multiple of 2π. Thethird coordinate is simply z itself. The coordinates r, θ and z are referred to as the cylindrical

coordinates.of the point P . We will not allow r to be negative when we utilize cylndrical

coordinates. Recall that we can always a value of θ ∈ [−π,π] by setting

θ =

⎧⎪⎪⎨⎪⎪⎩arccos

µx√x2+y2

¶if y ≥ 0,

− arccosµ

x√x2+y2

¶if y < 0.

Example 1 Determine the cylindrical coordinates of P if


a) P =¡1,√3, 2¢, b) P = (−1,−1, 2) in Cartesian coordinates

Solution

a) We have

r =

r12 +

³√3´2=√4 = 2,

and

cos (θ) =1

2, sin (θ) =

√3

2.

Thus, we can set θ = π/3 or θ = π/3 + 2nπ, where n is an arbitrary integer. Therefore, thecylndrical coordinates of P are r = 2, θ = π/3+2nπ and z = 2, where n is an arbitrary integer.b) We have

r =

q12 + (−1)2 =

√2,

and

cos (θ) =−1√2, sin (θ) =

−1√2.

We can set

θ = − arccosµ−1√

2

¶= −3π

4.

¤

Triple Integrals in Cylindrical Coordinates

Assume that D is a region in R3 that consists of points whose cylindrical coordinates satisfyinequalities of the form

r1 (θ) ≤ r ≤ r2 (θ) , α ≤ θ ≤ β and z1 (r, θ) ≤ z ≤ z2 (r, θ) .

ThenZ Z ZD

f (x, y, z) dxdydz =

Z θ=θ2

θ=θ1

Z r=r2(θ)

r=r1(θ)

Z z2(r,θ)

z1(r,θ)

f (r cos (θ) , r sin (θ) , z) rdzdrdθ.

Indeed, if D2 denotes the region in the xy-plane consisting of the points whose polar coordinates

satisfy the inequalities

r1 (θ) ≤ r ≤ r2 (θ) , α ≤ θ ≤ β,

thenZ Z ZD

f (x, y, z) dxdydz =

Z ZD2

ÃZ Z z2(r,θ)

z1(r,θ)

f (x, y, z) dz

!dxdy

=

Z θ=θ2

θ=θ1

Z r=r2(θ)

r=r1(θ)

ÃZ Z z2(r,θ)

z1(r,θ)

f (r cos (θ) , r sin (θ) , z) dz

!rdrdθ

=

Z θ=θ2

θ=θ1

Z r=r2(θ)

r=r1(θ)

Z z2(r,θ)

z1(r,θ)

f (r cos (θ) , r sin (θ) , z) rdzdrdθ

Thus, the use of cylindrical coordinates in triple integrals amounts to transforming to polar

coordinates in the xy-plane.


Example 2 Let D be the region that is bounded by the paraboloid z = x2 + y2 and the planez = 9. Use cylindrical coordinates to evaluateZ Z Z

D

x2zdV.

Figure 1

Solution

The equation of the paraboloid in cylindrical coordinates is z = r2. The projection of D to the

xy-plane is the disk that is bounded by the circle x2+ y2 = 9, i.e.,the circle of radius 3 centeredat the origin. Therefore,Z Z Z

D

x2zdV =

Z θ=2π

θ=0

Z 3

r=0

Z z=9

z=r2

¡r2 cos2 (θ)

¢zrdzdrdθ

=

Z 2π

θ=0

Z 3

r=0

r3 cos2 (θ)

µZ 9

z=r2zdz

¶drdθ.

We have Z 9

z=r2zdz =

1

2z2¯z=9z=r2

=81

2− r

4

2.

Therefore,Z 2π

θ=0

Z 3

r=0

r3 cos2 (θ)

µZ 9

z=r2zdz

¶drdθ =

Z 2π

θ=0

Z 3

r=0

r3 cos2 (θ)

µ81

2− r

4

2

¶drdθ

=

µZ 2π

θ=0

cos2 (θ) dθ

¶µZ 3

r=0

r3µ81

2− r

4

2

¶dr

¶.

We have Z 2π

θ=0

cos2 (θ) dθ =θ

2+1

2cos (θ) sin (θ)

¯2π0

= π,

and Z 3

r=0

r3µ81

2− r

4

2

¶dr =

Z 3

0

µ81

2r3 − 1

2r7¶dr

=81

8r4 − r

8

16

¯30

=81

8

¡34¢− 38

16=6561

16


Therefore, Z Z ZD

x2zdV =6561

16π.

¤

Example 3 Let D be the region that is bounded by the cone z2 = x2 + y2 and the cylinderr = 2. Use cylindrical coordinates to evaluateZ Z Z

D

e−√x2+y2dV.

Figure 2

Solution

Since x2+y2 = r2, in cylindrical coordinates the equation of the cone is z2 = r2. Thus, the conehas two parts, z = ±r. The projection of D onto the xy-plane is the disk of radius 2 centered

at the origin. Therefore,Z Z ZD

e−√x2+y2dV =

Z θ=2π

θ=0

Z 2

r=0

µZ z=r

z=−−re−rrdz

¶drdθ =

Z θ=2π

θ=0

Z 2

r=0

e−rrµZ z=r

z=−−rdz

¶drdθ

=

Z θ=2π

θ=0

Z 2

r=0

e−rr (2r) drdθ

=

ÃZ θ=2π

θ=0

dθ

!µ2

Z 2

r=0

e−rr2dr¶

We have Z 2

0

e−rr2dr = 2− 10e−2

(confirm via integration by parts). Therefore,Z Z ZD

e−√x2+y2dV = 4π

¡2− 10e−2¢ = 8π − 40π

e2

¤

Spherical Coordinates

Let P = (x, y, z) 6= (0, 0, 0). We set ρ to be the distance of P from the origin:

ρ =px2 + y2 + z2.


The angle φ is the angle between the position vector−−→OP of the point P and the postive z-axis

determined so that 0 ≤ φ ≤ π. Thus,

z = ρ cos (φ)

and

φ = arccos

µz

ρ

¶.

The distance of the projection Q of P onto the xy-plane from the origin is ρ sin (φ). In terms ofthe polar coordinates of that point, we have

x = ρ sin (φ) cos (θ) and y = ρ sin (φ) sin (θ) .

Here the polar angle θ is not unique and it is determined up to an integer multiple of 2π.

Figure 3

Example 4 Let P =¡9, 3√3, 6¢. Determine the spherical cooordinates of P .

Solution

We have

ρ =

r92 +

³3√3´2+ 62 = 12.

Therefore,

cos (φ) =z

ρ=6

12=1

2.

Since 0 ≤ φ ≤ π, we have φ = π/3. Thus,

x = ρ sin (φ) cos (θ) and y = ρ sin (φ) sin (θ)

so that

x = (12)

Ã√3

2

!cos (θ) = 6

√3 cos (θ) and y = 6

√3 sin (θ) .

Therefore,

cos (θ) =9

6√3=

3

2√3=

√3

2,

and

sin (θ) =3√3

6√3=1

2.

Thus, we can set θ = π/6. Therefore,

ρ = 12, φ = π/3 and θ = π/6

are spherical coordinates for P . ¤


Example 5 Let S be the surface that is the graph of the equation

z =px2 + y2

is Cartesian coordinates. Describe S as the graph of an equation in spherical coordinates and

identify the surface geometrically.

Solution

We have

x2 + y2 = (ρ sin (φ) cos (θ))2 + (ρ sin (φ) sin (θ))2

= ρ2 sin2 (φ) cos2 (θ) + ρ2 sin2 (φ) cos2 (θ)

= ρ2 sin2 (φ) .

Thus, px2 + y2 = ρ sin (φ) .

Therefore,

z =px2 + y2 ⇒ ρ cos (φ) = ρ sin (φ)⇒ cos (φ) = sin (φ) .

Since 0 ≤ φ ≤ π, we have φ = π/4. Thus, a point is on S if the angle φ between its positionvector and the positive z-axis is π/4. Since there are no other restriction on the spherical

coordinates of the points of S we identify S as a cone that has its vertex at the origin, the z-axis

forms its axis, and the opening with the z-axis is π/4. ¤

Figure 4: The cone φ = π/4

Triple Integrals in Spherical Coordinates

Assume that D is a region in R3 that consists of points whose spherical coordinates satisfyinequalities of the form

α ≤ θ ≤ β, γ ≤ φ ≤ δ, ρ1 (θ,φ) ≤ ρ ≤ ρ2 (θ,φ) .

Then Z Z ZD

f (x, y, z) dxdydz

=

Z θ=β

θ=α

Z φ=δ

φ=γ

Z ρ=ρ2(θ,φ)

ρ=ρ1(θ,φ)

f (ρ sin (φ) cos (θ) , ρ sin (φ) sin (θ) , ρ cos (φ)) ρ2 sin (φ) dρdφdθ

(the order of θ and φ may be interchanged). We are assuming that f and the functions ρ1 (θ,φ)and ρ2 (θ,φ) are continuous. The volume element dV = dxdydz in Cartesian cooordinates is

replaced by the expression

ρ2 sin (φ) dρdφdθ.


We will not prove the above statement. On the other hand, the expression is plausible: If

∆ρ, ∆φ and ∆θ small, lets consider a region G whose spherical coordinates are between ρ

and ρ +∆ρ, φ and φ +∆φ, θ and θ +∆θ, respectively. The region G is the counterpart of a

rectangular box when we deal with spherical coordinates. The volume of G is approximately

(∆ρ) (ρ∆φ) (ρ sin (φ)∆θ) = ρ2 sin (φ)∆ρ∆φ∆θ,

as illustrated in Figure 5.

Figure 5

Thus, the integral of f on such a small region is approximately

f (ρ sin (φ) cos (θ) , ρ sin (φ) sin (θ) , ρ cos (φ)) ρ2 sin (φ)∆ρ∆φ∆θ.

Example 6 Let D be the region that is bounded by the sphere of radius 4 centered at the

origin and the xy-plane. Evaluate Z Z ZD

z2dV.


Figure 6

Solution

We will express the integral in spherical coordinates:

Z Z ZD

z2dV =

Z θ=2π

θ=0

Z π/2

φ=0

µZ ρ=4

ρ=0

ρ2 cos2 (φ) ρ2 sin (φ) dρ

¶dφdθ

=

Z θ=2π

θ=0

Z π/2

φ=0

cos2 (φ) sin (φ)

µZ ρ=4

ρ=0

ρ4dρ

¶dφdθ

=

Z θ=2π

θ=0

Z π/2

φ=0

cos2 (φ) sin (φ)

µ45

5

¶dφdθ

=45

5(2π)

µZ u=0

u=1

−u2du¶

=45

5(2π)

µ1

3

¶=2048

15π

¤

Example 7 Let D be the region inside the cone φ = π/4 and the sphere ρ = 3. EvaluateZ Z ZD

zdV.

Figure 7

Solution

We express the integral in spherical coordinates:


Z Z ZD

zdV =

Z θ=2π

θ=0

Z φ=π/4

φ=0

Z ρ=3

ρ=0

ρ cos (φ) ρ2 sin (φ) dρdφdθ

=

ÃZ θ=2π

θ=0

dθ

!ÃZ φ=π/4

φ=0

cos (φ) sin (φ) dφ

!µZ ρ=3

ρ=0

ρ3dρ

¶

= 2π

Ã−Z u=

√2/2

u=1

udu

!µ34

4

¶

=34π

2

Ã1

2u2¯1√2/2

!

=34π

2

µ1

2− 14

¶=81

8π

¤

Problems

In problems 1-4 determine the Cartesian coordinates of the point with the given cylindrical

coordinates (r, θ, z):

1. (2,π/4, 1)2. (4,π/6, 2)

3. (3, 2π/3, 4)4. (2,−π/6, 4)

In problems 5-8 determine the cylindrical coordinates (r, θ, z) of the point with the given Carte-sian coordinates, such that −π < θ ≤ π:

5. (−1, 1, 4)6.¡2,−2√3,−1¢ 7. (1,−1, 2)

8.¡1,√3, 2¢

In problems 9-13 make use of cylindrical coordinates to evaluate the given triple integral (Note

that the symbol dxdydz stands for dV in Cartesian coordinates and does not specify the order

of integration):

9. Z Z ZD

px2 + y2dxdydz,

where D is inside the cylinder x2 + y2 = 16, between the planes z = −5 and z = 4.

5-5

0z

5

5

y

0

x0

-5-5

10. Z Z ZD

¡x3 + xy2

¢dxdydz,


where D is the region in the first octant that lies below the paraboloid z = 1− x2 − y2.

00

1

z

1

x y1 0

11. Z Z ZD

ezdxdydz,

where D is bounded by the paraboloid z = 1+x2+y2, the cylinder x2+y2 = 5 and the xy-plane.

0

1

2

z

6

y

0

2

x

-2 0-2

12. Z Z ZD

x2dxdydz,

where D is the region that is within the cylinder x2 + y2 = 1, above the xy-plane, and belowthe cone z2 = 4x2 + 4y2.

01

z

2

1

y

0

x0

-1 -1

13. Find the volume of the region within both the cylinder x2 + y2 = 1 and the spherex2 + y2 + z2 = 4.

2

2

0

-1

-2

1

z

2

1

0

y -1

-2

1

x0

-1-2

In problems 14-17 determine the Cartesian coordinates of the point with the given spherical

coordinates (ρ, θ,φ):


14. (1, 0, 0)15. (2,π/3,π/4)

16. (5,π,π/2)17. (4, 3π/4,π/3)

In problems 18-21 determine the spherical coordinates (ρ, θ,φ) of the point with the givenCartesian coordinates, such that −π < θ ≤ π:

18.¡1,√3, 2√3¢

19. (0,−1,−1)20.

¡0,√3, 1¢

21.¡−1, 1,√6¢

In problems 22-29 make use of cylindrical coordinates to evaluate the given triple integral (Note

that the symbol dxdydz stands for dV in Cartesian coordinates and does not specify the order

of integration):

22. Z Z ZD

¡9− x2 − y2¢ dxdydz,

where D is the region above the xy-plane that is bounded by the sphere x2 + y2 + z2 = 9.

1

x0

0

y

2

2

-2-2

0

z

2

3

23. Z Z ZD

zdxdydz,

where D is in the first octant, between the spheres ρ = 1 andρ = 2.

2

y1

0

00

x

1

2

z 1

2

24. Z Z ZD

e√x2+y2+z2dxdydz,

where D is in the first octant and in the sphere ρ = 3.

3

2x

z

2

1

1

00

0

1 y

2

3

3


25. Z Z ZD

x2dxdydz,

where D is the region that is bounded by the xz-plane and the hemispheres y =√9− x2 − z2

and y =√16− x2 − z2.

2

0z

-2

-4-4

-2

x

0

2

4 02

y

4

4

26. Z Z ZD

x2zdxdydz,

where D is the region between the spheres ρ = 2 and ρ = 4 and above the cone φ = π/3.

0

2

4

z

y

0

42

x0-2-4

27. Z Z ZD

dxdydz,

where D is the region bounded by the sphere ρ = 3 and the cones φ = π/6 and φ = π/3.

0

2

4

4

z

2

y

0

-2 42

x0

-4 -2-4

28. Z Z ZD

zdxdydz,

where D is the region that is above the cone φ = π/3 and below the sphere ρ = 4 cos (φ).

13.7. CHANGE OF VARIABLES IN MULTIPLE INTEGRALS 179

0

2

4

z

y

0

4

x

20-2-4

29. Find the volume of the region D within the sphere x2 + y2 + z2 = 4, above the xy-plane,and below the cone z =

px2 + y2.

022

1z

2

xy

00

-2-2

13.7 Change of Variables in Multiple Integrals

(the General Case)

Definition 1 Assume that D and D∗ are regions in the plane and that T is a function that isdefined on D∗ and has values in D. We will express this fact by writing

T : D∗ → D

and refer to T as a transformation from D∗ into D. If we denote the points in D by (x, y)and the points in D∗ by (u, v), so that x, y and u, v are Cartesian coordinates, we will say thatD is a region in the xy-plane and D∗ is a region in the uv-plane. If

T (u, v) = (x (u, v) , y (u, v)) ,

we will refer to the functions x (u, v) and y (u, v) as the coordinate functions that define thetransformation T . The transformation is from D∗ onto D if each point of D is the value of T

at some point in D∗. The transformation is one-one if the values of T at distinct points of D∗

are distinct points of D.

Example 1 Let θ be a fixed angle and set

T (u, v) = (cos (θ)u− sin (θ) v, sin (θ)u+ cos (θ) v)

for each (u, v) ∈ R2. Thus, the coordinate functions are

x (u, v) = cos (θ)u− sin (θ) v and y (u, v) = sin (θ)u+ cos (θ) v

Geometrically, T describes the rotation through the angle θ.


For example, if we set θ = π/3, we have

T (u, v) =³cos³π3

´u− sin

³π3

´v, sin

³π3

´u+ cos

³π4

´v´

=

Ã1

2u−√3

2v,

√3

2u+

1

2v

!.

Let D∗ be the sector of the unit disk in the uv-plane consisting of points with polar coordinates(r, θ) such that 0 ≤ r ≤ 1 and −π/3 ≤ θ ≤ 0. The image of D∗ under T is D, the sector of theunit disk in the xy-plane consisting of points with polar coordinates (r, θ) such that 0 ≤ r ≤ 1and 0 ≤ θ ≤ π/3. Figure 1 displays D∗ and D. ¤

1u

1

v

D

Π3

1x

y

D

Π3

Figure 1

Example 2 We can view the relationship between Cartesian and polar coordinates within the

framework of functions from R2 into R2.

For given r > 0 and θ ∈ R, the corresponding Cartesian coordinates are

x = r cos (θ) and y = r sin (θ) .

Let’s consider the Cartesian θr-plane. Thus, we have the θ and r as the orthogonal axes, and

any point (θ, r) in that plane is assigned the point

T (r, θ) = (r cos (θ) , r sin (θ))

in the Cartesian xy-plane. This assignment defines a transformation from R2 into R2 ( fromthe θr-plane into the xy-plane).

For example, let D∗ is the rectangle consisting of the points (θ, r) such that π/4 ≤ θ ≤ π/2 and1 ≤ r ≤ 2 in the the θr-plane, and let D be the image of D∗ under T . D is bounded by the rays

in the xy-plane that correspond to θ = π/4 and θ = π/2, and the arcs of the circles with radius1 and 2. Figure 2 displays D∗ and D. ¤


Π4

Π2

Θ

1

2

r

D

1 2x

1

2

y

D

Figure 2

Similarly, we can define transformations in three variables: If x (u, v, w), y (u, v, w) , z (u, v,w)define functions of u, v and w, the transformation

T (u, u,w) = (x (u, v, w) , y (u, v, w) , z (u, v, w))

is a function that maps a region in the uvw-space onto a region in the xyz-space.

Example 3 Let ρ, θ and φ be spherical coordinates in R3 so that

x = ρ sin (φ) cos (θ) , y = ρ sin (φ) sin (θ) and z = ρ cos (φ) .

Here ρ is the distance of the point (x, y, z) from the origin, i.e.,

ρ =px2 + y2 + z2,

φ is the angle between the xi+ yj+ zk and k (0 ≤ φ ≤ π), and θ is a polar angle of the point

(x, y, 0) (we can assume that θ is between 0 and 2π).

We can define the transformation from R3 (the ρφθ-space) into R3 (the xyz-space) by the rule

T (ρ,φ, θ) = (ρ sin (φ) cos (θ) , ρ sin (φ) sin (θ) , ρ cos (φ)) .

If D∗ is the box in the ρφθ-space consisting of points (ρ,φ, θ) such that ρ1 ≤ ρ ≤ ρ2, φ1 ≤φ ≤ φ2, θ1 ≤ θ ≤ θ2, the image of D

∗ under T in the xyz-space is bounded by the spheres

ρ = ρ1, ρ = ρ2, the cones φ = φ1, φ = φ2 and the planes θ = θ1, θ = θ2. ¤

Definition 2 If T (u, v) = (x (u, v) , y (u, v)) is a transformation from D∗ into D the Jacobian

of x and y with respect to u and v is

∂ (x, y)

∂ (u, v)=

¯¯ ∂x

∂u

∂x

∂v∂y

∂u

∂y

∂v

¯¯ = ∂x

∂u

∂y

∂v− ∂x

∂v

∂y

∂u


Theorem 1 Assume that T (u, v) = (x (u, v) , y (u, v)) is a transformation that mapsD∗ onto Din a one-one manner (i.e., the images of distinct points are distinct), and its coordinate functions

are differentiable. ThenZ ZD

f (x, y) dxdy =

Z ZD∗f (x (u, v) , y (u, v))

¯∂ (x, y)

∂ (u, v)

¯dudv

for any function f that is continuous on D.

A plausibility Argument for the Theorem

u

v

u, v u u, v

u, v v u u, v v

x

y

Tu, v

Tu u, v

Tu, v v

Figure 3

D can be covered by the images of rectangles in the uv-plane. Consider such a rectangle with

vertices (u, v) , (u+∆u, v) , (u, v +∆v) , (u+∆u, v +∆v). The image of this rectangle can beapproximated by a parallelogram with vertices T (u, v) , T (u+∆u, v) and T (u, v +∆v), asillustrated in Figure 3. This parallelogram is spanned by the vectors

T (u+∆u, v)− T (u, v) and T (u, v +∆v)− T (u, v) .

Assuming that ∆u and ∆v are small,

T (u+∆u, v)− T (u, v) = (x (u+∆u, v)− x (u, v) , y (u+∆u, v)− y (u, v))∼=

µ∂x

∂u(u, v)∆u,

∂y

∂u∆u

¶= ∆u

µ∂x

∂u(u, v) ,

∂y

∂u(u, v)

¶.

Similarly,

T (u, v +∆v)− T (u, v) ∼= ∆vµ∂x

∂v(u, v) ,

∂y

∂v(u, v)

¶.

Thus, the area of the parallelogram that is spanned by the vectors T (u+∆u, v)− T (u, v) and


T (u, v +∆v)− T (u, v) can be approximated by¯¯∆u

µ∂x

∂u(u, v) ,

∂y

∂u(u, v) , 0

¶×∆v

µ∂x

∂v(u, v) ,

∂y

∂v(u, v) , 0

¶¯¯

= ∆u∆v

¯¯¯

¯¯¯

i j k∂x

∂u(u, v)

∂y

∂u(u, v) 0

∂x

∂v(u, v)

∂y

∂v(u, v) 0

¯¯¯

¯¯¯

= ∆u∆v

¯∂x

∂u(u, v)

∂y

∂v(u, v)− ∂y

∂u(u, v)

∂x

∂v(u, v)

¯= ∆u∆v

¯∂x

∂u(u, v)

∂y

∂v(u, v)− ∂x

∂v(u, v)

∂y

∂u(u, v)

¯

= ∆u∆v

¯¯¯¯ ∂x

∂u

∂x

∂v∂y

∂u

∂y

∂v

¯¯¯¯ = ∆u∆v ¯∂ (x, y)

∂ (u, v)

¯.

Therefore, the integral of f on such a parallelogram can be approximated by

f (x (u, v) , y (u, v))

¯∂ (x, y)

∂ (u, v)

¯∆u∆v

Since we can approximate Z ZD

f (x, y) dxdy

by the sum of such integrals, it is indeed plausible thatZ ZD

f (x, y) dxdy =

Z ZD∗f (x (u, v) , y (u, v))

¯∂ (x, y)

∂ (u, v)

¯dudv

¥


ex2+4y2dxdy,

where D is the region that is bounded by the ellipse x2 + 4y2 = 1.

Solution

-1 1

-1/2

1/2

x

y

Figure 4

Let’s set u = x and v = 2y so that x = u and y = v/2. Thus,

x2 + 4y2 = 1⇔ u2 + v2 = 1.


The transformation T (u, v) = (u, v/2) maps the disk D∗ that is bounded by the unit circleu2 + v2 = 1 in a one-one manner onto D.

-1 1

-1

1

u

v

Figure 5

We have∂x

∂u= 1,

∂x

∂v= 0,

∂y

∂u= 0,

∂y

∂v=1

2.

Therefore,∂ (x, y)

∂ (u, v)=

∂x

∂u

∂y

∂v− ∂x

∂v

∂y

∂u=1

2.

Thus Z ZD

ex2+4y2dxdy =

Z ZD∗eu

2+v2¯∂ (x, y)

∂ (u, v)

¯dudv =

Z ZD∗eu

2+v2µ1

2

¶dudv.

Let’s transform to polar coordinates in the uv-plane so that u = r cos (θ) and v = r sin (θ).Therefore, Z Z

D

ex2+4y2dxdy =

1

2

Z ZD∗eu

2+v2dudv =1

2

Z 2π

θ=0

Z r=1

r=0

er2

2drdθ

= 2π

Z 1

0

er2

rdr.

We set w = r2 so that dw = 2rdr:

2π

Z 1

0

er2

rdr = π

Z 1

0

ewdw = π (e− 1) .

Therefore, Z ZD

ex2+4y2dxdy = π (e− 1) .

¤

Example 5 We can derive the expression for double integrals in polar coordinates from the

general theorem.

We have x = r cos (θ) and y = r sin (θ). Therefore,

∂x

∂r= cos (θ) ,

∂x

∂θ= −r sin (θ) , ∂y

∂r= sin (θ) ,

∂y

∂θ= r cos (θ) .

Thus,

∂ (x, y)

∂ (r, θ)=

∂x

∂r

∂y

∂θ− ∂x

∂θ

∂y

∂r

= cos (θ) r cos (θ)− (−r sin (θ)) sin (θ)= r cos2 (θ) + r sin2 (θ) = r.


Therefore, ZD

f (x, y) dxdy =

Z ZD∗f (r cos (θ) , r sin (θ))

¯∂ (x, y)

∂ (r, θ)

¯drdθ

=

Z ZD∗f (r cos (θ) , r sin (θ)) rdrdθ.

¤There is a counterpart of the theorem on the change of the variables in double integrals for the

change of variables in triple integrals.

Definition 3 If T (u, v, w) = (x (u, v, w) , y (u, v, w) , z (u, v, w)) is a transformation from D∗

into D the Jacobian of x,y and z with respect to u,v and w isx (u, v,w)

∂ (x, y, z)

∂ (u, v, w)=

¯¯¯∂x

∂u

∂x

∂v

∂x

∂w∂y

∂u

∂y

∂v

∂y

∂w∂x

∂u

∂z

∂v

∂z

∂w

¯¯¯

Theorem 2 Assume that T (u, v, w) = (x (u, v, w) , y (u, v,w) , z (u, v, w)) is a transformationthat maps D∗ onto D in a one-one manner (i.e., the images of distinct points are distinct), and

its coordinate functions are differentiable. ThenZ ZD

f (x, y, z) dxdy =

Z ZD∗f (x (u, v,w) , y (u, v, w) , z (u, v, w))

¯∂ (x, y, z)

∂ (u, v, w)

¯dudvdw

for any function f that is continuous on D.

Just as in the case of two variables, the above theorem is plausible: The volume of the image of a

rectangular box with vertices at (u, v, w), (u+∆u, v,w) , (u, v +∆v, w) and (u, v,w +∆w) canbe approximated by the volume of a parallelepiped with vertices at T (u, v, w), T (u+∆u, v, w),T (u, v +∆v, w) and T (u, v, w +∆w). That parallelepiped is spanned by the vectors

T (u+∆u, v, w)−T (u, v, w) , T (u, v +∆v,w)−T (u, v, w) and T (u, v, w +∆w)−T (u, v, w) .If ∆u, ∆v and ∆w are small, these vectors can be approximated by

∆u

µ∂x

∂u(u, v,w) ,

∂y

∂u(u, v,w) ,

∂z

∂u(u, v,w)

¶, ∆v

µ∂x

∂v(u, v) ,

∂y

∂v(u, v) ,

∂z

∂v(u, v)

¶and

∆w

µ∂x

∂w(u, v) ,

∂y

∂w(u, v) ,

∂y

∂w(u, v)

¶.

Thus, we can approximate the volume of the image of a rectangular box with vertices at (u, v, w),(u+∆u, v,w) , (u, v +∆v,w) by the scalar triple product of these vectors:¯

¯¯∂x

∂u

∂y

∂u

∂z

∂u∂x

∂v

∂y

∂v

∂z

∂v∂x

∂w

∂y

∂w

∂y

∂w

¯¯¯∆u ∆v∆w =

∂ (x, y, z)

∂ (u, v, w)∆u ∆v∆w.

Example 6 We can derive the expression for triple integrals in spherical coordinates from the

above theorem:


We have

x = ρ sin (φ) cos (θ) , y = ρ sin (φ) sin (θ) and z = ρ cos (φ) .

Thus,

∂x

∂ρ= sin (φ) cos (θ) ,

∂x

∂φ= ρ cos (φ) cos (θ) ,

∂x

∂θ= −ρ sin (φ) sin (θ) ,

∂y

∂ρ= sin (φ) sin (θ) ,

∂y

∂φ= ρ cos (φ) sin (θ) ,

∂y

∂θ= ρ sin (φ) cos (θ) ,

∂z

∂ρ= cos (φ) ,

∂z

∂φ= −ρ sin (φ) , ∂z

∂θ= 0.

Therefore,

∂ (x, y, z)

∂ (ρ,φ, θ)=

¯¯ sin (φ) cos (θ) ρ cos (φ) cos (θ) −ρ sin (φ) sin (θ)sin (φ) sin (θ) ρ cos (φ) sin (θ) ρ sin (φ) cos (θ)cos (φ) −ρ sin (φ) 0

¯¯

= cos (φ)

¯ρ cos (φ) cos (θ) −ρ sin (φ) sin (θ)ρ cos (φ) sin (θ) ρ sin (φ) cos (θ)

¯+ρ sin (φ)

¯sin (φ) cos (θ) −ρ sin (φ) sin (θ)sin (φ) sin (θ) ρ sin (φ) cos (θ)

¯= cos (φ)

¡ρ2 sin (φ) cos (φ) cos2 (θ) + ρ2 sin (φ) cos (φ) sin2 (θ)

¢+ρ sin (φ)

¡sin2 (φ) cos2 (θ) + ρ sin2 (φ) sin2 (θ)

¢= ρ2 cos2 (φ) sin (φ) + ρ2 sin3 (φ)

= ρ2 sin (φ)¡cos2 (φ) + sin2 (φ)

¢= ρ2 sin (φ) .

Thus,Z ZD

f (x, y, z) dxdy =

Z ZD∗f (x (ρ,φ, θ) , y (ρ,φ, θ) , z (ρ,φ, θ))

¯∂ (x, y, z)

∂ (ρ,φ, θ)

¯dρdφdθ

=

Z ZD∗f (x (ρ,φ, θ) , y (ρ,φ, θ) , z (ρ,φ, θ)) ρ2 sin (φ) dρdφdθ.

¤

Chapter 14

Vector Analysis

In this chapter we will discuss line integrals and surface integrals that correspond to impor-

tant topics from Physics, such as the work done by a force field on a particle as it moves along

a trajectory and the flux across a surface of a vector field such as the velocity field of fluid flow

or an electrostatic field. The chapter will conclude with three major theorems that have many

applications: The theorems of Green, Gauss and Stokes.

14.1 Vector Fields, Divergence and Curl

Vector Fields

A two-dimensional vector field is a function that assigns to each point (x, y) in a subset Dof the plane a two-dimensional vector F(x, y). Thus,

F(x, y) =M (x, y) i+N(x, y)j,

where M and N are real-valued functions defined on D. We can visualize such a vector field by

attaching the vector F (x, y) to the point (x, y). Computer algebra systems have special routinesfor such visualization.

A three-dimensional vector field is a function that assigns to each point (x, y, z) in a subsetD of R3 a three-dimensional vector F(x, y, z). Thus,

F(x, y, z) =M (x, y, z) i+N(x, y, z)j+ P (x, y, z)k,

where M , N and P are real-valued functions defined on D. We can visualize F (x, y, z) as avector attached to (x, y, z)

Example 1 Let

F (x, y) = r (x, y) = xi+ yj.

Figure 1 shows the vectors F (x, y) for a selection of the points (x, y). ¤

187

188 CHAPTER 14. VECTOR ANALYSIS

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

Figure 1: The vector field F (x, y) = r (x, y) = xi+ yj

Example 2 Let

F (x, y) = ur (x, y) =r (x, y)

||r (x, y)|| =1p

x2 + y2(xi+ yj)

for each (x, y) 6= (0, 0).

Thus, F assigns to each point other than the origin the unit vector in the direction of its position

vector. We can set r (x, y) = ||r (x, y)|| so that

F (x, y) = ur (x, y) =r

r(x, y) .

¤

Example 3 Let

S (x, y) = −yi+xj

Note that

S (x, y) · (xi+ yj) = −yx+ xy = 0,so that S (x, y) is orthogonal to the position vector r of (x, y). We also have

||S (x, y)|| =py2 + x2 = ||xi+ yj|| ,

so that the length of S (x, y) is constant on each circle centered at the origin. Figure 2 showssome of the vectors F (x, y). We may refer to S as a spin field. ¤

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

x

y

Figure 2: The spin field S

14.1. VECTOR FIELDS, DIVERGENCE AND CURL 189

Example 4 (A gravitational field) Newton’s inverse-square law says that the magnitude

of gravitational attraction between two objects of mass M and m is

GMm

d2,

where G is a universal constant and d is the distance between the objects. Thus, the force on

an object of unit mass at the point (x, y, z) due to the mass M at the origin is

F (x, y, z) = − GM

||r (x, y, z)||2ur (x, y, z) ,

where r (x, y, z) = xi+ yj+ zk 6= 0 and

ur (x, y, z) =r

||r|| (x, y, z)

is the unit vector in the direction of r. This is the gravitational field due to the mass M .

We can express F (x, y, z) as

F (x, y, z) = − GM

||r (x, y, z)||2ur (x, y, z) = −GM

||r (x, y, z)||2µr

||r|| (x, y, z)¶= − GM

||r (x, y, z)||3 r (x, y, z) .

If we set r (x, y, z) = ||r (x, y, z)||,

F (x, y, z) = − GM

r2 (x, y, z)ur (x, y, z) = − GM

r3 (x, y, z)r (x, y, z) = − GM

(x2 + y2 + z2)3/2

(xi+ yj+ zk)

¤

Example 5 (An electrostatic field) ByCoulomb’s law, the force per unit charge at (x, y, z)due to the charge q at the origin is

E (x, y, z) =q

4πεr2ur,

where ur is the unit vector in the radial direction. E (x, y, z) points towards the origin if q < 0and points away from the origin if q > 0.¤

The fields of examples 1, 2, 11 and 4 are radial fields: The value of such a field at (x, y) or(x, y, z) is of the form

C

rnr,

where r is the position vector of (x, y) or (x, y, z), r = ||r||, C is a constant and n is a positive

integer.

Recall that the gradient of a scalar function f (x, y) is defined by the expression

∇f (x, y) = ∂f

∂xi+

∂f

∂yj.

Thus, ∇f is a vector field in the plane. We noted that ∇f (x, y) is orthogonal to the level curveof f that passes through (x, y). Similarly, if f is a scalar function of x, y and z,

∇f (x, y, z) = ∂f

∂xi+

∂f

∂yj+

∂f

∂zk

defines a vector field in three dimensions. The vector ∇f (x, y, z) is orthogonal to the levelsurface of f that passes through (x, y, z).Examples 1, 2, 11 and 4 are examples of gradient fields:


Example 6

a) Let F (x, y) = r, where r = xi+ yj, as in Example 1. Show that

F (x, y) =∇µr2

2

¶=∇

µ1

2

¡x2 + y2

¢¶.

b) Let

F (x.y) = ur =r

r,

as in Example 2. Show that

F (x, y) =∇rc) Let

F (x, y, z) =C

r3r,

where C is a constant, as in examples 11 and 4. Show that

F (x, y, z) =∇µ−Cr

¶.

Solution

a)

∇µ1

2r2¶=1

2

∂

∂x

¡x2 + y2

¢i+

∂

∂y

¡x2 + y2

¢j =

1

2(2x) i+

1

2(2y) j = xi+ yj = r

b)

∇r = ∂r

∂xi+

∂r

∂yj =

∂

∂x

¡x2 + y2

¢102i+

∂

∂y

¡x2 + y2

¢1/2j

=1

2

¡x2 + y2

¢−1/2(2x) i+

1

2

¡x2 + y2

¢−1/2(2y) j =

x

ri+

y

rj =

r

r= ur

c) Since

∇µ−Cr

¶= −C∇

µ1

r

¶,

it is sufficient to show that

∇µ1

r

¶= − r

r3

Indeed,

∇µ1

r

¶=

∂

∂xr−1i+

∂

∂yr−1j

=

µ∂

∂x

¡x2 + y2 + z2

¢−1/2¶i+

µ∂

∂y

¡x2 + y2 + z2

¢−1/2¶j+

µ∂

∂z

¡x2 + y2 + z2

¢−1/2¶k

=

µ−12

¡x2 + y2 + z2

¢−3/2(2x)

¶i+

µ−12

¡x2 + y2 + z2

¢−3/2(2y)

¶j

+

µ−12

¡x2 + y2 + z2

¢−3/2(2z)

¶k

= − x

(x2 + y2 + z2)3/2i+− y

(x2 + y2 + z2)3/2j+− z

(x2 + y2 + z2)3/2k

= − 1

(x2 + y2 + z2)3/2

(xi+ yj+ zk) = − 1r3r.


¤

Definition 1 A field is conservative if it is the gradient of a scalar function. Thus, the field

F is conservative if there exists a scalar function f such that F =∇f . Such an f is referred toas a potential for F.

In Example 6 we showed that gravitational fields and electronic fields are conservative fields.

Remark 1 If f is a potential for F, then f + C is also a potential for F if C is an arbitrary

constant. Indeed,

∇ (f + C) = ∇F,since ∇C = 0. ♦

The Divergence and Curl of a Vector Field

Definition 2 The divergence of the vector field

F (x, y) =M (x, y) i+N(x, y)j

is the scalar function∂M

∂x(x, y) +

∂N

∂y(x, y) .

We will denote the divergence of F as div F so that

div F (x, y) =∂M

∂x(x, y) +

∂N

∂y(x, y) .

Let

∇ =∂

∂xi+

∂

∂yj

be the symbolic vector that is used in denoting the gradient of a scalar function. Usually, we

will denote the divergence of a vector field by the symbolic dot product

∇ · F (x, y) =µ

∂

∂xi+

∂

∂yj

¶· (M (x, y) i+N(x, y)j) =

∂M

∂x(x, y) +

∂N

∂y(x, y) .

Similarly, if F(x, y, z) =M (x, y, z) i+N(x, y, z)j+P (x, y, z)k is a vector field in R3, we define

the divergence of F by setting

divF (x, y, z) =∂M

∂x+

∂N

∂y+

∂P

∂z.

In terms of the symbolic vector

∇ =∂

∂xi+

∂

∂yj+

∂

∂zk,

divF (x, y, z) =∇ · F (x, y, z) =µ

∂

∂xi+

∂

∂yj+

∂

∂zk

¶· (M (x, y, z) i+N(x, y, z)j+ P (x, y, z)k)

=∂M

∂x+

∂N

∂y+

∂P

∂z.

Example 7


a) Let F (x, y) = r = xi+ yj, as in Example 1. Then

divF (x, y) =∇ · F (x, y) = ∂

∂x(x) +

∂

∂y(y) = 1 + 1 = 2.

b) S (x, y) = −yi+xj be the spin field of Example 3. Then

∇ · S (x, y) = ∂

∂x(−y) + ∂

∂y(x) = 0.

Later in this chapter we will see that the divergence of a vector field will be a measure of the

pointwise expansion or contraction of a vector field. The vector field of part a) “expands” with

respect to the origin. The vector field of part b) does not contract or expand.¤

Definition 3 The curl of a three-dimensional vector field F(x, y, z) is the vector fieldthat is defined by the symbolic cross product

∇×F (x, y, z) =µ

∂

∂xi+

∂

∂yj+

∂

∂zk

¶× (M (x, y, z) i+N(x, y, z)j+ P (x, y, z)k)

=

¯¯ i j k

∂

∂x

∂

∂y

∂

∂zM N P

¯¯ = µ∂P

∂y− ∂N

∂z

¶i−

µ∂P

∂x− ∂M

∂z

¶j+

µ∂N

∂x− ∂M

∂y

¶k

If F (x, y) =M (x, y) i+N(x, y)j is a two-dimensional vector field, we can consider F as a vectorfield in three dimensions by setting

F (x, y, z) =M (x, y) i+N(x, y)j+ 0k.

With this understanding,

∇× F (x, y) =

¯¯ i j k

∂

∂x

∂

∂y

∂

∂zM (x, y) N (x, y) 0

¯¯ = −∂N

∂zi+

∂M

∂zj+

µ∂N

∂x− ∂M

∂y

¶k

=

µ∂N

∂x− ∂M

∂y

¶k

Thus, the curl of a vector field in the xy-plane is always orthogonal to the xy-plane.

Example 8 Let ωS (x, y) = −ωyi+ ωxj, where S is the spin field of Example 3. Calculate

∇× (ωS).Solution

We have

∇× (ωS) (x, y) =µ

∂

∂x(ωx)− ∂

∂y(−ωy)

¶k = 2ωk

Thus, the curl of ωS at any point is a vector of magnitude |2ω| that is orthogonal to the xy-plane. We can imagine that ωS is the velocity field of a fictitious fluid that rotates about the

origin, and we may refer to 2ω as the angular velocity. Later in this chapter we will see that thecurl of a vector field at a point is a measure of the rotation of the vector field about that point.

¤

Definition 4 A vector field F such that ∇×F = 0 is said to be irrotational.


Proposition 1 A gradient field is irrotational:

∇× (∇f) = 0

Proof

∇× (∇f) =

¯¯¯i j k∂

∂x

∂

∂y

∂

∂z∂f

∂x

∂f

∂y

∂f

∂z

¯¯¯

=

µ∂

∂y

µ∂f

∂z

¶− ∂

∂z

µ∂f

∂y

¶¶i−

µ∂

∂x

µ∂f

∂z

¶− ∂

∂z

µ∂f

∂x

¶¶j

+

µ∂

∂x

µ∂f

∂y

¶− ∂

∂y

µ∂f

∂x

¶¶k = 0

¥

Example 9 Let

F (x, y, z) =r

r3,

where r = xi+ yj+ zk and r =px2 + y2 + z2.

In Example 6 we showed that F is a gradient field. Therefore, F is irrotational. In particular,

gravitational fields and electrostatic fields are irrotational. Later in this chapter we will see that

any irrotational vector field F is a gradient field if F satisfies certain smoothness conditions. ¤

Problems

In problems 1-6 determine

a) The divergence of F,

b) The curl of F.

1.

F(x, y, z) = xyzi− x2yk2.

F (x, y, z) = e−xxy2i+ e−xx2yj

3.

F (x, y, z) = cos (xz) j− sin (xy)k4.

F (x, y, z) = arctan³yx

´i+ arctan

µx

y

¶j

5.

F (x, y, z) = 2xyi+¡x2 + 2yz

¢j+ y2k

6.

F (x, y, z) = ln (x) i+ ln (xy) j+ ln (xyz)k


14.2 Line Integrals

The Integral of a Scalar Function with respect to arc length

Let C be a curve in the plane that is parametrized by the path σ : [a, b] → R2. Assumethat f is a real-valued function of two variables. You may imagine that C is a wire, and

f (σ (t)) = f (x (t) , y (t)) is the lineal density of the wire. How can we calculate the mass of thewire? It is reasonable to approximate mass by sums of the form

nXk=1

f (σ (tk))∆sk =nXk=1

f (x (tk) , y (tk))∆sk,

where t0, t1, . . . , tn−1, tn is a partition of [a, b] and ∆sk is the length of the part of C corre-

sponding to the interval [tk−1, tk]. Thus

∆sk =

Z tk

tk−1||σ0 (t)|| dt.

By the Mean Value Theorem for integrals,Z tk

tk−1||σ0 (t)|| dt = σ0 (t∗k) (tk − tk−1) = σ0 (t∗k)∆tk,

where t∗k is some point between tk−1 and tk. Therefore,

nXk=1

f (σ (tk))∆sk =nXk=1

f (σ (tk)) ||σ0 (t∗k)||∆tk

This is almost a Riemann sum for the integralZ b

a

f (σ (t)) ||σ0 (t)|| dt

and approximates the integral with desired accuracy provided that f and σ0 (t) are continuous.This leads to the following definition:

Definition 1 The integral of the scalar function f with respect to arc length on the

curve C that is parametrized by the function σ : [a, b]→ R2 is defined asZ b

a

f (σ (t)) ||σ0 (t)|| dt

Symbolically,

ds =ds

dtdt =

¯¯dσ

dt

¯¯dt,

so that we can express the integral as ZC

fds.

In terms of the coordinate functions x (t) and y (t) of σ (t),

ZC

fds =

Z b

a

f (x (t) , y (t))

sµdx

dt

¶2+

µdy

dt

¶2dt.

14.2. LINE INTEGRALS 195

Example 1 Evaluate ZC

x2y2ds,

where C is the semicircle x2 + y2 = 4 that is parametrized so that C is traversed from (2, 0) to(−2, 0).Solution

We can set

σ (t) = (2 cos (t) , 2 sin (t)) , 0 ≤ t ≤ π.

Then σ parametrizes C as required.

2 1 1 2x

1

2y

Figure 1

Therefore,

σ0 (t) = −2 sin (t) i+ 2 cos (t) j⇒ ||σ0 (t)|| =q4 sin2 (t) + 4 cos2 (t) = 2.

Thus, ZC

x2y2ds =

Z π

0

¡4 cos2 (t)

¢ ¡4 sin2 (t)

¢ ||σ0 (t)|| dt = 16Z π

0

cos2 (t) sin2 (t) (2) dt

= 32

Z π

0

cos2 (t) sin2 (t) dt.

As we saw in Section 7.3,Zcos2 (t) sin2 (t) dt =

1

8x− 1

8sin (x) cos3 (x) +

1

8sin3 (x) cos (x) .

Therefore,

32

Z π

0

cos2 (t) sin2 (t) dt = 4x− 4 sin (x) cos3 (x) + 4 sin3 (x) cos (x) .¯π0= 4π

¤A curve C can be parametrized by different functions and may be traversed from point A to

point B or vice versa. Thus, there is a notion of "the orientation" of a curve. Assume that

C is a curve in the plane that is parametrized initially by σ : [a, b] → R2 and σ (t) traces Cfrom "the initial point" P1 to "the terminal point" P2 as t increases from a to b. We

will say that σ : [α,β] → R2 is an orientation preserving parametrization of C if σ (τ)traces C from the initial point P1 to the terminal point P2 as the new parameter τ increases

from α to β. We will say that σ : [α,β]→ R2 is an orientation reversing parametrizationof C if σ (τ) traces C from the terminal point P2 to the iniital point P1 as the new parameter

τ increases from α to β. The positive orientation of C is the orientation provided by the


initial parametrization σ and the negative orientation of C is the orientation provided by

any orientation reversing parametrization.You can find the precise defintions at the end of this

section. The line integral of a scalar function is does not change if the curve is parametrized by

any orientation preserving or orientation reversing parametrization:

Proposition 1 If C is parametrized by σ : [a, b] → R2 and σ : [α,β] → R2 is an orientationpreserving or orientation reversing parametrization of C, thenZ

C

fds =

Z t=b

t=a

f (σ (t))

¯¯dσ

dt

¯¯dt =

Z τ=β

τ=α

f (σ (τ))

¯¯dσ

dτ

¯¯dτ

You can find the proof of Proposition 1 at the end of this section.

Example 2 Let P1 = (1, 2) and P2 = (2, 4), and let C be the directed line segment P1P2. We

can parametrize C by

σ (t) = OP1 + tP1P2 = (1, 2) + t (1, 2) = (1 + t, 2 + 2t) , 0 ≤ t ≤ 1.

If we set

σ (τ) = OP1 + 2τP1P2 = (1, 2) + 2τ (1, 2) = (1 + 2τ , 2 + 4τ) , 0 ≤ τ ≤ 12,

then σ is an orientation preserving parametrization of C.

If we set

σ∗ (μ) = OP2 + μτP2P1 = (2, 4) + μ (−1,−2) = (2− μ, 4− 2μ) , 0 ≤ μ ≤ 1,

then σ∗ is an orientation reversing parametrization of C. Confirm that the statements of Propo-sition 1 are valid in the special case Z

C

sin³π2(x− y)

´ds

Solution

We havedσ

dt= (1, 2) so that

¯¯dσ

dt

¯¯=√1 + 4 =

√5.

Therefore ZC

sin³π2(x− y)

´ds =

Z 1

0

sin³π2((1 + t)− (2 + 2t))

´ ¯¯dσdt

¯¯dt

=

Z 1

0

sin³π2(−t− 1)

´√5dt

=√5

Z 1

0

− sin³π2(t+ 1)

´dt.

We set

u =π

2(t+ 1)⇒ du =

π

2dt.

Thus,

√5

Z 1

0

− sin³π2(t+ 1)

´dt =

2√5

π

Z π

π/2

− sin (u) du = 2√5

π

³cos (u)|u=πu=π/2

´= −2

√5

π.


Now let’s use the orientation preserving parametrization σ. We have

dσ

dτ= (2, 4) so that

¯¯dσ

dt

¯¯=√4 + 16 =

√20 = 2

√5.

Therefore ZC

sin³π2(x− y)

´ds =

Z 1/2

0

sin³π2((1 + 2τ)− (2 + 4τ))

´ ¯¯dσdτ

¯¯dτ

=

Z 1/2

0

sin³π2(−2τ − 1)

´2√5dτ

= −2√5

Z 1/2

0

sin³π2(2τ + 1)

´dτ .

We set

u =π

2(2τ + 1)⇒ du = πdτ .

Thus

2√5

Z 1/2

0

− sin³π2(2τ + 1)

´dτ =

2√5

π

Z π

π/2

− sin (u) du = −2√5

π,

as in the case of the parametrization by σ.

σ∗ (μ) = OP2 + μτP2P1 = (2, 4) + μ (−1,−2) = (2− μ, 4− 2μ) , 0 ≤ μ ≤ 1,

Now let’s use the orientation reversing parametrization σ∗. We have

dσ∗

dμ= (−1,−2) so that

¯¯dσ∗

dμ

¯¯=√5.

ThereforeZ 1

0

sin³π2((2− μ)− (4− 2μ))

´ ¯¯dσ∗dμ

¯¯dμ =

Z 1

0

sin³π2(μ− 2)

´√5dμ

=√5

Z 1

0

− sin³π2(2− μ)

´dμ.

We set

u =π

2(2− μ)⇒ du = −π

2dμ.

Thus

√5

Z 1

0

− sin³π2(2− μ)

´dμ =

2√5

π

Z π/2

π

sin (u) du =2√5

π.

Therefore Z 1

0

sin (σ∗ (μ))¯¯dσ∗

dμ

¯¯dμ = −

ZC

fds,

as predicted by Proposition 1


The Line Integral of a Vector Field in the Plane

Assume that F (x, y) = M (x, y) i + N (x, y) j is a two-dimensional force field, σ : [a, b] → R2,and a particle is at σ (t) = (x (t) , y (t)) at time t. Let C be the curve that is parametrized by σ.How should we calculate the work done by the force F as the particle moves from (x (a) , y (a))to (x (b) , y (b)) along C? Our starting point is the definition of the work done by a constantforce F on a particle that moves along a line. If the displacement of the particle is represented

by the vector w, the work done is

F ·w = ||F|| ||w|| cos (θ) ,

where θ is the angle between F and w. Let us subdivide the interval [a, b] into subintervals bya partition

P = a = t0 < t1 < · · · < tk−1 < tk < · · · < tn−1 < tn = b .As usual we set ∆tk = tk − tk−1.Let ∆σk = σ (tk)−σ (tk−1). If the norm of the partition P is

small, we can approximate the work done by F as the particle moves from σ (tk−1) to σ (tk) by

F (σ (tk)) ·∆σk = F (σ (tk)) · (σ (tk)− σ (tk−1)) ∼= F (σ (tk)) · σ (tk)− σ (tk−1)∆tk

∆tk

∼= F (σ (tk)) · dσdt(tk)∆tk

Therefore, the total work done is approximately

nXk=1

F (σ (tk)) · dσdt(tk)∆tk

This is a Riemann sum for the integralZ b

a

F (σ (t)) · dσdtdt.

We will define the work done by the force F on the particle as it moves along the

curve C that is parametrized by σ by this integral. In more general terms, this is a line integral:

Definition 2 Assume that the curve C in the plane is parametrized by σ : [a, b] → R2 and Fis a continuous vector field in the plane. The line integral of F on C isZ

C

F·dσ =Z b

a


The notationRCF·dσ is appropriate since the line integral is approximated by sums of the form

nXk=1

F (σ (tk)) ·∆σk.

Symbolically,

dσ =dσ

dtdt.


ΣtΣ't

FΣt

Figure 2

Example 3 LetF (x, y) = 2xi+yj and assume that C is parametrized by σ (t) = (cosh (t) , sinh (t)) ,where 0 ≤ t ≤ 3. Calculate Z

C

F·dσ

Solution

Figure 3 shows the curve C.

1 3 6 9x

2

4

6

8

y

Figure 3

We havedσ

dt= sinh (t) i+ cosh (t) j.

Therefore, ZC

F·dσ =Z 3

0

(2 cosh (t) i+ sinh (t) j) · (sinh (t) i+ cosh (t) j) dt

=

Z 3

0

(2 cosh (t) sinh (t) + sinh (t) cosh (t)) dt

=

Z 3

0

3 cosh (t) sinh (t) dt.

If we set u = cosh (t) then du = sinh (t) dt, so thatZ 3

0

3 cosh (t) sinh (t) dt. = 3

Z cosh(3)

cosh(0)

udu = 3

Ãu2

2

¯cosh(3)1

!=3

2

¡cosh2 (3)− 1¢

¤


We stated that the integral of a scalar function with respect to arc length does not depend on

an orientation preserving or orientation reversing parametrization of C. In the case of the line

integral of a vector field, an orientation reversing reparametrization introduces a minus sign:

Proposition 2 Assume that the curve C is parametrized by σ : [a, b]→ R2. If σ : [α,β]→ R2is an orientation preserving parametrization of C, thenZ β

α

F (σ (τ)) · dσ (τ)dτ

dτ =

Z b

a


If σ : [α,β]→ R2 is an orientation reversing parametrization of C, then

Z β

α


dτ = −Z b

a

F (σ (t)) · dσdtdt

You can find the proof of Proposition 2 at the end of this section.

Remark Assume that the curve C is parametrized by σ : [a, b]→ R2. and that σ : [α,β]→ R2is an orientation reversing parametrization of C. We will setZ

−CF · d σ =

Z β

α


dτ .

By Proposition 2, Z−CF · dσ = −

Z b

a

F (σ (t)) · dσdtdt = −

ZC

F · dσ.

If C is a piecewise smooth curve that can be expressed as a sequence of smooth curves C1, C2, . . . , Cm,

we will write

C = C1 + C2 + · · ·+ Cm.With this understanding, we setZ

C

F·dσ =ZC1+C2+···+Cm

F·dσ =ZC1

F·dσ +ZC2

F·dσ + · · ·+ZCm

F·dσ

♦

Example 4 Let

F (x, y) = −yi+ xj.Evaluate Z

C1+C2

F · dσ,

where C1 is the line segment from (−1, 0) to (1, 0) on the x-axis, and C2 is the semicircle ofradius 1 centered at the origin traversed in the counterclockwise direction

Solution

1 1x

1y

C1

C2

Figure 4


We can parametrize the line segment by σ1 (t) = (t, 0), where t ∈ [−1, 1]. The semicircle can beparametrized by σ2 (t) = (cos (t) , sin (t)), where t ∈ [0,π]. Therefore,Z

C1

F · d σ =Z 1

−1F (σ (t)) · dσ

dtdt =

Z 1

−1tj · idt = 0,

andZC2

F · dσ =Z π

0

F (σ (t)) · dσdtdt =

Z π

0

(− sin (t) i+ cos (t) j) · (− sin (t) i+ cos (t) j) dt

=

Z π

0

¡sin2 (t) + cos2 (t)

¢dt =

Z π

0

1dt = π.

Thus, ZC1+C2

F · d σ =ZC1

F ·dσ +ZC2

F·dσ = π.

¤

The Line Integral as an Integral with respect to Arc Length

We can express the line integral of F on the curve C as the integral of the tangential

component of F with respect to arc length (Recall that T (t) denotes the unit tangentto C at σ (t)):

ZC

F · dσ =Z b

a


Z b

a

F (σ (t)) ·dσ

dt¯¯dσ

dt

¯¯ ¯¯dσdt

¯¯dt =

Z b

a

F (σ (t)) ·T (t)¯¯dσ

dt

¯¯dt

=

Z b

a

F (σ (t)) ·T (t) dsdtdt

=

ZC

F ·Tds.

Thus, we have the following fact:

Proposition 3 ZC

F · d σ =ZC

F ·Tds.

Example 5 Let

F (x, y) = −yi+ xjand let C be the circle of radius 2 that is centered at the origin and is traversed counterclockwise.

Evaluate the line integral of F on C as an integral with respect to arc length.

Solution

We can parametrize C via the function σ (t) = (2 cos (t) , 2 sin (t)), where 0 ≤ t ≤ 2π. We haveσ0 (t) = −2 sin (t) i+ 2 cos (t) j,

so that

F (σ (t)) = F (2 cos (t) , 2 sin (t)) = −2 sin (t) i+ 2 cos (t) j,


ds = ||σ0 (t)|| dt =q4 sin2 (t) + 4 cos2 (t)dt = 2dt,

and

T (t) =σ0 (t)||σ0 (t)|| = − sin (t) i+ cos (t) j.

Therefore,ZC

F · d σ =ZC

F ·Tds =Z 2π

0

F (σ (t)) ·T (t) ds

=

Z 2π

0

(−2 sin (t) i+ 2 cos (t) j) · (− sin (t) i+ cos (t) j) 2dt

=

Z 2π

0

¡4 sin2 (t) + 4 cos2 (t)

¢dt

=

Z 2π

0

4dt = 8π.

¤

The Differential Form Notation

There is still another useful way to express a line integral. If F (x, y) = M (x, y) i + N (x, y) jand the curve C is parametrized by σ (t) = (x (t) , y (t)) , where a ≤ t ≤ b,Z

C

F · dσ =Z b

a

(M (x (t) , y (t)) i+N (x (t) , y (t)) j) ·µdx

dti+

dy

dtj

¶dt

=

Z b

a

µM (x (t) , y (t))

dx

dt+N (x (t) , y (t))

dy

dt

¶dt

=

Z b

a

M (x (t) , y (t))dx

dtdt+

Z b

a

N (x (t) , y (t))dy

dtdt.

Using the formalism,

dx =dx

dtdt and dy =

dy

dtdt,

we can express the integrals asZ b

a

M (x (t) , y (t)) dx+

Z b

a

N (x (t) , y (t)) dy.

This motivates the notation ZC

Mdx+Ndy

for the line integral of F =M i+N j on the curve C.

Definition 3 If M and N are functions of x and y, and C is a smooth curve in the plane that

is parametrized by σ : [a, b]→ R2 we setZC

Mdx+Ndy =

Z b

a

M (x (t) , y (t))dx

dtdt+

Z b

a

N (x (t) , y (t))dy

dtdt.


Thus, ZC

F · d σ =ZC

Mdx+Ndy

if F =M i+N j. The expression Mdx+Ndy is referred to as a differential form, and we willrefer to the above expression for a line integral as the integral of a differential form on the

curve C.

Example 6 Evaluate the line integral of F (x, y) = y3i+x2j on the ellipse that is parametrizedby

σ (t) = (4 cos (t) , sin (t)) , 0 ≤ t ≤ 2π,by expressing the line integral as the integral of a differential form.

4-4

-1

1

x

y

Figure 5

Solution

ZC

F·dσ =ZC

y3dx+ x2dy =

Z 2π

0

µsin3 (t)

dx

dt+ 16 cos2 (t)

dy

dt

¶dt

=

Z 2π

0

¡sin3 (t) (−4 sin (x)) + 16 cos2 (t) (cos (t))¢ dt

=

Z 2π

0

¡−4 sin4 (t) + 16 cos3 (t)¢ dt = −3π(Check: You will have to refresh your memory with respect to the integrals of powers of sines

and cosines as we discussed in Section 7.3). ¤

Example 7 Let C be the boundary of the square [−1, 1] × [−1, 1] that is traversed in thecounterclockwise direction. Calculate Z

C

−ydx+ xdy.

1-1

1

-1

x

y

Figure 6


Solution

We can express C as C1 + C2 + C3 + C4,as indicated in Figure 4.We can parametrize C1 by setting x (t) = t and y (t) = −1, where −1 ≤ t ≤ 1. Then,Z

C1

−ydx+ xdy =Z 1

−1

µdx

dt+ tdy

dt

¶dt =

Z 1

−1dt = 2

We can express C3 as −C3, where −C3 is parametrized by setting x (t) = t and y (t) = 1, where−1 ≤ t ≤ 1. Then,Z

C3

−ydx+ xdy = −Z−C3−ydx+ xdy = −

Z 1

−1

µ−dxdt+ tdy

dt

¶dt =

Z 1

−1dt = 2

We can parametrize C2 by setting x (t) = 1 and y (t) = t, where −1 ≤ t ≤ 1. Then,ZC2

−ydx+ xdy =Z 1

−1

µ−tdxdt+dy

dt

¶dt =

Z 1

−1dt = 2

Similarly, C4 = −C2, where −C2 is parametrized by x (t) = −1 and y (t) = t, −1 ≤ t ≤ 1.Then, Z

C4

−ydx+ xdy = −Z−C2−ydx+ xdy = −

Z 1

−1

µ−dydt

¶dt =

Z 1

−1dt = 2

Therefore, ZC

−ydx+ xdy. =4Xk=1

ZCk

−ydx+ xdy = 2 + 2 + 2 + 2 = 8.

¤

Curves in R3

Our discussion of integrals with respect to arc length and line integrals extends to curves in

three dimensions. Thus, assume that σ : [a, b]→ R3 is a smooth function and parametrizes thecurve C. If f is a continuous scalar function of three variables, we define the integral of f with

respect to arc length on C as ZC

fds =

Z b

a

f (σ (t)) ||σ0 (t)|| dt.

If σ (t) = (x (t) , y (t) , z (t)),ZC

fds =

Z b

a

f (x (t) , y (t) , z (t))

sµdx

dt

¶2+

µdy

dt

¶2+

µdz

dt

¶2dt.

As in the case of curves in the plane, the above integral is the same for orientation preserving

and orientation reversing parametrizations of C.

If F is a vector field inR3 so that

F(x, y, z) =M (x, y, z) i+N (x, y, z) j+ P (x, y, z)k,

the line integral of F on the curve C isZC

F·dσ =Z b

a



In the differential form notation,ZC

F·dσ =ZC

Mdx+Ndy + Pdz

=

Z b

a

µM (x (t)) , y (t) , z (t)

dx

dt+N (x (t)) , y (t) , z (t)

dy

dt+ P (x (t)) , y (t) , z (t)

dz

dt

¶dt.

The line integral can be expressed as an integral with respect to arc length:ZC

F·d σ =ZC

F ·Tds,

where T denotes the unit tangent to the curve C. As in the case of two-dimensional vector

fields, ZC

F·dσ

is unchanged by an orientation preserving parametrization of C and multiplied by −1 if theparametrization reverses the orientation.

Example 8 Let C be the helix that is parametrized by

σ (t) = (cos (t) , sin (t) , t) , 0 ≤ t ≤ 4π,

and let F (x, y, z) = −yi+ xj+zk. Evaluatea) Z

C

zds,

b) ZC

F · dσ

Solution

The picture shows the helix.

z

yx

Figure 7

a) We have

σ0 (t) = − sin (t) i+ cos (t) j+ kTherefore,

ds = ||σ0 (t)|| =qsin2 (t) + cos2 (t) + 1 =

√1 + 1 =

√2.


Thus, ZC

zds =

Z 4π

0

t ||σ0 (t)|| dt =Z 4π

0

t√2dt =

√2

Ãt2

2

¯4π0

!=√2

µ16π2

2

¶= 8√2π2

b) ZC

F · dσ =Z 4π

0

(− sin (t) i+ cos (t) j+ tk) · (− sin (t) i+ cos (t) j+ k) dt

=

Z 4π

0

¡sin2 (t) + cos2 (t) + t

¢dt

=

Z 4π

0

(1 + t) dt = t+t2

2

¯4π0

= 4π +16π2

2= 4π + 8π2.

¤

Precise Definitions and Proofs

Let’s begin by stating the precise definitions of orientation preserving and orientation reversing

parametrizations of a curve:

Definition 4 Assume that C is a curve in the plane that is parametrized by σ : [a, b] → R2.We say that σ : [α,β] → R2 is an orientation preserving parametrization of C if there

exists a differentiable increasing function h : [α,β] → [a, b] such that σ (τ) = σ (h (τ)) foreach τ ∈ [α,β]. The function σ is an orientation reversing parametrization of C if thereexists a differentiable decreasing function h : [α,β]→ [a, b] such that σ (τ) = σ (h (τ)) for eachτ ∈ [α,β].Now let’s prove Proposition 1:

Assume that C is a curve in the plane that is parametrized by σ : [a, b] → R2 and thatσ : [α,β]→ R2 is an orientation preserving or reversing parametrization of C. ThenZ b

a

f (σ (t))

¯¯dσ

dt

¯¯dt =

Z β

α

f (σ (τ))

¯¯dσ

dt

¯¯dτ .

Proof

Assume that σ (τ) = σ (h (τ)), where h is an increasing or decreasing and differentiable functionfrom [α;β] onto [a, b]. By the chain rule,

dσ

dτ=dh

dτ

dσ

dt(h (τ)) .

Therefore, ¯¯dσ

dτ

¯¯=

¯dh

dτ

¯ ¯¯dσ

dt(h (τ))

¯¯Thus, Z β

α

f (σ (h (τ)))

¯¯dσ

dτ

¯¯dτ =

Z β

α

f (σ (h (τ)))

¯dh

dτ

¯ ¯¯dσ

dt(h (t))

¯¯dτ

If dh/dτ > 0, Z β

α

f (σ (h (τ)))

¯dh

dτ

¯ ¯¯dσ

dt

¯¯dτ =

Z β

α

f (σ (h (τ)))dh

dτ

¯¯dσ

dt

¯¯dτ .


By the substitution rule,

Z β

α

f (σ (h (τ)))dh

dτ

¯¯dσ

dt

¯¯dτ =

Z β

α

f (σ (h (τ)))

¯¯dσ

dt

¯¯dh

dτdτ =

Z h(β)

h(α)

f (t)

¯¯dσ

dt

¯¯dt

=

Z b

a

f (t)

¯¯dσ

dt

¯¯dt.

Thus, Z β

α

f (σ (h (τ)))

¯¯dσ

dτ

¯¯dτ =

Z b

a

f (t)

¯¯dσ

dt

¯¯dt,

as claimed.

Similarly, if dh/dτ < 0,

Z β

α

f (σ (h (τ)))

¯¯dσ

dτ

¯¯dτ =

Z β

α

f (σ (h (τ)))

¯dh

dτ

¯ ¯¯dσ

dt

¯¯dτ

= −Z β

α

f (σ (h (τ)))dh

dτ

¯¯dσ

dt

¯¯dτ

=

Z α

β

f (σ (h (τ)))dh

dτ

¯¯dσ

dt

¯¯dτ

=

Z h(α)

h(β)

f (σ (t))

¯¯dσ

dt

¯¯dt =

Z b

a

f (σ (t))

¯¯dσ

dt

¯¯dt.

¥

Now we will prove Proposition 2 with regard to the effect of reparametrization on the line

integral of a vector field. Recall the statement:

Assume that the curve C is parametrized by σ : [a, b]→ R2. If σ : [α,β]→ R2 is an orientationpreserving parametrization of C, then

Z β

α


dτ =

Z b

a


If σ : [α,β]→ R2 is an orientation reversing parametrization of C, then

Z β

α


dτ = −Z b

a


Proof

a) By the chain rule,

dσ (h (τ))

dτ=dh

dτ

dσ

dt(h (τ)) .


Therefore, Z β

α


dτ =

Z β

α

F (σ (h (τ))) · dσ (h (τ))dτ

dτ

=

Z β

α

F (σ (h (τ))) ·µdh

dτ

dσ

dt(h (τ))

¶dτ

=

Z β

α

F (σ (h (τ))) · dσdt(h (τ))

dh

dτdτ

=

Z h(β)

h(α)


=

Z b

a

F (σ (t)) · dσdtdt,

by the substitution rule.

b) As in part a)Z β

α


dτ =

Z β

α

F (σ (h (τ))) · dσdt(h (τ))

dh

dτdτ

=

Z h(β)

h(α)


Since h (τ) is a decreasing function, we have h (α) = b and h (β) = a. Therefore,Z h(β)

h(α)


Z a

b

F (σ (t)) · dσdtdt = −

Z b

a

F (σ (t)) · dσdtdt,

as claimed. ¥

Problems

In problems 1-5 evaluate the given integral of a scalar function with respect to arc length:

1. ZC

y3ds,

where C is parametrized by

σ (t) =¡t3, t

¢, 0 ≤ t ≤ 2.

2. ZC

xy2ds,

where C is parametrized by

σ (t) = (2 cos (t) , 2 sin (t)) , −π2≤ t ≤ π

2.

3. ZC

xeyds,

where C is the line segment traversed from (2, 1) to (4, 5) .

4. ZC

y

x2 + y2ds,


where C is the semicircle of radius 2 that is centered at (4, 3) and traversed from (6, 3) to (2, 3)in the counterclockwise direction.

5. ZC

(x− 2) e(y−3)(z−4)ds,

where C is the line segment from (2, 3, 4) to (3, 5, 7).

In problems 6-10 evaluate the line integral ZC

F·dσ

6.

F (x, y) = x2i+ y2j

and C is the line segment from (1, 2) to (3, 4).

7.

F (x, y) = yi− xjand C is the part of the unit circle traversed from (0, 1) to (−1, 0) in the counterclockwisedirection.

8.

F (x, y) = ln (y) i− exjand C is parametrized by

σ (t) =¡ln (t) , t3

¢, 1 ≤ t ≤ e.

9.

F (x, y, z) = cos (y) i+ sin (z) j+ xk

and C is parametrized by

σ (t) = (cos (t) , t, t) ,π

4≤ t ≤ π

2

10.

F (x, y) = xyi+ (x− y) jand C = C1 + C2, where C1 is the line segment from (0, 0) to (2, 0) and C2 is the line segmentfrom (2, 0) to (3, 2).

In problems 11 and 12 evaluate ZC

F ·Tds

11.

F (x, y) = −2xyi+ (y + 1) jand C is the part of the circle of radius 2 centered at the origin traversed from (−2, 0) to (0, 2)in the clockwise direction.

12.

F (x, y, z) = ex+yi+ xzj+ yk

and C is the line segment from (1, 2, 3) to (−1,−2,−3) .



a) Express the line integral ZC

F·dσ

in the differential form notation,

b) Evaluate the expression that you obtained in part a).

13.

F (x, y) = −xyi+ 1

x2 + 1j

and C is parametrized by

σ (t) =¡t, t2

¢, −4 ≤ t ≤ −1

14.

F (x, y) = yi− xjand C is the part of the unit circle that is traversed from (0,−1) to (0, 1) in the counterclockwisedirection.

In problems 15-18 evaluate the given line integral:

15. ZC

3x2dx− 2y3dy

where C is the part of the unit circle traversed from (1, 0) to (0, 1).

16. ZC

exdx+ eydy,

where C is the part of the ellipse x2 + 4y2 = 4 traversed from (0, 1) to (2, 0) in the clockwisedirection.

Hint: Parametrize C by a function of the form (a cos (θ) , b sin (θ)).

17. ZC

− sin (x) dx+ cos (x) dy,

where C is the part of the parabola y = x2 traversed from (0, 0) to¡π,π2

¢.

18. ZC

x3dx+ y2dy + zdz,

where C is the line segment from the origin to (2, 3, 4).

14.3 Line Integrals of Conservative Vector Fields

In this section we will see that the line integral of the gradient of a scalar function on a curve

is simply the difference between the values of the function at the endpoints of the curve. Thus,

such a line integral does not depend on the particular path that connects two given points. We

will discuss conditions under which a vector field is the gradient of a scalar function.

14.3. LINE INTEGRALS OF CONSERVATIVE VECTOR FIELDS 211

The Fundamental Theorem for Line Integrals

Theorem 1 Assume that the initial point of a curve C in the plane is (x1, y1) and its terminalpoint is (x2, y2). Then Z

C

∇f (x, y) · dσ = f (x2, y2)− f (x1, y1) .

Proof

Assume that σ (t) = (x (t) , y (t)), where a ≤ t ≤ b, parametrizes C. ThenZC

∇f · dσ =Z b

a

∇f (σ (t)) · dσdtdt

=

Z b

a

µ∂f

∂x(x (t) , y (t)) i+

∂f

∂y(x (t) , y (t)) j

¶·µdx

dti+

dy

dtj

¶dt

=

Z b

a

µ∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt

¶dt.

By the chain rule,

∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt=d

dtf (x (t) , y (t)) .

By the Fundamental Theorem of Calculus,Z b

a

d

dtf (x (t) , y (t)) dt = f (x (b) , y (b))− f (x (a) , y (a)) = f (x2, y2)− f (x1, y1) .

Therefore, ZC

∇f · dσ = f (x2, y2)− f (x1, y1) ,as claimed. ¥

Remark 1 Theorem 1can be expressed in the differential form notation: We haveZC

∇f · dσ=Z b

a

µ∂f

∂x(x (t) , y (t))

dx

dt+

∂f

∂y(x (t) , y (t))

dy

dt

¶dt

=

ZC

∂f

∂xdx+

∂f

∂ydy,

so that ZC

∂f

∂xdx+

∂f

∂ydy = f (x2, y2)− f (x1, y1) .

Since

df =∂f

∂xdx+

∂f

∂ydy,

we can also write ZC

df = f (x2, y2)− f (x1, y1) .Thus the line integral of the differential of a scalar function f on C is the difference between the

values of f at the endpoints of C. A similar statement is valid for gradient fields in R3: Assumethat the initial point of a curve C in R3 is (x1, y1,z1) and its terminal point is (x2, y2, z2). ThenZ

C

∇f · dσ =ZC

df = f (x2, y2, z2)− f (x1, y1, z1) .

♦


Definition 1 We will say that the line integralZC

F · dσ

is independent of path in the region D if its value is the same for all curves in D that have

the same initial point and terminal point.

Thus, the line integral of a gradient field is independent of path.

Example 1 Let

r (x, y) = xi+ yj

be the position vector of the point (x, y), r (x, y) =px2 + y2 = ||r (x, y)|| and

ur (x, y) =r

r(x, y) =

xpx2 + y2

i+yp

x2 + y2j.

Thus, ur is the unit vector in the radial direction. Show thatZC

ur (x, y) · dσ

is independent of path in any region that does not contain the origin. Determine such a line

integral if C joins the point (x1, y1) to (x2, y2) and does not pass through the origin.

Solution

In Example 6 of Section 14.1 we showed that

∇r (x, y) = ur (x, y) .

By Theorem 1,ZC

ur (x, y) · dσ =ZC

∇r (x, y) · dσ = r (x2, y2)− r (x1, y1)

=qx22 + y

22 −

qx21 + y

21 .

¤

Example 2 Let

r = r (x, y, z) = xi+ yj+ zk and r = r (x, y, z) = ||r|| =px2 + y2 + z2,

and

F (x, y, z) = −GMr3

r

be the gravitational field due to the massM at the origin, as we discussed in Section 15.1. Show

that ZCr

F (x, y, z) · dσ

is independent of path in any region that does not contain the origin. Determine such a line

integral if C joins the point (x1, y1, z1) to (x2, y2, z1) and does not pass through the origin.


Solution

As we showed in Example 6 of Section 14.1,

∇µ1

r

¶= − r

r3.

Thus,

F (x, y, z) = −GMr3

r =∇µGM

r

¶.

Therefore, ZC

−GMr3

r · dσ =ZC

∇µGM

r

¶· dσ= GMp

x22 + y22 + z

22

− GMpx21 + y

21 + z

21 .

Thus, the work done by the gravitational field in moving a unit mass from one point to another

depends only on the distances of the points from the mass M , irrespective of the path that is

followed. ¤

Definition 2 A vector field F is said to be conservative in a region D if F is the gradient of

a scalar function f in D. Such a scalar function is referred to as a potential for F.

Thus, the field of Example 1 and the gravitational field that was discussed in Example 2 are

conservative fields in a region that does not contain the origin.

By Theorem 1, the line integral of a field F on a curve C depends only on the endpoints of C

if F is conservative in a region that contains C. In particular, if C is closed,ZC

F·dσ =ZC

∇f · dσ = 0.

Remark 2 (Conservation of Energy) In Physics and some fields of engineering, a function

φ is referred to as a potential for F if F = −∇φ. The adjective “conservative” has its origin inmechanics. Assume that F = −∇φ is a force field in R3 and σ (t) is the position of an object attime t. The work done by F on the object as it moves from the point P1 to P2 along the curve

C that is parametrized by σ (t) isZC

F · d σ =ZC

−∇φ · dσ = − (φ (P2)− φ (P1)) = φ (P1)− φ (P2) .

By Newton’s second law of motion, if the mass of the object is m, a (t) is its accelerationand v (t) is its velocity at time t, then

F = ma (t) = mdv

dt.

If P1 = σ (t1) and P2 = σ (t1),ZC

F · d σ =Z t2

t1

mdv

dt· dσdtdt = m

Z t2

t1

dv

dt· vdt

= m

Z t2

t1

1

2

d

dt(v·v) dt

= m

Z t2

t1

1

2

d

dt||v||2 dt

=1

2m ||v (t2)||2 − 1

2m ||v (t1)||2 .


Thus,

φ (P1)− φ (P2) =1

2m ||v (t2)||2 − 1

2m ||v (t1)||2 ,

so that

φ (P1) +1

2m ||v (t1)||2 = φ (P2) +

1

2m ||v (t2)||2 .

The quantity φ (P ) is called the potential energy at P and the quantity

1

2m ||v (t)||2

is the kinetic energy at time t. Their sum is the total energy. Therefore, the above

equality says that the total energy is conserved as the object moves from a point P1 to a point

P2 in a force field that has a potential. ♦

Conditions for a Field to be Conservative

Given a vector field F, we would like to be able to determine whether F is conservative by

examining the components of F. Let’s begin with a vector field

F (x, y) =M (x, y) i+N (x, y) j

in the plane. Assume that F (x, y) =∇f (x, y) for each (x, y) in the region D. Thus,

M (x, y) =∂f

∂x(x, y) and N (x, y) =

∂f

∂y(x, y)

for each (x, y) ∈ D. Therefore,

∂M

∂y=

∂2f

∂y∂xand

∂N

∂x=

∂2f

∂x∂y.

If we assume that the second partial derivatives of f are continuous in D, we have

∂2f

∂y∂x(x, y) =

∂2f

∂x∂y(x, y) ,

so that∂M

∂y(x, y) =

∂N

∂x(x, y)

for each (x, y) ∈ D.Let’s record this important fact:

Proposition 1 A necessary condition for F (x, y) = M (x, y) i + N (x, y) j to be conservativeand to have a potential function with continuous second-order partial derivatives in the region

Dis that∂M

∂y(x, y) =

∂N

∂x(x, y)

for each (x, y) ∈ D.

Remark 3 (Caution) The above condition is necessary but not sufficient for a vector field to

be conservative. Indeed, let


F(x, y) = − y

x2 + y2i+

x

x2 + y2j.

We have∂

∂y

µ− y

x2 + y2

¶= −

¡x2 + y2

¢− y (2y)(x2 + y2)

2 =−x2 + y2(x2 + y2)

2

and∂

∂x

µx

x2 + y2

¶=

¡x2 + y2

¢− x (2x)(x2 + y2)

2 =−x2 + y2(x2 + y2)

2 .

Thus,∂

∂y

µ− y

x2 + y2

¶=

∂

∂x

µx

x2 + y2

¶for each (x, y) 6= (0, 0). If F were conservative in D =

©(x, y) ∈ R2 : (x, y) 6= 0ª, the line integral

of F would have been 0 around any closed curve in D. Let C be the unit circle so that C can

be parametrized by σ (t) = (cos (t) , sin (t)), where 0 ≤ t ≤ 2π. We have

F (cos (t) , sin (t)) = − sin (t)

cos2 (t) + sin2 (t)i+

cos (t)

cos2 (t) + sin2 (t)j

= − sin (t) i+ cos (t) jand

dσ

dt= − sin (t) i+ cos (t) j.

Therefore, ZC

F · dσ =Z 2π

0

(− sin (t) i+ cos (t) j) · (− sin (t) i+ cos (t) j) dt

=

Z 2π

0


¢dt =

Z 2π

0

dt = 2π 6= 0.

Therefore F is not conservative. As we will see in the following sections, the problem is that the

vector field is not defined at (0, 0) so that the necessary conditions are not satisfied at (0, 0):The origin forms a ”hole" in the region D. ♦Even though the condition that is stated in Proposition 1is not sufficient for the existence of a

potential function, at least we can rule out the existence of such a function if the condition is

not satisfied. We can go ahead and try to find a potential function if the condition is satisfied.

Example 3 Let

F (x, y) = x2y2i+ 2xyj

Show that F is not conservative.

Solution

We have∂

∂y

¡x2y2

¢= 2x2y and

∂

∂x(2xy) = 2y.

Now,

2x2y = 2y ⇔ y¡x2 − 1¢ = 0⇔ y = 0 or x = ±1.

Thus, the necessary condition for the existence of a potential is satisfied only on the x-axis and

on the vertical lines x = ±1. This implies that the necessary condition is not satisfied in anyopen region D. Therefore F does not have a potential function in any open region. ¤


Example 4 Let

F (x, y) =x

x2 + y2i+

y

x2 + y2j

a) Determine a potential functions for F .

b) Evaluate the line integral of F on any curve that joins (1, 0) to (0,√e) and does not pass

through the origin.

Solution

a) We have

∂

∂y

µx

x2 + y2

¶=

−2xy(x2 + y2)

2 ,

and∂

∂x

µy

x2 + y2

¶=

−2xy(x2 + y2)

2 .

Therefore,∂

∂y

µx

x2 + y2

¶=

∂

∂x

µy

x2 + y2

¶if (x, y) 6= (0, 0). Thus the necessary condition for the existence of a potential function is satisfiedeverywhere except at the origin. We saw that in such a case a potential function need not exist

in a region that contains the origin(2). Let’s try to find a potential in a region that excludes

the origin. We have F (x, y) =∇f (x, y) if and only if∂f

∂x=

x

x2 + y2and

∂f

∂y=

y

x2 + y2

By the first condition

f (x, y) =

Z∂f

∂xdx =

Zx

x2 + y2dx.

If we set u = x2 + y2.∂u

∂x= 2x.

We keep y constant and set du = 2xdx. Since any function of y is treated as a constant, thereexists g (y) such thatZ

x

x2 + y2dx =

1

2

Z1

udu =

1

2ln (|u|) + g (y) = 1

2ln¡x2 + y2

¢+ g (y) .

Thus,

f (x, y) =1

2ln¡x2 + y2

¢+ g (y)

Therefore,∂f

∂y=

1

2 (x2 + y2)(2y) +

dg (y)

dy=

y

x2 + y2+dg (y)

dy.

Thus,y

x2 + y2+dg (y)

dy=

∂f

∂y=

y

x2 + y2

This implies thatdg (y)

dy= 0


so that g is a constant K. Therefore,

f (x, y) =1

2ln¡x2 + y2

¢+K.

b) By part a),the line integral of F on a curve that that joins (1, 0) to (0,√e) and does not pass

through the origin is

f¡0,√e¢− f (1, 0) = 1

2ln (e)− 1

2ln (1) =

1

2.

¤Now let’s consider conditions for the existence of a potential function for a vector field in three

dimensions. Let

F (x, y, z) =M (x, y, z) i+N (x, y, z) j+ P (x, y, z)k

and assume that F =∇f . Then

M =∂f

∂x, N =

∂f

∂yand P =

∂f

∂z.

Assuming that f has continuous second-order partial derivatives,

∂M

∂y=

∂2f

∂y∂x=

∂2f

∂x∂y=

∂N

∂x,

∂M

∂z=

∂2f

∂z∂x=

∂2f

∂x∂z=

∂P

∂x,

∂N

∂z=

∂2f

∂z∂y=

∂2f

∂y∂z=

∂P

∂y.

Let’s record these facts:

Proposition 2 The necessary conditions for

F (x, y, z) =M (x, y) i+N (x, y) j+ P (x, y, z)k

to be conservative and to have a potential function with continuous second-order partial deriv-

atives in the region D are the equalities

∂M

∂y=

∂N

∂x,∂M

∂z=

∂P

∂x,∂N

∂z=

∂P

∂y

for each (x, y, z) ∈ D.These equalities are equivalent to the condition that ∇×F = 0 since

∇×F =

¯¯ i j k

∂

∂x

∂

∂y

∂

∂zM N P

¯¯ = µ∂P

∂y− ∂N

∂z

¶i+

µ∂M

∂z− ∂P

∂x

¶j+

µ∂N

∂x− ∂M

∂y

¶k

In the next chapter we will examine the sufficiency of this condition for the existence of a

potential function.

Remark 4 If F (x, y) =M (x, y) i+N (x, y) j is a two-dimensional vector field, we define∇×Fas

∇× (M (x, y) i+N (x, y) j+ 0k) =

µ∂N

∂x− ∂M

∂y

¶k.

Therefore, the condition that is stated in Proposition 1 can be considered to be a special case

of Proposition 2. ♦


Example 5 Let

F (x, y, z) = −2xe−x2−y2 sin (z) i− 2ye−x2−y2 sin (z) j+ e−x2−y2 cos (z)k.

a) Show that F satisfies the necessary condition for the existence

b) Determine a potential function for F.

c) Determine the line integral of F on any curve C that connects (0, 0, 0) to (1, 1,π/2).

Solution

a)

∇×F (x, y, z) =

¯¯ i j k

∂

∂x

∂

∂y

∂

∂z

−2xe−x2−y2 sin (z) −2ye−x2−y2 sin (z) e−x2−y2 cos (z)

¯¯

=

µ∂

∂y

³e−x

2−y2 cos (z)´− ∂

∂z

³−2ye−x2−y2 sin (z)

´¶i

−µ

∂

∂x

³e−x

2−y2 cos (z)´− ∂

∂z

³−2xe−x2−y2 sin (z)

´¶j

+

µ∂

∂x

³−2ye−x2−y2 sin (z)

´− ∂

∂y

³−2xe−x2−y2 sin (z)

´¶k

=³−2ye−x2−y2 cos (z) + 2ye−x2−y2 cos (z)

´i

−³−2xe−x2−y2 cos (z) + 2xe−x2−y2 cos (z)

´j

+³4xye−x

2−y2 sin (z)− 4xye−x2−y2 sin (z)´k

= 0.

b) We have F =∇f if∂f

∂x(x, y, z) = −2xe−x2−y2 sin (z) ,

∂f

∂y(x, y, z) = −2ye−x2−y2 sin (z) ,

∂f

∂z(x, y, z) = e−x

2−y2 cos (z) .

By the first equation,

f (x, y, z) =

Z−2xe−x2−y2 sin (z) dx

= e−y2

sin (z)

Z−2xe−x2dx

= e−y2

sin (z) e−x2

+ g (y, z) = e−x2−y2 sin (z) + g (y, z) ,

since any function of x and y has to be treated as a constant when we integrate with respect to

x.

Therefore,

−2ye−x2−y2 sin (z) = ∂f

∂y(x, y, z) = −2ye−x2−y2 sin (z) + ∂g (y, z)

∂y,


so that∂g (y, z)

∂y= 0⇒ g (y, z) = h (z) ,

for some function h of z. Thus

f (x, y, z) = e−x2−y2 sin (z) + g (y, z) = e−x

2−y2 sin (z) + h (z) .

Therefore,

e−x2−y2 cos (z) =

∂f

∂z= e−x

2−y2 cos (z) +dh

dz.

Thus,dh

dz= 0⇒ h (z) = K,

where K is a constant. Therefore,

f (x, y, z) = e−x2−y2 sin (z) +K.

c) ZC

F · dσ = f (1, 1,π)− f (0, 0, 0) = e−2 sin (π/2)− sin (0) = e−2.

¤

Problems

In problems 1-4

a) Check whether the conditions that are necessary for the vector field F to be conservative are

satisfied,

b) If your response to a) is in the affirmative, find a a potential f for F.

1.

F (x, y) = (2x− 3y) i+ (−3x+ 4y − 8) j2.

F (x, y) = ex cos (y) i+ ex sin (y) j

3.

F (x, y) = ex sin (y) i+ ex cos (y) j

4.

F (x, y, z) = − cos (z) e−x−yi− cos (z) e−x−yj− sin (z) e−x−yk

In problems 5-7,

a) Find a potential f for F,

b) Make use of f to evaluate the line integralZC

F · dσ

5.

F (x, y) = xy2i+ x2yj

and C is an arbitrary curve from (0, 1) to (2, 1).

6.

F (x, y) = (yexy − 1) i+ xexyj


and C is an arbitrary curve from (0, 1) to (4, ln (2)).

7.

F (x, y, z) = yzi+ xzj+ (xy + 2z)k

and C is an arbitrary curve from (1, 0,−2) to (4, 6, 3).

In problems 8 and 9 evaluate the line integralZC

M (x, y) dx+N (x, y) dy

by finding a function f (x, y) such that

df =M (x, y) dx+N (x, y) dy.

8. ZC

y

x2 + y2dx− x

x2 + y2dy

where C is a curve that is in the upper half-plane (y > 0) and joins (0, 1) to (1, 1).

9. ZC

y2

1 + x2dx+ 2y arctan (x) dy

where C is an arbitrary curve from (0, 0) to (1, 2) .

10. Evaluate ZC

y2 cos (z) dx+ 2xy cos (z) dy − xy2 sin (z) dz,

where C is an arbitrary curve from (0, 0, 0) to (1, 1,π) by finding f (x, y, z) such that

df = y2 cos (z) dx+ 2xy cos (z) dy − xy2 sin (z) dz

14.4 Parametrized Surfaces and Tangent Planes

In this section we will introduce parametrizes surfaces. Graphs of functions of two variables are

special cases. A surface such as a sphere cannot be expressed as the graph of a single function of

two variables. In such a case the parametric representation of the surface is very useful. We will

be able to calculate normal vectors to such surfaces. That will turn out to be very important

in the calculation of surface integrals in the next section.

Parametrized Surfaces

Assume that Φ (u, v) = (x (u, v) , y (u, v) , z (u, v)) for each (u, v) in a region D ⊂ R2. Thus Φis a function of two variables and has values in the three-dimensional space R3. We will use thenotation Φ : D ⊂ R2 → R3 to refer to such a function. The symbols u and v are referred to asparameters, and D is the parameter domain. We will identify the position vector

Φ (u, v) = x (u, v) i+ y (u, v) j+ z (u, v)k

with the point Φ (u, v), so that we may refer to the coordinate functions as the componentfunctions of Φ. The surface M that is parametrized by the function Φ is the image of

Φ, i.e., S is the set of all points (x, y, z) in R3 such that x = x (u, v), y = y (u, v) and z = z (u, v)where (u, v) varies in D.

14.4. PARAMETRIZED SURFACES AND TANGENT PLANES 221

Example 1 Let

Φ (u, v) = (cosh(u) cos(v), 2 cosh (u) sin (v) , 4 sinh (u)) ,

where 1 ≤ u ≤ 1 and 0 ≤ v ≤ 2π.Thus Φ : D → R3, where the parameter domain D is the rectangle [−1, 1] × [0, 2π] in theuv-plane. The parameters are u and v. The coordinate functions of Φ are

x (u, v) = cosh(u) cos(v), y (u, v) = 2 cosh (u) sin (v) , and z (u, v) = 4 sinh (u) .

The surface M that is parametrized by Φ is shown in Figure 1.

-10

0z

10

5

4

y

0 2

x0

-5 -2

-4

Figure 1

The picture gives the impression that M is a hyperboloid of one sheet. Indeed,

16x2 (u, v) + 8y2 (u, v)− z2 (u, v)= 16 cosh2(u) cos2(v) + 16 cosh2 (u) sin2 (v)− 16 sinh2 (u)= 16 cosh2(u)

¡cos2 (v) + sin2 (v)

¢− 16 sinh2 (u)= 16 cosh2(u)− 16 sin2 (u)= 16

¡cosh2 (u)− sinh2 (u)¢ = 16.

Thus, the points(x, y, z) on the surface M satisfy the equation

16x2 + 8y2 − z2 = 16.The intersection of M with a horizontal plane is an ellipse (or empty) The intersection of M

with planes of the form x = constant or y = constant are hyperbolas (or empty). ¤The graph of a scalar function of two variables can be considered to be a parametrized surface:

If f (x, y) is defined for each (x, y) ∈ D ⊂ R2, we setΦ (x, y) = (x, y, f (x, y)) .

Thus, x and y are the parameters. The parameter domain coincides with the domain of f .

Example 2 Let f (x, y) = x2 − y2.The graph of f can be considered to be the surface M that is parametrized by the function

Φ : R2 → R3, whereΦ (x, y) =

¡x, y, x2 − y2¢ .

The associated vector-valued function is

Φ (x, y) = xi+ yj+¡x2 − y2¢k.


Figure 2 shows S. ¤

5

-10

0z

10

5

y

0

x

0

-5 -5

Figure 2

Assume that a surface S is the graph of the equation ρ = f (φ, θ), where ρ, φ and θ are sphericalcoordinates. Thus,

x (φ, θ) = ρ sin (φ) cos (θ) = f (φ, θ) sin (φ) cos (θ) ,

y (φ, θ) = ρ sin (φ) sin (θ) = f (φ, θ) sin (φ) sin (θ) ,

z (φ, θ) = ρ cos (φ) = f (φ, θ) cos (φ) ,

where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π. Therefore, the surface M is parametrized by the function

Ω (φ, θ) = (f (φ, θ) sin (φ) cos (θ) , f (φ, θ) sin (φ) sin (θ) , f (φ, θ) cos (φ)) .

Example 3 Let M be the sphere of radius 2 centered at the origin. In spherical coordinates,

M

The sphere can be parametrized by

Ω (φ, θ) = (2 sin (φ) cos (θ) , 2 sin (φ) sin (θ) , 2 cos (φ)) ,

where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π. Figure 3 shows S. ¤

-22

0z

2

2

y

0

x

0

-2 -2

Figure 3

Example 4 Let M be the cone φ = φ0. Determine a parametrization of S.

Solution

We set

Ω (ρ, θ) = (ρ sin (φ0) cos (θ) , ρ sin (φ0) sin (θ) , ρ cos (φ0)) ,

where ρ ≥ 0 and 0 ≤ θ ≤ 2π. Figure 4 shows the cone φ = π/6. ¤


0

1

2

2

z

3

1

y

0

x0

-2 -1

Figure 4

Assume that r, θ and z are cylindrical coordinates, so that

r = x2 + y2, cos (θ) =x

r, sin (θ) =

y

r.

If the surface S is the graph of z = g (r, θ), we can view S as the surface that is parametrizedby

Φ (r, θ) = (r cos (θ) , r sin (θ) , g (r, θ)) .

Example 5 Let M be the surface that is described by the equation z = r2 in cylindrical coor-dinates. Determine a parametrization of M .

Solution

The surface S can be parametrized by the function

Φ (r, θ) =¡r cos (θ) , r sin (θ) , r2

¢,

where r ≥ 0 and 0 ≤ θ ≤ 2π. Note that M is a paraboloid. Figure 5 shows the paraboloid. ¤

02

2z

4

2

y

0

x0

-2 -2

Figure 5

Example 6 Determine a parametrization of the cylinder: r = 2.

Solution

The cylinder can be parametrized by

Φ (θ, z) = (2 cos (θ) , 2 sin (θ) , z) ,

where 0 ≤ θ ≤ 2π and −∞ < z < +∞. Figure 6 shows the cylinder. ¤


-52

0z

5

2

y

0

x0

-2 -2

Figure 6

Surfaces of Revolution can be expressed parametrically. Assume that we form a surface

S by rotating the graph of z = f (x) on the interval [a, b] about the x-axis. If the point

(x, 0, f (x)) is rotated by angle θ the resulting point is (x, f (x) sin (θ) , f (x) cos (θ)). Thus, Scan be parametrized by the function

Ω (x, θ) = (x, f (x) sin (θ) , f (x) cos (θ)) ,

where a ≤ x ≤ b and 0 ≤ θ ≤ 2π.

Example 7 LetM be the surface generated by revolving the graph of z = 1+x2 on the interval[1, 2] about the x-axis. Determine a parametrization of M .

Solution

We can set

Ω (x, θ) =¡x,¡1 + x2

¢sin (θ) ,

¡1 + x2

¢cos (θ)

¢,

where 1 ≤ x ≤ 2 and 0 ≤ θ ≤ 2π. Then M is parametrized by the function Ω. Figure 7 showsS. ¤

-4

-2

0

2

4

z

42

y

0-2

2.0-4

x

1.81.61.41.21.0

Figure 7

Assume that we rotate z = f (x) about the z-axis. The resulting surface of revolution can beparametrized by the function

(x cos (θ) , x sin (θ) , f (x)) , 0 ≤ θ ≤ 2π.For a given x, we have p

y2 + x2 = x.

This is the radius of the circular cross section. The point (x cos (θ) , x sin (θ) , f (x)) is at anangle θ from the x-direction.

Example 8 LetM be the surface generated by revolving the graph of z = 1+x2 on the interval[1, 2] about the z-axis. Determine a parametrization of M.


Solution

We can set

Ω (x, θ) =¡x cos (θ) , x sin (θ) , 1 + x2

¢,

where 1 ≤ x ≤ 2 and 0 ≤ θ ≤ 2π. Then S is parametrized by the function Ω. Figure 8 showsM . ¤

22

3

4

z

5

12

y

0 1

x0-1

-1-2 -2

Figure 8

Normal Vectors, Tangent Planes and Orientation

Assume that the surface M is parametrized by the function Φ : D ⊂ R2 → R3, where

Φ (u, v) = (x (u, v) , y (u, v) , z (u, v))

for each (u, v) ∈ D. Given a point (u0, v0) ∈ D, the restriction of Φ to the line v = v0, i.e., thefunction

u→ Φ (u, v0)parametrizes a curve on the surfaceM . We will refer to this curve as the u-coordinate curve

on M that passes through Φ (u0, v0). Similarly, the v-coordinate curve on M that passes

through Φ (u0, v0) is parametrized by the restriction of Φ to the line u = u0, i.e., the function

v → Φ (u0, v) .

Figure 9: Coordinate Curves on a Surface

We will define the partial derivatives of Φ as the vector-valued functions

∂uΦ (u, v) =∂Φ

∂u(u, v) =

∂x

∂u(u, v) i+

∂y

∂u(u, v) j+

∂z

∂u(u, v)k,


and

∂vΦ (u, v) =∂Φ

∂v(u, v) =

∂x

∂v(u, v) i+

∂y

∂v(u, v) j+

∂z

∂v(u, v)k.

Thus, the vectors ∂uΦ (u0, v0) and ∂vΦ (u0, v0) are tangent to the u-coordinate curve and thev-coordinate curve, respectively, that pass through Φ (u0, v0).

Definition 1 Assume that ∂uΦ (u0, v0) and ∂vΦ (u0, v0) exist and N (u0, v0) = ∂uΦ (u0, v0)×∂vΦ (u0, v0) 6= 0. In this case we will say that the surface M that is parametrized by Φ is

smooth at Φ (u0, v0). The vector N (u0, v0) is the normal to the surface M at Φ(u0, v0).The tangent plane to S at P0 = Φ(u0, v0) is the set of points P such that

N (u0, v0) ·−−→P0P = 0.

Figure 10: Tangent and normal vectors

Example 9 The sphere M of radius ρ0 can be parametrized by the function

Φ (φ, θ) = (ρ0 sin (φ) cos (θ) , ρ0 sin (φ) sin (θ) , ρ0 cos (φ)) ,

where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π (spherical coordinates).a) Identify the coordinate curves on M

b) Determine the Norrmal to M at Φ (φ, θ) .c) Set the radius to be 1 and determine the plane that is tangent to M at Φ (π/4,π/6)

Solution

a) If φ = φ0, the corresponding θ-coordinate curve on the sphere is parametrized by the function

θ → Φ (φ0, θ) = (ρ0 sin (φ0) cos (θ) , ρ0 sin (φ0) sin (θ) , ρ0 cos (φ0)) ,where 0 ≤ θ ≤ 2π. Thus, z has the constant value ρ0 cos (φ0) and we would expect that thiscurve is a circle on the plane z = ρ0 cos (φ0) that is centered at (0, 0, ρ0 cos (φ0)). Indeed,thedistance of Φ (φ0, θ) from (0, 0, ρ0 cos (φ0)) isq

(ρ0 sin (φ0) cos (θ))2+ (ρ0 sin (φ0) sin (θ)) =

qρ20 sin

2 (φ0) cos2 (θ) + ρ20 sin

2 (φ0) sin2 (θ)

= ρ0 sin (φ0)

qcos2 (θ) + sin2 (θ) = ρ0 sin (φ0) .

Therefore, a φ-coordinate curve that passes through Φ (φ0, θ0) is a circle of radius ρ0 sin (φ0).For example, if φ0 = π/2, the circle is “the equator" on the xy-plane, centered at the origin andhaving radius ρ0. If φ0 = 0, the circle degenerates to the “north pole" (0, 0, ρ0).


As for the φ-coordinate curves, if θ = θ0, the corresponding coordinate curve on the sphere is

parametrized by the function

φ→ Φ (φ, θ0) = (ρ0 sin (φ) cos (θ0) , ρ0 sin (φ) sin (θ0) , ρ0 cos (φ)) ,

where 0 ≤ φ ≤ π. The curve is a semicircle of radius ρ0 in the plane θ = θ0 that is centered at

the origin ("a meridian").

z

y x

Figure 11: Coordinate curves on a sphere

b) We have

∂φΦ (φ, θ) = ∂φ (ρ0 sin (φ) cos (θ)) i+ ∂φ (ρ0 sin (φ) sin (θ)) j+ ∂φ (ρ0 cos (φ))k

= ρ0 cos (φ) cos (θ) i+ ρ0 cos (φ) sin (θ) j− ρ0 sin (φ)k,

and

∂θΦ (φ, θ) = ∂θ (ρ0 sin (φ) cos (θ)) i+ ∂θ (ρ0 sin (φ) sin (θ)) j+ ∂θ (ρ0 cos (φ))k

= −ρ0 sin (φ) sin (θ) i+ ρ0 sin (φ) cos (θ) j.

Therefore,

N (φ, θ) = ∂φΦ (φ, θ)× ∂θΦ (φ, θ)

=

¯¯ i j k

ρ0 cos (φ) cos (θ) ρ0 cos (φ) sin (θ) −ρ0 sin (φ)−ρ0 sin (φ) sin (θ) ρ0 sin (φ) cos (θ) 0

¯¯

=

¯ρ0 cos (φ) sin (θ) −ρ0 sin (φ)ρ0 sin (φ) cos (θ) 0

¯i−

¯ρ0 cos (φ) cos (θ) −ρ0 sin (φ)−ρ0 sin (φ) sin (θ) 0

¯j

+

¯ρ0 cos (φ) cos (θ) ρ0 cos (φ) sin (θ)−ρ0 sin (φ) sin (θ) ρ0 sin (φ) cos (θ)

¯k

= ρ20 sin2 (φ) cos (θ) i+ρ20 sin

2 (φ) sin (θ) j

+¡ρ20 sin (φ) cos (φ) cos

2 (θ) + ρ20 sin (φ) cos (θ) sin2 (θ)

¢k

= ρ20 sin2 (φ) cos (θ) i+ ρ20 sin

2 (φ) sin (θ) j+ ρ20 sin (φ) cos (φ)k

= ρ0 sin (φ) (ρ0 sin (φ) cos (θ) i+ ρ0 sin (φ) sin (θ) j+ ρ0 cos (φ)k)

= ρ0 sin (φ)Φ (φ, θ) .

Thus, the normal N (φ, θ) is in the radial direction, as we would have expected.


c) In particular, if ρ0 = 1,

N (π/4,π/6) = sin2 (π/4) cos (π/6) i+ sin2 (π/4) sin (π/6) j+ sin (π/4) cos (π/4)k

=

µ1

2

¶Ã√3

2

!i+

µ1

2

¶µ1

2

¶j+

Ã√2

2

!Ã√2

2

!k

=

√3

4i+1

4j+

1

2k,

and

Φ (π/4,π/6) = (sin (π/4) cos (π/6) , sin (π/4) sin (π/6) , cos (π/4))

=

ÃÃ√2

2

!Ã√3

2

!,

Ã√2

2

!µ1

2

¶,

√2

2

!

=

Ã√6

4,

√2

4,

√2

2

!.

Therefore, the tangent plane to the sphere at Φ (π/4,π/6) is the graph of the equation

√3

4

Ãx−√6

4

!+1

4

Ãy −√2

4

!j +

1

2

Ãz −√2

2

!= 0.

¤

With reference to the above example, the normal N (φ, θ) points towards the exterior of thesphere. if we interchange the order the parameters φ and θ we have N (φ, θ) = −N (φ, θ)(confirm), so that N (φ, θ) points towards the interior of the sphere. The sphere is an exampleof an "orientable surface". Intuitively, if a surface M is orientable we can tell the difference

between "inside" and "outside" on M . More precisely, a surface is orientable provided that

N (u, v) is continuous on D if (u, v) varies in "the parameter domain" D. The normal vectormay point outward, and we may label that case as "the positive orientation" of the surface.

Then we label the case when the normal points inward as "the negative orientation" of the

surface. Thus, in the case of the sphere of the above example, N (φ, θ) = −N (φ, θ) specifiesthe negative orientation of the surface. In the general case, Assume that M is parametrized

initially by Φ : D → R3. If N (u, v) is the normal determined by the parametrization Φ and

determines "the positive orientation" of M , we say that Ω : D∗ → R2 is a positivelyoriented parametrization of M if there is a one-one correspondence between the parameter

regions D and D∗ such that normal N∗ (u∗, v∗) determined by Ω points in the same directionas the corresponding N (u, v). If Ω : D∗ → R2 is a parametrization of M so that N∗ (u∗, v∗)determined by Ω points in the direction of −N (u, v), we say that Ω is a parametrization of Mwith negative orientation, The precise definition of the orientation of surface is given at the

end of this section. The informal intuitive understanding of orientability and the positive or

negative orientation of a surface will be adequate for our purposes. All of the previous examples

are orientable surfaces. In any case, let’s note that there are nonorientable surfaces:

Example 10 A Mobius band is not orientable. This is a surface that you can construct by

identifying two opposite sides of a rectangle so that A matches A0 and B matches B0 is in thepicture:


A

A'B

B'

Figure 12

The result is a surface that looks like this:

Figure 13

The entire surface does not have two distinct sides that can be identified as "inside" and "outside.

If you imagine that a bug that does not fall off the surface starts crawling on one side of a small

patch the surface, it will eventually end up on the other side of the same patch. The normal

vector N at the beginning is twisted on the way and comes back as −N. ¤It is worth noting the special case of a surface that is the graph of a function of two

variables. Assume that the surface M is the graph of z = f (x, y), where (x, y) ∈ D ⊂ R2.Then S can be parametrized by the function Φ : D→ R3, where

Φ (x, y) = (x, y, f (x, y)) .

In this case, the x-coordinate curve that passes through the point Φ (x0, y0) = (x0, y0, f (x0, y0))can be parametrized by the function

x→ Φ (x, y0) = (x, y0, f (x, y0)) ,

just as we defined in Section 12.5. Similarly, the y-coordinate curve that passes through the

point Φ (x0, y0) can be parametrized by the function

y → Φ (x0, y) = (x0, y, f (x0, y)) ,

just as we defined in Section 12.5.

We have

∂xΦ (x0, y0) = (1, 0, ∂xf (x0, y0))

and

∂yΦ (x0, y0) = (0, 1, ∂yf (x0, y0)) .


Therefore,

N (x0, y0) =

µi+

∂f

∂x(x0, y0)k

¶×µj+

∂f

∂y(x0, y0)k

¶

=

¯¯ i j k

1 0 fx (x0, y0)0 1 fy (x0, y0)

¯¯

= −∂xf (x0, y0) i− ∂yf (x0, y0) j+ k,

as in Section 12.5 (Remark 1 of Section 12.5). Thus, we can express the tangent plane as the

graph of the equation

N(x0, y0) · ((x− x0) i+ (y − y0) j+(z − z0)k) = 0,

i.e.,

−∂xf (x0, y0) (x− x0)− ∂yf (x0, y0) (y − y0) + (z − f (x0, y0)) = 0.Thus,

z = f (x0, y0) + fx (x0, y0) (x− x0) + fy (x0, y0) (y − y0) ,as before. ♦

Example 11 Let f (x, y) = x2 − y2 and let M be the graph of f .

a) Identify the coordinate curves on M that pass through (2, 1, 1), and determine vectors thatare tangent to these curves at (2, 1, 1) .b) Determine a normal vector to M at (2, 1, 1) and the tangent plane at that point.

Solution

Figure 14 shows the graph of f.

-20

5

0z

20

5

y

0

x0

-5 -5

Figure 14

a) Let

Φ (x, y) = (x, y, f (x, y)) =¡x, y, x2 − y2¢ .

The x-coordinate curve that passes through (2, 1, 1) is parametrized by the function

x→ Φ ¡x, 1, x2 − 1¢ = ¡x, 1, x2 − 1¢ .Thus, the curve is in the plane y = 1 and its projection on the xz-plane is the parabola z = x2−1.The y-coordinate curve that passes through (2, 1, 1) is parametrized by the function

y → Φ ¡2, y, 4− y2¢ = ¡2, y, 4− y2¢ .Thus, the curve is in the plane x = 2 and its projection on the yz-plane is the parabola z = 4−y2.


b) We have

∂xΦ (x, y) = (1, 0, 2x) and ∂yΦ (x, y) = (0, 1,−2y) .Therefore,

∂xΦ (2, 1) = (1, 0, 4) and ∂yΦ (2, 1) = (0, 1,−2.)These vectors are tangent to the x- and y-coordinate curves, respectively, at (2, 1, 1). We have

N (x, y) = ∂xΦ (2, 1)× ∂yΦ (2, 1) =

¯¯ i j k

1 0 40 1 −2

¯¯

=

¯0 41 −2

¯i−

¯1 40 −2

¯j+

¯1 00 1

¯k

= −4i+ 2j+ k.Therefore, the tangent plane to the graph of f at (2, 1, 1) is the graph of the equation

−4 (x− 2) + 2 (y − 1) + (z − 1) = 0¤

Example 12 Assume that the surface M is parametrized by

Φ (θ, z) = (2 cos (θ) , 2 sin (θ) , z) ,

where 0 ≤ θ ≤ 2π and −∞ < z < +∞ (cylindrical coordinates).

a) Identify the coordinate curves on M .

b) Determine the tangent plane to M at Φ (π/4, 1) .

Solution

Figure 15 shows the surface M . Note that M is a cylinder whose axis is along the z-axis, since

a point (x, y, z) on M satisfies the equation

x2 + y2 = (2 cos (θ))2+ (2 sin (θ))

2= 4.

-52

0z

5

2

y

0

x0

-2 -2

Figure 15

a) If θ = θ0, then

z → Φ (θ0, z) = (2 cos (θ0) , 2 sin (θ0) , z)parametrizes a line that is parallel to the z-axis and passes through the point (x0, y0, 0), wherex0 = 2 cos (θ0) and y0 = 2 sin (θ0).If z = z0,

θ → Φ (θ, z0) = (2 cos (θ) , 2 sin (θ) , z0)parametrizes a circle of radius 2 that is in the plane z = z0 and centered at (0, 0, z0) .


b) We have

∂θΦ (θ, z) = −2 sin (θ) i+ 2 cos (θ) j, and ∂zΦ (θ, z) = k.

Therefore,

N (θ, z) = ∂θΦ (θ, z)× ∂zΦ (θ, z) =

¯¯ i j k

−2 sin (θ) 2 cos (θ) 00 0 1

¯¯

= 2 cos (θ) i+ 2 sin (θ) j

We have

Φ (π/4, 1) =³2 cos

³π4

´, 2 sin

³π4

´, 1´=³√2,√2, 1´,

and

N (π/4, 1) =√2i+

√2j.

Therefore, the equation of the plane that is tangent to M at¡√2,√2, 1¢is

√2³x−√2´+√2³y −√2´= 0.

¤

Example 13 Assume that a > b > 0. Let

Φ (θ,φ) = ((a+ b cos (φ)) cos (θ) , (a+ b cos (φ)) sin (θ) , b sin (φ)) ,

where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ 2π. The surface M that is parametrized by Φ is a torus. S is

generated by revolving a disk of radius b that is perpendicular to the xy-plane, about the z-axis,

with its center at a distance a from the z-axis. Figure 16 shows M when a = 4 and b = 1.

-1

0

1

z

yx

Figure 16

If φ = φ0, the corresponding θ-coordinate curve is parametrized by the function

θ → Φ (θ,φ0) = ((a+ b cos (φ0)) cos (θ) , (a+ b cos (φ0)) sin (θ) , b sin (φ0)) ,

where 0 ≤ θ ≤ 2π. We notice that the z-coordinate has the constant value b sin (φ0), and thatthe curve is a circle of radius (a+ b cos (φ0)) centered at (0, 0, b sin (φ0)). For example, if φ0 = 0,the curve is the circle of radius a + b in the xy-plane that is centered at the origin. If φ0 = π,

the curve is the circle of radius a− b in the xy-plane that is centered at the origin.If θ = θ0, the corresponding φ-coordinate curve is parametrized by the function

φ→ Φ (θ0,φ) = ((a+ b cos (φ)) cos (θ0) , (a+ b cos (φ)) sin (θ0) , b sin (φ)) ,


where 0 ≤ φ ≤ 2π. This is the circle of radius b centered at (a cos (θ0) , a sin (θ0) , 0). in theplane θ = θ0. Indeed,

Φ (θ0,φ)− (a cos (θ0) , a sin (θ0) , 0)= ((a+ b cos (φ)) cos (θ0) , (a+ b cos (φ)) sin (θ0) , b sin (φ))− (a cos (θ0) , a sin (θ0) , 0)= (b cos (φ)) cos (θ0) , b cos (φ) sin (θ0), b sin (φ)) ,

so that

||Φ (θ0,φ)− (a cos (θ0) , a sin (θ0) , 0)||2= b2 cos2 (φ) cos2 (θ0) + b

2 cos2 (φ) sin2 θ0) + b2 sin2 (φ)

= b2 cos2 (φ) + b2 sin2 (φ) = b2.

¤

Example 14 Assume that the torus M is parametrized by

Φ (θ,φ) = ((2 + cos (φ)) cos (θ) , (2 + cos (φ)) sin (θ) , sin (φ)) ,

where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ 2π.a) Identify the coordinate curves that pass through Φ (0,π/3).b) Determine the tangent plane to M at Φ (0,π/3).

Solution

a) Note that

Φ (0,π/3) =

Ã5

2, 0,

√3

2

!.

We have

Φ (θ,π/3) = ((2 + cos (π/3)) cos (θ) , (2 + cos (π/3)) sin (θ) , sin (π/3))

=

Ã5

2cos (θ) ,

5

2sin (θ) ,

√3

2

!.

This is a circle of radius 5/2 in the plane z =√3/2 that is centered at

¡0, 0,√3/2¢.

We also have

Φ (0,φ) = (2 + cos (φ) , 0, sin (φ)) .

This coordinate curve is the circle of radius 1 in the xz-plane that is centered at (2, 0, 0) Indeed,

||Φ(0,φ)− (2, 0, 0)||2 = cos2 (φ) + sin2 (φ) = 1.b)

∂Φ

∂θ(θ,φ) =

∂

∂θ((2 + cos (φ)) cos (θ)) i+

∂

∂θ((2 + cos (φ)) sin (θ)) j+

∂

∂θsin (φ)k

= − (2 + cos (φ)) sin (θ) i+ (2 + cos (φ)) cos (θ) j,

∂Φ

∂φ(θ,φ) =

∂

∂φ((2 + cos (φ)) cos (θ)) i+

∂

∂φ((2 + cos (φ)) sin (θ)) j+

∂

∂φsin (φ)k

= − sin (φ) cos (θ) i− sin (φ) sin (θ) j+ cos (φ)k.


Therefore,

∂Φ

∂θ

³0,π

3

´= − (2 + cos (φ)) sin (θ) i+ (2 + cos (φ)) cos (θ) j|θ=0,φ=π/3

= −µ2 +

1

2

¶sin (0) i+

µ2 +

1

2

¶cos (0) j =

5

2j,

and

∂Φ

∂φ

³0,π

3

´= − sin (φ) cos (θ) i− sin (φ) sin (θ) j+ cos (φ)k|θ=0,φ=π/3

= −√3

2i+

1

2k.

Thus,

N³0,π

3

´=

∂Φ

∂θ

³0,π

3

´× ∂Φ

∂φ

³0,π

3

´=5

2j×

Ã−√3

2i+1

2k

!=5√3

4k+

5

4i.

Note that ¯¯N³0,π

3

´¯¯=

r75

16+25

16=10

4,

so that a unit vector in the direction of N (0,π/3) is

n =1

10

³5i+ 5

√3k´=1

2i+

√3

2k.

We have

Φ³0,π

3

´= Φ (0,π/3) =

Ã5

2, 0,

√3

2

!.

The required tangent plane consists of points P = (x, y, z) such thatµx− 5

2

¶µ1

2

¶+

Ãz −√3

2

!Ã√3

2

!= 0.

¤

The Orientation of a Surface and an Expression for the Normal Vector

(optional)

We have

N (u, v) =∂Φ

∂u(u, v)× ∂Φ

∂v(u, v)

=

¯¯ i j k

xu yu zuxv yv zv

¯¯

=

¯yu zuyv zv

¯i−

¯xu zuxv zv

¯j+

¯xu yuxv yv

¯k

= (yuzv − yvzu) i+ (xvzu − xuzv) j+ (xuyv − xvyu)k.


We introduce the notation

∂ (y, z)

∂ (u, v)=

¯yu zuyv zv

¯,∂ (z, x)

∂ (u, v)=

¯zu xuzv xv

¯,∂ (x, y)

∂ (u, v)=

¯xu yuxv yv

¯(read, Jacobian of y and z with respect to u and v, etc.). Thus, we can express N (u, v)as

N (u, v) =∂ (y, z)

∂ (u, v)i+

∂ (z, x)

∂ (u, v)j+

∂ (x, y)

∂ (u, v)k.

Example 15 Let ρ, φ and θ be spherical coordinates. For a given φ0, the cone φ = φ0 can be

parametrized by the function

Φ (ρ, θ) = (ρ sin (φ0) cos (θ) , ρ sin (φ0) sin (θ) , ρ cos (φ0)) ,

where ρ ≥ 0 and 0 ≤ θ ≤ 2π, as in Example 11.a) Use the above expression to determine N (ρ, θ).b) Determine the tangent plane to the cone at Φ (2,π/2).

0

1

2

2

z

3

1

y

0

x

0

-2 -1

Figure 17

Solution

a) We have

∂ (y, z)

∂ (ρ, θ)=

¯∂ρy (ρ, θ) ∂ρz (ρ.θ)∂θy (ρ.θ) ∂θz (ρ, θ)

¯=

¯∂ρ (ρ sin (φ0) sin (θ)) ∂ρ (ρ cos (φ0))∂θ (ρ sin (φ0) sin (θ)) ∂θ (ρ cos (φ0))

¯=

¯sin (φ0) sin (θ) cos (φ0)ρ sin (φ0) cos (θ) 0

¯= −ρ sin (φ0) cos (φ0) cos (θ) ,

∂ (z, x)

∂ (ρ, θ)=

¯∂ρ (ρ cos (φ0)) ∂ρ (ρ sin (φ0) cos (θ))∂θ (ρ cos (φ0)) ∂θ (ρ sin (φ0) cos (θ))

¯=

¯cos (φ0) sin (φ0 sin (φ))0 −ρ sin (φ0) sin (θ)

¯= −ρ sin (φ0) cos (φ0) sin (θ) ,


∂ (x, y)

∂ (ρ, θ)=

¯∂ρ (ρ sin (φ0) cos (θ)) ∂ρ (ρ sin (φ0) sin (θ))∂θ (ρ sin (φ0 cos (θ))) ∂θ (ρ sin (φ0) sin (θ))

¯=

¯sin (φ0) cos (θ) sin (φ0) sin (θ)−ρ sin (φ0) sin (θ) ρ sin (φ0) cos (θ)

¯= ρ sin2 (φ0) cos

2 (θ) + ρ sin2 (φ0) sin2 (θ)

= ρ sin2 (φ0) .

Therefore,

N (ρ, θ) =∂ (y, z)

∂ (ρ, θ)i+

∂ (z, x)

∂ (ρ, θ)j+

∂ (x, y)

∂ (ρ.θ)k

= −ρ sin (φ0) cos (φ0) cos (θ) i− ρ sin(φ0) cos (φ0) sin (θ) j+ ρ sin2 (φ0)k.

¤Let T : D∗ ⊂ R2 → D ⊂ R2 be the transformation such that

T (u∗, v∗) = (u (u∗, v∗) , v (u∗, v∗)) ,

Recall that the Jacobian of T is

∂ (u, v)

∂ (u∗, v∗)=

¯¯ ∂u

∂u∗∂u

∂v∗∂v

∂u∗∂v

∂v∗

¯¯ .

Definition 2 Assume that the orientable surface S is parametrized by Φ : D ⊂ R2 → R3. Wesay that the function Φ∗ : D∗ ⊂ R2 → R3 is an orientation-preserving parametrizationof S if there is a smooth (i.e., with continuous partial derivatives) one-one transformation T :D∗ → D such that Φ∗= ΦT and the Jacobian of T is positive on D∗. The parametrization Φ∗is said to be an orientation-reversing parametrization of S if the Jacobian of T is negative

on D∗.

Problems

In the following problems, let S be the surface that is parametrized by the given function.

a) Determine the Normal to S at the given point P0.

b) Determine the plane that is tangent to S at P0.

1.

Φ (u, v) = (v, 3 cos (u) , 3 sin (u)) , −4 ≤ v ≤ 4, 0 ≤ u ≤ 2π,P0 = Φ (π/6, 1)

Note that S is part of the cylinder y2 + z2 = 9 that has as its axis the x-axis.

-2

0z

2

2

4

y

0 2

x0

-2 -2-4


2.

Φ (u, v) =¡u2, u cos (v) , u sin (v)

¢, u ≥ 0, 0 ≤ v ≤ 2π,

P0 = Φ (3,π/3)

Note that S is part of the paraboloid: x = y2 + z2.

-4

-2

4

0

2

z

4

2

y

0

-2 15

x

105-4 0

3.

Φ (r, θ) =¡3r cos (θ) , r2, 2r sin (θ)

¢, r ≥ 0, 0 ≤ θ ≤ π,

P0 = Φ (2,π/4)

Note that S is part of the elliptic paraboloid

x2

9+z2

4= y

-5

-10

0

15

z

5

10

x

0

y5

100

4.

Φ(u, v) = (cosh(u) cos(v), sinh (u) , cosh (u) sin (v)) , 0 ≤ v ≤ 2π, −2 ≤ u ≤ 2,P0 = Φ (1,π/2)

Note that S is part of the hyperboloid

x2 + z2 − y2 = 1

-4

-2

0z

2

-5

4

x

010

y05

-10


5.

Φ (ρ, θ) =³ρ sin

³π6

´cos (θ) , ρ sin

³π6

´sin (θ) , ρ cos

³π6

´´=

Ã1

2ρ cos (θ) ,

1

2ρ sin (θ) ,

√3

2ρ

!

where 0 ≤ ρ ≤ 4 and 0 ≤ θ ≤ 2π,

P0 = Φ (2,π/3)

Note that S is part of the cone φ = π/6 (ρ,φ and θ are spherical coordinates).

0

1

2

2

z

3

y

0

2

x0

-2 -2

6.

Φ (φ, θ) =

µ2 sin (φ) cos (θ) , sin (φ) sin (θ) ,

1

3cos (φ)

¶,

where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π,P = Φ (π/4,π/2)

Note that S is the ellipsoid

x2

4+ y2 + 9z2 = 1.

-0.2

0.0

0.2

1

z

y

02

x

0-1 -2

7.

Ω (x, θ) = (x, ex sin (θ) , ex cos (θ)) ,

where 0 ≤ x ≤ 2, 0 ≤ θ ≤ 2π,P = Ω (1,π/6)

Note that S is part of the surface of revolution that is obtained by revolving the graph of z = ex

about the x-axis.

14.5. SURFACE INTEGRALS 239

-5

0

5

z

5

y

02

x

-5 10

8.

Ω (x, θ) = (x cos (θ) , x sin (θ) , ex) ,

where 0 ≤ x ≤ 2 and 0 ≤ θ ≤ 2π,P = Ω (1,π/2)

Note that S is part of the surface of revolution that is obtained by revolving the graph of z = ex

about the z-axis.

2

y

0

2

-2

4z

x

6

0-22

14.5 Surface Integrals

In this section we will compute the area of a parametrized surface and define the integrals of

scalar functions and vector fields on such a surface. We will consider only orientable surfaces,

as we discussed in Section 14.4.

Surface Area

We will motivate the definition of the area of a surface. Assume that the surface M is parame-

trized by the function Φ : D ⊂ R2 → R3, such that

Φ(u, v) = (x (u, v) , y (u, v) , z (u, v)) , (u, v) ∈ D.

We assume that Φ is smooth, i.e., the tangent vectors

∂Φ

∂u(u, v) =

∂x

∂u(u, v) i+

∂y

∂u(u, v) j+

∂z

∂u(u, v)k

and∂Φ

∂v(u, v) =

∂x

∂v(u, v) i+

∂y

∂v(u, v) j+

∂z

∂v(u, v)k

are continuous on D, and the normal vector

N (u, v) =∂Φ

∂u(u, v)× ∂Φ

∂v(u, v) 6= 0


for each (u, v) ∈ D. Let (u, v) be a point in the interior of D and assume that ∆u and ∆v arepositive numbers that are small enough so that the rectangle D (u, v,∆u,∆v) determined bythe points (u, v) and (u+∆u, v +∆v) is contained in the interior of D. The vectors

∆u∂Φ

∂u(u, v) = ∆u

∂x

∂u(u, v) i+∆u

∂y

∂u(u, v) j+∆u

∂z

∂u(u, v)k,

and

∆v∂Φ

∂v(u, v) = ∆v

∂x

∂v(u, v) i+∆v

∂y

∂v(u, v) j+∆v

∂z

∂v(u, v)k

are tangent to S at Φ (u, v). Since ∆u and ∆v are small,

Φ(u+∆u, v)−Φ(u, v) ∼= ∆u∂Φ∂u

(u, v)

and

Φ(u, v +∆v)−Φ(u, v) ∼= ∆v∂Φ∂v

(u, v) .

Figure 1

It is reasonable to approximate the area of the part of M corresponding to the rectangle

D (u, v,∆u,∆v) by the area of the part of the tangent plane at Φ(u, v) that is spanned by

∆u∂Φ

∂u(u, v) and ∆v

∂Φ

∂v(u, v) .

Recall that this area can be expressed as¯¯∆u

∂Φ

∂u(u, v)× ∆v

∂Φ

∂v(u, v)

¯¯=

¯¯∂Φ

∂u(u, v)× ∂Φ

∂v(u, v)

¯¯∆u∆v

= ||N (u, v)||∆u∆v.Therefore, it makes sense to define the area of the surface as the integral of the magnitude of

the normal vector:

Definition 1 Assume that the surface M is parametrized by the smooth function Φ : D ⊂R2 → R3. The area of M isZ Z

D

||N (u, v)|| dudv =Z Z

D

¯¯∂Φ

∂u(u, v)× ∂Φ

∂v(u, v)

¯¯dudv.


Example 1 Confirm that the area of a sphere of radius ρ0 is 4πρ20.

-22

0z

2

2

y

0

x0

-2 -2

Figure 2

Solution

We can center the sphere M at the origin and parametrize it by the function


where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2. As in Example 9 of Section 14.4,

N (φ, θ) = ρ20 sin (φ) (sin (φ) cos (θ) i+ sin (φ) sin (θ) j+ cos (φ)k)

Therefore,

||N (φ, θ)|| = ρ20 sin (φ)

qsin2 (φ) cos2 (θ) + sin2 (φ) sin2 (θ) + cos2 (φ)

= ρ20 sin (φ)qsin2 (φ)

¡cos2 (θ) + sin2 (θ)

¢+ cos2 (φ)

= ρ20 sin (φ)

qsin2 (φ) + cos2 (φ)

= ρ20 sin (φ) .

Let D = [0,π]× [0, 2π] denote the domain of Φ in the φθ-plane. The area of the sphere isZ ZD

||N (φ, θ)|| dφdθ =Z Z

D

ρ20 sin (φ) dφdθ = ρ20

Z φ=π

φ=0

Z θ=2π

θ=0

sin (φ) dθdφ

= ρ20

Z φ=π

φ=0

2π sin (φ) dφ

= 2πρ20 (− cos (φ)|π0 )= 2πρ20 (2) = 4πρ

20,

as claimed. ¤

Remark 1 Assume that M is the graph of f : D → R, where D is a region in the plane. M

can be parametrized by Φ : D→ R3, where

Φ (x, y) = (x, y, f (x, y))

for each (x, y) ∈ D. In Section 14.4 we noted that

N (x, y) = −∂f∂x(x, y) i− ∂f

∂y(x, y) j+ k.


Therefore,

||N (x, y)|| =sµ

∂f

∂x

¶2+

µ∂f

∂y

¶2+ 1.

Therefore, the area of S can be expressed asZ ZD

sµ∂f

∂x

¶2+

µ∂f

∂y

¶2+ 1 dxdy.

♦

Example 2 Let f (x, y) = x2 − y2, and let M be the part of the graph of f corresponding to

the unit disk in the xy-plane (i.e., the disk of radius 1 centered at the origin). Determine the

area of M.

5

-10

0z

10

5

y

0

x0

-5 -5

Figure 3

Solution

We have∂f

∂x(x, y) = 2x and

∂f

∂y(x, y) = −2y.

Therefore,sµ∂f

∂x

¶2(x, y) +

µ∂f

∂y(x, y)

¶2(x, y) + 1 =

q(2x)2 + (−2y)2 + 1 =

p4x2 + 4y2 + 1

Thus, the area of S is Z ZD

p4x2 + 4y2 + 1dxdy,

whereD is the unit disk. In polar coordinates r and θ,Z ZD

p4x2 + 4y2 + 1dxdy =

Z 2π

θ=0

Z 1

r=0

p4r2 + 1rdrdθ

= 2π

Z 1

r=0

p4r2 + 1rdr

= 2π

Ã1

12

¡4r2 + 1

¢3/2 ¯10

!= 2π

µ5

12

√5− 1

12

¶Example 3 Let M be the part of the cylinder that is parametrized by

Φ (θ, z) = (a cos (θ) , a sin (θ) , z) ,

where 0 ≤ θ ≤ 2π and 0 ≤ z ≤ h. Determine the area of S.


-52

0z

5

2

y

0

x0

-2 -2

Figure 4

Solution

As in Example 11 of Section 14.4,

N (θ, z) = a cos (θ) i+ a sin (θ) j.

Therefore,

||N (θ, z)|| =qa2 cos2 (θ) + a2 sin2 (θ) = a.

Thus, the area of S is Z 2π

θ=0

Z h

z=0

adzdθ = 2πah,

as it should be (the perimeter of the base times height). ¤

Example 4 Assume that the torus M is parametrized by


where 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ 2π. Determine the area of S.

x

z

y

Figure 5

Solution

We have

∂Φ

∂θ(θ,φ) =

∂

∂θ((a+ b cos (φ)) cos (θ)) i+

∂

∂θ((a+ b cos (φ)) sin (θ)) j+ b

∂

∂θsin (φ)k

= − (a+ b cos (φ)) sin (θ) i+ (a+ b cos (φ)) cos (θ) j,and

∂Φ

∂φ(θ,φ) =

∂

∂φ((a+ b cos (φ)) cos (θ)) i+

∂

∂φ((a+ b cos (φ)) sin (θ)) +ג b

∂

∂φsin (φ)k

= −b sin (φ) cos (θ) i− b sin (φ) sin (θ) j+ b cos (φ)k.


Therefore,

N (θ,φ) =∂Φ

∂θ(θ,φ)× ∂Φ

∂φ(θ,φ)

=

¯¯ i j k

− (a+ b cos (φ)) sin (θ) (a+ b cos (φ)) cos (θ) 0−b sin (φ) cos (θ) −b sin (φ) sin (θ) b cos (φ)

¯¯

= b (a+ b cos (φ)) (cos (θ) cos (φ) i+ sin (θ) cos (φ) j+ sin (φ)k)

Thus,

||N (θ,φ)||2 = b2 (a+ b cos (φ))2 ¡cos2 (θ) cos2 (φ) + sin2 (θ) cos2 (φ) + sin2 (φ)¢= b2 (a+ b cos (φ))

2 ¡cos2 (φ) + sin2 (φ)

¢= b2 (a+ b cos (φ))2 .

Therefore,

||N (θ,φ)|| = b (a+ b cos (φ))Thus, the area of the torus isZ Z

D

||N (θ,φ)|| dθdφ =Z Z

D

b (a+ b cos (φ)) dθdφ,

where D is the square [0, 2π]× [0, 2π] in the φθ-plane. Therefore,Z ZD

b (a+ b cos (φ)) dθdφ =

Z φ=2π

φ=0

Z θ=2π

θ=0

b (a+ b cos (φ)) dθdφ

= 2πb

Z φ=2π

φ=0

(a+ b cos (φ)) dφ

= 2πb³aφ+ b sin (φ)|2π0

´= 2πb2πa = 4πab.

¤

Surface Integrals of Scalar Functions

Definition 2 Assume that the (orientable) surface M is parametrized by the function Φ : D ⊂R2 → R3, and that the scalar function f is continuous on M . We define the integral of f on Mas

Z ZD

f (Φ (u, v)) ||N (u, v)|| dudv.

We will denote the integral of f on M as Z ZM

fdS.

Symbolically, dS is “the element of area" ||N (u, v)|| dudv. You can imagine that dS representsthe area of an “infinitesimal" portion of the surface M . If the surface M is a thin shell, and f

is its mass density per unit area, the integralZ ZM

fdS


yields the total mass of the shell. Soon we will discuss other contexts for the appearance of

surface integrals.

Example 5 Let M be the hemisphere sphere of radius 2 that is parametrized by

Φ (φ, θ) = (2 sin (φ) cos (θ) , 2 sin (φ) sin (θ) , 2 cos (φ)) ,

where 0 ≤ φ ≤ π/2 and 0 ≤ θ ≤ 2π Assume that M is a thin shell with mass density

f (x, y, z) = 4− z.Determine the total mass of M .

Solution

Let D = [0,π]× [0, 2π] denote the domain of Φ in the φθ-plane. As in Example 1,||N (φ, θ)|| = 22 sin (φ) = 4 sin (φ) .

Therefore, the mass of M isZ ZM

fdS =

Z(4− z) ds =

Z(4− 2 cos (φ)) ||N (φ, θ)|| dφdθ

=

Z ZD

(4− 2 cos (φ)) (4 sin (φ)) dφdθ

= 8

Z φ=π/2

φ=0

Z θ=2π

θ=0

(2 sin (φ)− cos (φ) sin (φ)) dφdθ

= 16π

Z φ=π/2

φ=0

(2 sin (φ)− cos (φ) sin (φ)) dφ

= 16π

Ã−2 cos (φ) + 1

2cos2 (φ)

¯π/20

!

= 16π

µ3

2

¶= 24π.

(we made use of the substitution u = cos (φ)). ¤The integral of a scalar function on a surface M is independent of the parametrization of M .

In particular, the calculation of the area of a surface does not change under an orientation

preserving or reversing parametrization of M . You can find the proof at the end of this section.

Remark 2 As we noted before, if M is the graph of the function g : D ⊂ R2 → R, then S canbe parametrized by

Φ (x, y) = (x, y, g (x, y)) , (x, y) ∈ D,N (x, y) = −∂g

∂x(x, y) i− ∂g

∂y(x, y) j+ k,

and

||N (x, y)|| =sµ

∂g

∂x

¶2+

µ∂g

∂y

¶2+ 1.

Therefore, the integralR R

Mfds takes the formZ Z

D

f (x, y, g (x, y)) ||N (x, y)|| dxdy =Z Z

D

f (x, y, g (x, y))

sµ∂g

∂x

¶2+

µ∂g

∂y

¶2+ 1dxdy.

The roles of the coordinates can be interchanged, of course. ♦


Example Let M be the part of the graph of z = g (x, y) = 4 − x2 − y2 over the xy-plane.Evaluate Z Z

M

1

1 + 4 (x2 + y2)dS.

20

2z

-2

4

y

0

x

0-22

Figure 7

Solution

We have

4− x2 − y = 0⇔ x2 + y2 = 4.

Thus, the intersection of M with the xy-plane is the circle x2 + y2 = 4, and M is the graph of

g over the disk D of radius 2 centered at the origin.

We have∂g

∂x(x, y) = −2x and ∂g

∂y(x, y) = −2y.

Therefore,Z ZM

1

1 + 4 (x2 + y2)dS =

Z ZD

1

1 + 4 (x2 + y2)

q(−2x)2 + (−2y2) + 1dxdy

=

Z ZD

1

1 + 4 (x2 + y2)

p4x2 + 4y2 + 1dxdy

=

Z ZD

1p1 + 4 (x2 + y2)

dxdy.

It is convenient to transform the integral to polar coordinates:Z ZD

1p1 + 4 (x2 + y2)

dxdy =

Z θ=2π

θ=0

Z r=2

r=0

1√1 + 4r2

rdrdθ

= 2π

Z r=2

r=0

1√1 + 4r2

rdr.

If we set u = 1 + 4r2 then du = 8rdr, so thatZ r=2

r=0

1√1 + 4r2

rdr =1

8

Z u=17

u=1

1√udu =

1

8

Z u=17

u=1

u−1/2du

=1

8

µ2u1/2

¯171

¶=1

4

³√17− 1

´.


Therefore, Z ZS

1

1 + 4 (x2 + y2)dS = 2π

Z r=2

r=0

1√1 + 4r2

rdr

= 2π

µ1

4

³√17− 1

´¶=

π

2

³√17− 1

´.

¤

Flux Integrals

In order to motivate the definition of a flux integral, let us imagine that v (x, y, z) is the velocityof a fluid particle at (x, y, z) ∈ R3, and that M is a surface in R3. We would like to calculatethe flow across S per unit time. Assume that a small patch of the surface is almost planar, has

area dS, and n is the unit normal to that piece which points in the direction of the flow. The

component of the velocity vector in the direction of n is v · n. If the mass density of the fluidhas the constant value δ, the mass of fluid that flows across that small patch is approximately

δv · ndS. Therefore it is reasonable to calculate the total flow across S per unit time asZ ZM

δv · ndS.

This leads to the definition of the flux integral of an arbitrary vector field over a surface:

Definition 3 Assume that F is a continuous vector field in R3 and M is a smooth surface that

is parametrized by the function Φ : D ⊂ R2 → R3. The flux integral of F over M is the

integral of the normal component of F on MS, i.e.Z ZM

F · ndS

where n is the unit normal to M.

The “vectorial element of area" is the symbol

dS = ndS,

so that the flux integral of F over M can be denoted as.Z ZM

F · dS.

If

N (u, v) =∂Φ

∂u(u, v)× ∂Φ

∂v(u, v) ,

then

n =N

||N|| and dS = ||N|| dudv.

Thus, Z ZM

F · ndS =Z Z

D

F (Φ (u, v)) · N (u, v)

||N (u, v)|| ||N (u, v)|| dudv

=

Z ZD

F (Φ (u, v)) ·N (u, v) dudv

=

Z ZD

F (Φ (u, v)) ·µ∂Φ

∂u(u, v)× ∂Φ

∂v(u, v)

¶dudv.


Example 6 Consider the cylinder of radius 2 whose axis is along the z-axis. Let M be the

portion of the cylinder between z = 0 and z = 4. As in Example 3, M can be parametrized

by Φ (θ, z) = (2 cos (θ) , 2 sin (θ) , z), where 0 ≤ θ ≤ 2π and 0 < z < 4. Let F (x, y, z) = xi.

Calculate the flux of F over M .

02

2z

4

2

y

0

x

0

-2 -2

Figure 8

Solution

We have

F (Φ (θ, z)) = F (2 cos (θ) , 2 sin (θ) , z) = 2 cos (θ) i.

As in Example 3,

N (θ, z) = 2 cos (θ) i+ 2 sin (θ) j.

Therefore,

F (Φ (θ, z)) ·N (θ, z) = (2 cos (θ) i) · (2 cos (θ) i+ 2 sin (θ) j) = 4 cos2 (θ) .

Thus, ZM

F · ndS =Z Z

D

F (Φ (θ, z)) ·N (θ, z) dθdz =Z 4

z=0

Z 2π

θ=0

4 cos2 (θ) .dθdz

= 4

Z 4

z=0

ÃZ θ=2π

θ=0

cos2 (θ) dθ

!dz

= 16

Z θ=2π

θ=0

cos2 (θ) dθ

= 16

Ã1

2cos (θ) sin (θ) +

1

2θ

¯2π0

!= 16π.

¤

Example 7 Let M be the sphere of radius ρ0 that is centered at the origin. M can be para-

metrized by


where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π.CalculateZ ZM

(xi+ yj+ zk) · dS


Solution

Z ZM

(xi+ yj+ zk) · dS =Z 2π

θ=0

Z π

φ=0

Φ (φ, θ) ·N (φ, θ) dφdθ,

where

N (φ, θ) = ρ20 sin (φ) (sin (φ) cos (θ) i+ sin (φ) sin (θ) j+ cos (φ)k)

= ρ0 sin (φ)Φ (φ, θ) ,

as in Example 1. Therefore,Z 2π

θ=0

Z π

φ=0

Φ (φ, θ) ·N (φ, θ) dφdθθ =Z 2π

θ=0

Z π

φ=0

Φ (φ, θ) · ρ0 sin (φ)Φ (φ, θ) dφdθ

= ρ0

Z 2π

θ=0

Z π

φ=0

sin (φ)Φ (φ, θ) ·Φ (φ, θ) dφdθ

= ρ0

Z 2π

θ=0

Z π

φ=0

sin (φ) ||Φ (φ, θ)||2 dφdθ

= ρ0

Z 2π

θ=0

Z π

φ=0

sin (φ) ρ20dφdθ

= ρ0¡4πρ20

¢= 4πρ30.

¤

Proposition 1 Assume that M is an orientable surface. The flux integralZM

F · ndS

is independent of an orientation-preserving parametrization ofM . The flux integral is multiplied

by −1 under an orientation-reversing parametrization of M.

The proof of Proposition 2can be found at the end of this section.

Remark 3 If M is the graph of the function g (x, y) over the domain D in the xy-plane, we

can parametrize M by

Φ (x, y) = (x, y, g (x, y)) ,

where (x, y) ∈ D. As we noted above,

N (x, y) = −gx (x, y) i− gy (x, y) j+ k,

and

||N (x, y)|| =sµ

∂g

∂x

¶2+

µ∂g

∂y

¶2+ 1.


Therefore,Z ZM

F · dS =Z Z

M

F · ndS

=

Z ZM

F · N

||N||dS

=

Z ZD

F (x, y, g (x, y)) · N (x, y)||N (x, y)|| ||N (x, y)|| dxdy

=

Z ZD

F (x, y, g (x, y)) ·N (x, y) dxdy

=

Z ZD

F (x, y, g (x.y)) ·µ−∂g∂x(x, y) i− ∂g

∂y(x, y) j+ k

¶dxdy..

♦

Example 8 Let F (x, y, z) = zk and letM be the hemisphere of radius a centered at the origin.

Calculate the flux integral of F across M in the direction of the outward normal.

Solution

The surface M is the graph of the function

g (x, y) =pa2 − x2 − y2,

where (x, y) is in the disk Da of radius a centered at the origin. A normal that points outwardis

−gx (x, y) i− gy (x, y) j+ k. = xpa2 − x2 − y2 i+

ypa2 − x2 − y2 j+ k

On the hemisphere M ,

F (x, y, z) = F (x, y, g (x, y)) = g (x, y)k =pa2 − x2 − y2k.

Therefore,Z ZM

F · dS =Z Z

Da

g (x, y)k · (−gx (x, y) i− gy (x, y) j+ k) dxdyZ ZDa

pa2 − x2 − y2k·

Ãxp

a2 − x2 − y2 i+yp

a2 − x2 − y2 j+ k!dxdy

=

Z ZDa

pa2 − x2 − y2dxdy.

In polar coordinates,Z ZDa

pa2 − x2 − y2dxdy =

Z 2π

θ=0

Z a

r=0

pa2 − r2rdrdθ

= 2π

µZ a

r=0

pa2 − r2rdr

¶= 2π

µa3

3

¶=2πa3

3.

¤


Example 9 Assume that the electrostatic field due to the charge q at the origin is

F (x, y, z) = qxi+ yj+ zk

(x2 + y2 + z2)3/2= q

r

||r||3

Calculate the flux of F out of the sphere M of radius ρ0 centered at the origin.

Solution

Z ZM

F · ndS =Z Z

M

qr

||r||3 ·r

||r||dS

= q

Z ZM

||r||2||r||4 dS = q

Z ZM

1

||r||2 dS =q

ρ20

Z ZM

dS =q

ρ20

¡4πρ20

¢= 4πq

¤

Example 10 Let

F (x, y, z) =−yi+ xjx2 + y2

.

Determine the flux of F in the direction of j across the rectangle M in the xz-plane that has

the vertices (1, 0, 0) and (3, 0, 2)

Solution

The unit normal to M is j. We haveZ ZM

F · ndS =Z 2

z=0

Z 3

x=1

1

xdxdz = 2 ln (3)

¤

Alternative Notations and some Proofs (Optional)

At the end of Section 14.4. we saw that

N (u, v) =∂ (y, z)

∂ (u, v)i+

∂ (z, x)

∂ (u, v)j+

∂ (x, y)

∂ (u, v)k,

where∂ (y, z)

∂ (u, v)=

¯yu zuyv zv

¯,∂ (z, x)

∂ (u, v)=

¯zu xuzv xv

¯,∂ (x, y)

∂ (u, v)=

¯xu yuxv yv

¯Therefore, we can express the area of the surface that is parametrized by (x (u, v) , y (u, v) , z (u, v)),where (u, v) ∈ D, as

Z ZD

||N (u, v)|| dudv =Z Z

D

sµ∂ (y, z)

∂ (u, v)

¶2+

µ∂ (z, x)

∂ (u, v)

¶2+

µ∂ (x, y)

∂ (u, v)

¶2dudv.

Example 11 Consider the cone that is parametrized by

Φ (ρ, θ) = (ρ sin (φ0) cos (θ) , ρ sin (φ0) sin (θ)) , ρ cos (φ0) .

Here φ0 is a given angle, ρ ≥ 0 and 0 ≤ θ ≤ 2π. Let M be the part of the cone of slant height

l, i.e., 0 ≤ ρ ≤ l. Determine the area of M .


0

1

2

2

z

3

1

y

0

x0

-2 -1

Figure 9

Solution

As in Example 14 of Section 14.4,

N (ρ, θ) =∂ (y, z)

∂ (ρ, θ)i+

∂ (z, x)

∂ (ρ, θ)j+

∂ (x, y)

∂ (ρ.θ)k

= −ρ sin (φ0) cos (φ0) cos (θ) i− ρ sin(φ0) cos (φ0) sin (θ) j+ ρ sin2 (φ0)k.

Therefore,

||N (ρ, θ)|| =qρ2 sin2 (φ0) cos

2 (φ0) cos2 (θ) + ρ2 sin2

¡φ20¢cos2 (φ0) sin

2 (θ) + ρ2 sin4 (φ0)

=

qρ2 sin2 (φ0) cos

2 (φ0) + ρ2 sin4 (φ0)

= ρ sin (φ0)

qcos2 (φ0) + sin

2 (φ) = ρ sin (φ0)

Thus, the area of S isZ 2π

θ=0

Z l

ρ=0

ρ sin (φ0) dρdθ = 2π sin (φ0)

Z l

ρ=0

ρdρ = 2π sin (φ0)

µl2

2

¶= πl2 sin (φ0) .

Note that the radius of the base is l sin (φ0), so that the above expression is π×slant height×radiusof the base. ¤

Proposition 2 The surface integral Z ZM

fdS

is independent of an orientation-preserving or orientation reversing parametrization of the sur-

face M .

Proof

Assume that M is parametrized by Φ,

Φ (u, v) = (x (u, v) , y (u, v) .z (u, v)) , for each (u, v) ∈ D,and

u = u (u∗, v∗) , v = v (u∗, v∗)

for each (u∗, v∗) ∈ D∗. Setx (u∗, v∗) = x (u (u∗, v∗) , v (u∗, v∗)) , y (u∗, v∗) = y (u (u∗, v∗) , v (u∗, v∗))z (u∗, v∗) = z (u (u∗, v∗) , v (u∗, v∗)) ,


and

Φ∗ (u∗, v∗) = (x (u∗, v∗) , y (u∗, v∗) .z (u∗, v∗)) , where (u∗, v∗) ∈ D∗.If we make use of the parametrization Φ∗, the surface integral can be expressed asZ Z

D∗f (Φ∗ (u∗, v∗)) ||N (u∗, v∗)|| du∗dv∗

=

Z ZD∗f (Φ∗ (u∗, v∗))

sµ∂ (y, z)

∂ (u∗, v∗)

¶2+

µ∂ (z, x)

∂ (u∗, v∗)

¶2+

µ∂ (x, y)

∂ (u∗, v∗)

¶2du∗dv∗

As a consequence of the chain rule,

∂ (y, z)

∂ (u∗, v∗)=

∂ (y, z)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗),

∂ (z, x)

∂ (u∗, v∗)=

∂ (z, x)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗),

∂ (x, y)

∂ (u∗, v∗)=

∂ (x, y)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗).

Therefore,Z ZD∗f (Φ∗ (u∗, v∗)) ||N (u∗, v∗)|| du∗dv∗

=

Z ZD∗

ZD∗f (Φ∗ (u∗, v∗))

sµ∂ (y, z)

∂ (u∗, v∗)

¶2+

µ∂ (z, x)

∂ (u∗, v∗)

¶2+

µ∂ (x, y)

∂ (u∗, v∗)

¶2du∗dv∗

=

Z ZD∗

ZD∗f (Φ∗ (u∗, v∗))

sµ∂ (y, z)

∂ (u, v)

¶2+

µ∂ (z, x)

∂ (u, v)

¶2+

µ∂ (x, y)

∂ (u, v)

¶2 ¯∂ (u, v)

∂ (u∗, v∗)

¯du∗dv∗

=

Z ZD

f (Φ (u, v))

sµ∂ (y, z)

∂ (u, v)

¶2+

µ∂ (z, x)

∂ (u, v)

¶2+

µ∂ (x, y)

∂ (u, v)

¶2dudv,

by the rule for the change of variables in double integrals. Note thatsµ∂ (y, z)

∂ (u, v)

¶2+

µ∂ (z, x)

∂ (u, v)

¶2+

µ∂ (x, y)

∂ (u, v)

¶2= ||N (u, v)|| .

Therefore, Z ZD∗||N (u∗, v∗)|| du∗dv∗ =

Z ZD

||N (u, v)|| dudv.

Thus, the surface integral Z ZM

fdS

can be calculated by making use of either parametrization of S, as claimed .¥The Differential Form Notation for a Flux Integral

As we saw before, if Φ (u, v) = (x (u, v) , y (u, v) , z (u, v)) then

N (u, v) =∂ (y, z)

∂ (u, v)i+

∂ (z, x)

∂ (u, v)j+

∂ (x, y)

∂ (u, v)k.

Therefore, if F (x, y, z) =M (x, y, z) i+N (x, y, z) j+ P (x, y, z)k then


F (Φ (u, v)) ·N (u, v) =M (Φ (u, v))∂ (y, z)

∂ (u, v)+N (Φ (u, v))

∂ (z, x)

∂ (u, v)+ P (Φ (u, v))

∂ (x, y)

∂ (u, v)

Therefore,Z ZM

F · ndS =Z Z

D

F (Φ (u, v)) ·N (u, v) dudv

=

Z ZD

µM (Φ (u, v))

∂ (y, z)

∂ (u, v)+N (Φ (u, v))

∂ (z, x)

∂ (u, v)+ P (Φ (u, v))

∂ (x, y)

∂ (u, v)

¶dudv.

Symbolically, we can set

dydz =∂ (y, z)

∂ (u, v)dudv, dzdx =

∂ (z, x)

∂ (u, v)dudv, dxdy =

∂ (x, y)

∂ (u, v)dudv,

and writeZ ZD

µM (Φ (u, v))

∂ (y, z)

∂ (u, v)+N (Φ (u, v))

∂ (z, x)

∂ (u, v)+ P (Φ (u, v))

∂ (x, y)

∂ (u, v)

¶dudv

=

Z ZM

Mdydz +Ndzdx+ Pdxdy.

This is the differential form notation for the flux integral of

F (x, y, z) =M (x, y, z) i+N (x, y, z) j+ P (x, y, z)k

over the surface M .

Example 12 Let M be the torus that is parametrized by


where a > b > 0, 0 ≤ θ ≤ 2π and 0 ≤ φ ≤ 2π. Determine the flux integralZ ZM

xdydz + ydzdx+ zdxdy

Solution

We have

xdydz = (a+ b cos (φ)) cos (θ)∂ (y, z)

∂ (θ,φ)dθdφ

= (a+ b cos (φ)) cos (θ) b (a+ b cos (φ)) cos (θ) cos (φ) dθdφ

= b (a+ b cos (φ))2 cos2 (θ) cos (φ) dθdφ,

ydzdx = (a+ b cos (φ)) sin (θ)∂ (z, x)

∂ (θ,φ)dθdφ

= b (a+ b cos (φ)) sin (θ) (a+ b cos (φ)) sin (θ) cos (φ) dθdφ

= b (a+ b cos (φ))2sin2 (θ) cos (φ) dθdφ,


and

zdxdy = z∂ (x, y)

∂ (u, v)dθdφ

= b sin (φ) (a+ b cos (φ)) sin (φ) dθdφ

= b (a+ b cos (φ)) sin2 (φ) dθdφ.

Therefore,


= b (a+ b cos (φ))¡(a+ b cos (φ)) cos2 (θ) cos (φ) + (a+ b cos (φ)) sin2 (θ) cos (φ) + sin2 (φ)

¢dθdφ

= b (a+ b cos (φ))¡(a+ b cos (φ)) cos (φ) + sin2 (φ)

¢dθdφ

Thus Z ZS


=

Z θ=2π

θ=0

Z φ=π

φ=0

b (a+ b cos (φ))¡(a+ b cos (φ)) cos (φ) + sin2 (φ)

¢dθdφ

= 2π

µb2aπ +

1

2baπ

¶= 2ab2π2 + π2ab.

¤

The Proof of Proposition 2:

Assume that M is an orientable surface. The flux integralZM

F · ndS

is independent of an orientation-preserving parametrization ofM . The flux integral is multiplied

by −1 under an orientation-reversing parametrization of M.As in the proof of Proposition 1, assume that M is parametrized by Φ,

Φ (u, v) = (x (u, v) , y (u, v) .z (u, v)) , for each (u, v) ∈ D,

and

u = u (u∗, v∗) , v = v (u∗, v∗)

for each (u∗, v∗) ∈ D∗. Set

x (u∗, v∗) = x (u (u∗, v∗) , v (u∗, v∗)) , y (u∗, v∗) = y (u (u∗, v∗) , v (u∗, v∗))z (u∗, v∗) = z (u (u∗, v∗) , v (u∗, v∗)) ,

and

Φ∗ (u∗, v∗) = (x (u∗, v∗) , y (u∗, v∗) .z (u∗, v∗)) , where (u∗, v∗) ∈ D∗.Let’s assume that the transformation (u∗, v∗)→ (u, v) is orientation preserving so that

∂ (u, v)

∂ (u∗, v∗)> 0


on D∗. We haveZ ZD∗(M (Φ∗ (u∗, v∗))

∂ (y, z)

∂ (u∗, v∗)+N (Φ∗ (u∗, v∗))

∂ (z, x)

∂ (u∗, v∗)

+P (Φ∗ (u∗, v∗))∂ (x, y)

∂ (u∗, v∗))du∗dv∗

=

Z ZD∗(M (Φ∗ (u∗, v∗))

∂ (y, z)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗)+N (Φ∗ (u∗, v∗))

∂ (z, x)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗)

+P (Φ∗ (u∗, v∗))∂ (x, y)

∂ (u, v)

∂ (u, v)

∂ (u∗, v∗))du∗dv∗

=

Z ZD∗

µM (Φ∗ (u∗, v∗))

∂ (y, z)

∂ (u, v)+N (Φ∗ (u∗, v∗))

∂ (z, x)

∂ (u, v)+ (Φ∗ (u∗, v∗))

∂ (x, y)

∂ (u, v)

¶∂ (u, v)

∂ (u∗, v∗)du∗dv∗

=

Z ZD

µM (Φ (u, v))

∂ (y, z)

∂ (u, v)+N (Φ (u, v))

∂ (z, x)

∂ (u, v)+ P (Φ (u, v))

∂ (x, y)

∂ (u, v)

¶dudv

=

Z ZM

F · ndS

by the rule for the change of variables in double integrals.

If the transformation is orientation reversing so that

∂ (u, v)

∂ (u∗, v∗)< 0

on D∗, we haveZ ZD∗

µM (Φ∗ (u∗, v∗))

∂ (y, z)

∂ (u, v)+N (Φ∗ (u∗, v∗))

∂ (z, x)

∂ (u, v)+ (Φ∗ (u∗, v∗))

∂ (x, y)

∂ (u, v)

¶∂ (u, v)

∂ (u∗, v∗)du∗dv∗

= −Z Z

D∗

µM (Φ∗ (u∗, v∗))

∂ (y, z)

∂ (u, v)+N (Φ∗ (u∗, v∗))

∂ (z, x)

∂ (u, v)+ (Φ∗ (u∗, v∗))

∂ (x, y)

∂ (u, v)

¶ ¯∂ (u, v)

∂ (u∗, v∗)

¯du∗dv∗

= −Z Z

D

µM (Φ (u, v))

∂ (y, z)

∂ (u, v)+N (Φ (u, v))

∂ (z, x)

∂ (u, v)+ P (Φ (u, v))

∂ (x, y)

∂ (u, v)

¶dudv

= −ZM

F · ndS.

¥

Problems

In problems 1-5, determine the area of the surface that is parametrized by the given function

(leave the relevant integral in a simplified form, if that is indicated)

1 (a cylinder)

Φ (u, v) = (v, 3 cos (u) , 3 sin (u)) , −4 ≤ v ≤ 4, 0 ≤ u ≤ 2π.

2 (a paraboloid)

Φ (u, v) =¡u2, u cos (v) , u sin (v)

¢, 0 ≤ u ≤ 3, 0 ≤ v ≤ 2π.


3 (a cone)

Φ (ρ, θ) =

Ã1

2ρ cos (θ) ,

1

2ρ sin (θ) ,

√3

2ρ

!where 0 ≤ ρ ≤ 4 and 0 ≤ θ ≤ 2π.4 (an ellipsoid) Let

Φ (φ, θ) =

µ2 sin (φ) cos (θ) , sin (φ) sin (θ) ,

1

3cos (φ)

¶,

where 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π. Express the area of the surface as an integral. Simplify theexpression as much as possible. Do not attempt to evaluate the integral.

5 (a surface of revolution)

Ω (x, θ) = (x, ex sin (θ) , ex cos (θ)) ,

where 0 ≤ x ≤ 2, 0 ≤ θ ≤ 2π.

6. Let S be the hemisphere that is parametrized by


where 0 ≤ φ ≤ π and 0 ≤ θ ≤ π EvaluateZ ZS

¡x2 + y2

¢dS

7. Let S be the surface that is parametrized by

Φ (u, v) = (v, 2 cos (u) , 2 sin (u)) , −4 ≤ v ≤ 4, 0 ≤ u ≤ π.

Evaluate Z ZS

y2dS

(this is part of the cylinder y2 + z2 = 1).8. Let S be the helicoid that is parametrized by

Φ (r, θ) = (r cos (θ) , r sin (θ) , θ) , 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π.

Evaluate Z ZS

px2 + y2dS

9. Let S be the part of the plane

x+ y + z = 1

in the first octant (i.e., x ≥ 0, y ≥ 0 and z ≥ 0). EvaluateZ ZS

xdS.


10. Let S be the part of the sphere

x2 + y2 + z2 = 1

that is in the first octant (i.e., x ≥ 0, y ≥ 0 and z ≥ 0). EvaluateZ ZS

z2dS.

11. Let S be the surface that is parametrized by

Φ (u, v) = (cos (u) , v, sin (u)) , −2 ≤ v ≤ 2, 0 ≤ u ≤ π

(this is part of the cylinder x2 + z2 = 1). EvaluateZ ZS

(xi+ zk) · dS.

12. Let S be the planar surface that is parametrized by

Φ (y, z) = (2, y, z) , 0 ≤ y ≤ 3, 0 ≤ z ≤ 4.Evaluate Z Z

S

(yi+ xj+ zk) · dS.

13. Let S be the hemisphere of radius 3 that is parametrized by


where 0 ≤ φ ≤ π and 0 ≤ θ ≤ π. EvaluateZ ZS

xi+ yj+ zk

(x2 + y2 + z2)· dS.

14. Let S be the portion of the plane

z = 1− 12x− 1

2y

in the first octant. Evaluate Z ZS

(xi− zk) ·dS,

if S is oriented so that the normal points upward.

15. Let S be the hemisphere

z =p4− x2 − y2.

Evaluate Z ZS

(xi+yj+ zk) · dS,


16. Let S be the portion of the paraboloid

z = 1− x2 − y2

14.6. GREEN’S THEOREM 259

above the xy-plane. Evaluate Z ZS

(yj+ k) · dS


17. Assume that S consists of the hemisphere S1

z =p4− x2 − y2

and the disk

S2 =©(x, y, 0) : x2 + y2 ≤ 4ª

Evaluate Z ZS

(yzi+ xzj+ xyk) · dS,

if S is oriented so that the normal points towards the exterior of the region enclosed by S.

14.6 Green’s Theorem

In this section we will discuss Green’s Theorem that establishes a link between a line integral on

a closed curve and a double integral and some of the applications of the theorem. For example,

Green’s theorem will be used to show that the necessary condition for the existence of a potential

for a vector field is also sufficient if the vector field is smooth in a domain without holes.

Green’s Theorem in Simply Connected Regions

Assume that the boundary of a region D in the plane is a simple closed curve C, i.e., a closed

curve without self intersections, and that C is oriented so that D is to the left as C is traversed.

In this case we will say that C is the positively oriented boundary of D and refer to D as

the interior of C. The region D is said to be simply connected. There are no holes in

D. You can imagine that any simple closed curve that lies in the interior of D can be shrunk to

a point in D without leaving D.

D

C

Figure 1: A simply connected region D and its positively oriented boundary C

Green’s Theorem relates a line integral on C to a double integral on D:

Theorem 1 (Green’s Theorem) Let the simple closed curve C be the positively oriented

boundary of the region D. We have

ZC

Mdx+Ndy =

Z ZD

µ∂N

∂x− ∂M

∂y

¶dxdy,


provided that the partial derivatives of M and N are continuous in D ∪ C.We will not give the proof of Green’s Theorem for an arbitrary region, but we will confirm the

statement of the theorem in the case of a region that is both of type I and type II. We will refer

to such a region as a simple region. The theorem will be confirmed by showing thatZC

Mdx = −Z Z

D

∂M

∂ydxdy,

and ZC

Ndy =

Z ZD

∂N

∂xdxdy.

Let’s establish the first equality. Since D is of type I, D can be expressed as the set of point

(x, y) such that a ≤ x ≤ b and f (x) ≤ y ≤ g (x), as illustrated in Figure 2.

x

y

y gx

C

y fx

C

a b

Figure 2

By Fubini’s Theorem, Z ZD

∂M

∂ydxdy =

Z b

a

ÃZ y=g(x)

y=f(x)

∂M

∂y(x, y) dy

!dx.

By the Fundamental Theorem of Calculus,Z y=g(x)

y=f(x)

∂M

∂y(x, y) dy =M (x, g (x))−M (x, f (x)) .

Therefore, Z ZD

∂M

∂ydxdy =

Z b

a

(M (x, g (x))−M (x, f (x))) dx

=

Z b

a

M (x, g (x)) dx−Z b

a

M (x, f (x)) dx.

Note that Z b

a

M (x, g (x)) dx = −ZC+

Mdx,

and Z b

a

M (x, f (x)) dx =

ZC−Mdx.


Therefore, ZC

Mdx =

ZC+

Mdx+

ZC−Mdx

= −Z b

a

M (x, g (x)) dx+

Z b

a

M (x, f (x)) dx

= −ÃZ b

a

M (x, g (x)) dx−Z b

a

M (x, f (x)) dx

!

= −Z Z

D

∂M

∂ydxdy,

as claimed.

Similarly, since D is also of type II, we can express D as the set of points (x, y) such thatc ≤ y < d and F (y) ≤ x < G (y), as illustrated in Figure 3.

x

y

x GyCx Fy C

a b

Figure 3

By Fubini’s Theorem, Z ZD

∂N

∂xdxdy =

Z d

c

ÃZ x=G(y)

x=F (x)

∂N

∂x(x, y) dx

!dy.

By the Fundamental Theorem of Calculus,Z x=G(y)

x=F (x)

∂N

∂x(x, y) dx = N (G(y), y)−N (F (y), y) .

Therefore, Z ZD

∂N

∂xdxdy =

Z d

c

(N (G(y), y)−N (F (y), y)) dy

=

Z d

c

N (G(y), y) dy −Z d

c

N (F (y), y) dy

Note that Z d

c

N (G(y), y) dy =

ZC+

Ndy,

and

−Z d

c

N (F (y), y) dy =

ZC−Ndy.


Therefore, ZC

Ndy =

ZC+

Ndy +

ZC−Ndy

=

Z d

c

N (G(y), y) dy −Z d

c

N (F (y), y) dy =

Z ZD

∂N

∂xdxdy,

as claimed. ¥

Remark Note that ZC

Mdx+Ndy =

ZC

(M i+N j) · dσ.Thus, Green’s Theorem can be stated asZ

C

F · dσ =Z Z

D

µ∂N

∂x− ∂M

∂y

¶dxdy,

where F =M i+Nj.

Example 1 Let D be the unit disk (x, y) : x2 + y2 ≤ 1 and let C be its positively oriented

boundary. Confirm Green’s Theorem for the line integralZC

ydx+ ln¡x2 + y2 + 1

¢dy.

Solution

Let’s evaluate the line integral directly. We can parametrize the unit circle C by σ (t) =(cos (t) , sin (t)), t ∈ [0, 2π]. Thus,Z

C

ydx+ ln¡x2 + y2 + 1

¢dy =

Z 2π

0

(sin (t) (− sin t) + ln (2) cos (t)) dt

=

Z 2π

0

¡− sin2 (t) + ln (2) cos (t)¢ dt = −π.By Green’s Theorem,Z

C

ydx+ ln¡x2 + y2 + 1

¢dy =

Z ZD

µ∂

∂xln¡x2 + y2 + 1

¢− ∂

∂y(y)

¶dxdy

=

Z ZD

µ1

x2 + y2 + 1(2x)− 1

¶dxdy.

In polar coordinates,Z ZD

µ1

x2 + y2 + 1(2x)− 1

¶dxdy =

Z 2π

θ=0

Z 1

r=0

µ2r cos (θ)

r2 + 1− 1¶rdrdθ

=

Z 1

r=0

2r2

r2 + 1

Z 2π

θ=0

cos (θ) dθdr −Z θ=2π

θ=0

Z 1

r=0

rdrdθ

= 0− 2πµ1

2

¶= −π.

Thus, we have confirmed thatZC

ydx+ ln¡x2 + y2 + 1

¢dy =

Z ZD

µ1

x2 + y2 + 1(2x)− 1

¶dxdy.

¤


Green’s Theorem in Multiply Connected Regions

Assume that D is a region whose boundary consists of the closed curves C1 and C2, as in Fig-

ure 4, with the orientations of the curves as indicated (if you imagine that you walk along the

boundary you should see the region D to your left). Such a region is doubly connected: You

can consider that there is a hole in the interior of C1 that is bounded by the inner curve C2.

x

y

C1C2

Figure 4

In this case Green’s Theorem takes the following form:ZC1

Mdx+Ndy +

ZC2

Mdx+Ndy =

Z ZD

µ∂N

∂x− ∂M

∂y

¶dxdy.

We can provide a plausibility argument as follows: Imagine that a cut is made in D so that the

resulting region D0 is bounded by the simple closed curve C1 + C3 + C2 + C4, as in Figure 5(you can imagine that the cut has width that can be made arbitrarily small, so that C3 and C4correspond to the same curve traversed in opposite directions).

x

y

C1C2

C3C4

Figure 5

Since Green’s Theorem is valid in the simply connected domain D0, we haveZC1

Mdx+Ndy+

ZC3

Mdx+Ndy+

ZC2

Mdx+Ndy+

ZC4

Mdx+Ndy =

Z ZD0

µ∂N

∂x− ∂M

∂y

¶dxdy

Since ZC3

Mdx+Ndy +

ZC4

Mdx+Ndy = 0

due to the opposite orientations of C3 and C4.ZC1

Mdx+Ndy +

ZC2

Mdx+Ndy =

Z ZD0

µ∂N

∂x− ∂M

∂y

¶dxdy


Since we can imagine that the width of the cut tends to 0, we have can replace the double

integral on D0 by the double integral on D. ThusZC1

Mdx+Ndy +

ZC2

Mdx+Ndy =

Z ZD

µ∂N

∂x− ∂M

∂y

¶dxdy

as claimed. ¥The above version of Green’s theorem can be generalized to "multiply connected regions"

with more than one hole by making cuts that enable us to use Green’s Theorem for a simply

connected region. For example, assume that the region D is bounded by the curves C1, C2 and

C3, with the orientations that are indicated the picture:

x

y

C1

C2C3

Figure 6

Then ZC1

Mdx+Ndy +

ZC2

Mdx+Ndy +

ZC3

Mdx+Ndy =

Z ZD

µ∂N

∂x− ∂M

∂y

¶dxdy.

Example 2 Let

F(x, y) = − y

x2 + y2i+

x

x2 + y2j

Assume that C1 is a simple closed curve that is the positively oriented boundary of the region

D and that the positively oriented (counterclockwise) circle C2 of radius a is in D.

x

y

C1

C2

Figure 7

a) Use Green’s Theorem to show thatZC1

F·dσ =ZC2

F·dσ

b) Use the result of part a) to evaluateRC1F · dσ.

Solution


a) We need to consider the negative orientation of the inner boundary C2 in order to apply

Green’s Theorem. ThusZC1

F ·dσ −ZC2

F ·dσ =Z Z

D

µ∂

∂x

µx

x2 + y2

¶− ∂

∂y

µ− y

x2 + y2

¶¶dxdy

=

Z ZD

Ã− x2 − y2(x2 + y2)

2 +x2 − y2(x2 + y2)

2

!dxdy = 0.

Therefore ZC1

F·dσ =ZC2

F·dσ,

as claimed.

b) We can parametrize C2 by σ (t) = (a cos (t) , a sin (t)), t ∈ [0, 2π]. By part a)ZC1

F·dσ =ZC2

F·dσ

=

ZC2

− y

x2 + y2dx+

x

x2 + y2dy

=

Z 2π

0

µ−a sin (t)

a2(−a sin (t)) + a cos (t)

a2(a cos (t))

¶dt

=

Z 2π

0


¢dt

=

Z 2π

0

1dt = 2π.

¤Green’s Theorem can be used to express the area of a region in terms of a line integral on its

boundary:

Proposition 1 Assume that D is the interior of C and that C is positively oriented. Then

Area of D =1

2

ZC

xdy − ydx.

Proof

1

2

ZC

xdy − ydx = 1

2

Z ZD

µ∂

∂x(x) +

∂

∂y(y)

¶dxdy

=1

2

Z ZD

(2) dxdy = Area of D.

¥

Interpretations of Green’s Theorem

In Section 14.1 we defined the curl of a two-dimensional vector field F (x, y) = M (x, y) i +N (x, y) j as

∇×F (x, y) =∇× (M (x, y) i+N (x, y) j+ 0k) =

µ∂N

∂x− ∂M

∂y

¶k.


Therefore, we can express Green’s Theorem,ZC

Mdx+Ndy =

Z ZD

µ∂N

∂x− ∂M

∂y

¶dxdy,

as ZC

F · dσ =Z Z

D

(∇×F) · k dxdy.

F

x

y

z

Figure 10

If F (x, y) is the velocity of a fluid particle at (x, y), the line integral is the circulation aroundC. Therefore, the circulation of the velocity field around C is equal to the integral

of the normal component of its curl over its interior D. We can think of the normal

component of curl of F at (x0, y0) as the average circulation of F at (x0, y0):

Proposition 2 Assume that Cr is a circle of radius r centered at (x0, y0) that is orientedcounterclockwise and Dr is the disk bounded by Cr. We have

limr→0

1

πr2

ZCr

F · dσ = (∇×F) (x0, y0) · k

(assuming that ∇×F is continuous).A plausibility argument for Proposition2:

By Green’s Theorem, ZCr

F · dσ =Z Z

Dr

∇×F · k dxdy.

By the continuity of ∇×F, if r is small,Z ZDr

∇×F · k dxdy ∼= (∇×F (x0, y0) · k) (area of Dr) = (∇×F (x0, y0) · k)¡πr2

¢.

Thus,

1

πr2

Z∂Dr

F ·dσ = 1

πr2

Z ZDr

∇×F ·k dS ∼= 1

πr2(∇×F (x0, y0) · k)

¡πr2

¢=∇×F (x0, y0) ·k.

This lends support to the assertion that

limr→0

1

πr2

Z∂Dr

F · dσ =∇×F (x0, y0) · k.

¥


Example 3 Let

F (x, y) = −ωyi+ ωxj,

so that the vector field represents circulation around the origin.

In Section 14.1 we saw that

∇×F (x, y) = 2ωk⇒∇×F (x, y) · k = 2ω.Let Cr be the circle of radius r that is centered at the origin. We have

1

πr2

ZCr

F · dσ = ω

πr2

Z 2π

0

r2 (− sin (t) i+ cos (t) j) · (− sin (t) + cos (t)) dt

=ω

π

Z 2π

0


¢dt

=ω

π(2π) = 2ω,

as well. ¤We can also interpret Green’s Theorem in terms of the flux of a vector field across

a simple closed curve C. If F (x, y) =M (x, y) i+N (x, y) j, the divergence of F at (x, y) is

∇ · F (x, y) = ∂M

∂x(x, y) +

∂N

∂y(x, y)

By Green’s Theorem,ZC

−Ndx+Mdy =Z Z

D

µ∂N

∂y+

∂M

∂x

¶dxdy =

Z ZD

∇ · F (x, y) dA

Let us reexamine the expression on the left hand side. Assume that C is parametrized by

σ (t) = (x (t) , y (t)), where t ∈ [a, b]:ZC

−Ndx+Mdy =Z b

a

(−N (x (t) , y (t))) dxdt+M (x (t) , y (t))

dy

dt)dt

=

Z b

a

(M (x, y) i+N (x, y) j) ·µdy

dti− dx

dtj

¶dt

=

Z b

a

F (x (t) , y (t)) ·µdy

dti− dx

dtj

¶dt

=

Z b

a

F (x (t) , y (t)) ·

µdy

dti− dx

dtj

¶sµ

dx

dt

¶2+

µdy

dt

¶2sµ

dx

dt

¶2+

µdy

dt

¶2dt.

Note that

n (t) =

µdy

dti− dx

dtj

¶sµ

dx

dt

¶2+

µdy

dt

¶2is orthogonal to the tangent

dσ

dt=dx

dti+

dy

dtj,


||n (t)|| = 1, and n (t) points towards the exterior of C. We will refer to n (t) as the unitnormal at (x (t) , y (t)) that points towards the exterior of C. Thus,

ZC

−Ndx+Mdy =Z b

a

F (x (t) , y (t)) ·

µdy

dti− dx

dtj

¶sµ

dx

dt

¶2+

µdy

dt

¶2sµ

dx

dt

¶2+

µdy

dt

¶2dt

=

Z b

a

F (x (t) , y (t)) · n (t) ds

=

ZC

F · nds

The quantity ZC

F · nds.is the outward flux of F across C. By Green’s Theorem,Z

C

F · nds =Z Z

D

∇ · F (x, y) dxdy

Thus, .the outward flux of F across C is equal to the integral of the divergence of F

over the interior of C.

x

y

nT F

Figure 11

We can interpret the divergence of F at (x0, y0) as the flux per unit area of F at

(x0, y0).

Proposition 3 Assume that Cr is a circle of radius r centered at (x0, y0) that is orientedcounterclockwise and Dr is the disk bounded by Cr. We have

limr→0

1

πr2

ZCr

F · nds =∇ · F (x0, y0)

(assuming that ∇ · F is continuous).A plausibility argument for Proposition2:

We have ZCr

F · nds =Z Z

Dr

∇ · F (x, y) dxdy

By continuity, if r is smallZ ZDr

∇ · F (x, y) dxdy =∇ · F (x0, y0)× (area of Dr) =∇ · F (x0, y0)×¡πr2

¢.


Therefore1

πr2

ZCr

F · nds ∼= 1

πr2∇ · F (x0, y0)×

¡πr2

¢=∇ · F (x0, y0)

if r is small. This supports the claim that

limr→0

1

πr2

ZCr

F · nds =∇ · F (x0, y0) .

¥

Irrotational and Incompressible Flow

If F is the velocity field of a fluid in motion, we say that the flow is incompressible If

∇ · F (x, y) = 0 for each (x, y) in a region D that is relevant to the flow. The flow is irro-

tational if we have ∇×F (x, y) = 0 for each(x, y) ∈ D. Assume that C is simply connected.

If C is a closed curve that lies in D and F =M i+N j, thenZC

−Ndx+Mdy =ZC

F · nds =Z Z

D

∇ · F (x, y) dA = 0.

Therefore, the vector field −N i+Mj is the gradient of a scalar function, say g (x, y). Thus,

−N (x, y) = ∂g

∂x(x, y) and M (x, y) =

∂g

∂y(x, y) .

Since we also have ∇×F (x, y) = 0 for each(x, y) ∈ D, F is the gradient of scalar function f :

M (x, y) =∂f

∂x(x, y) and N (x, y) =

∂f

∂y(x, y) .

Therefore,∂f

∂x(x, y) =

∂g

∂y(x, y) and

∂f

∂y(x, y) = −∂g

∂x(x, y)

Also note that

∂2f

∂x2+

∂2f

∂y2=

∂M

∂x+

∂N

∂y=

∂2g

∂x∂y+

∂

∂y

µ−∂g∂x

¶= 0,

∂2g

∂x2+

∂2g

∂y2=

∂

∂x(−N) + ∂M

∂y= − ∂2f

∂x∂y+

∂

∂y

µ∂f

∂x

¶= 0.

Thus, if a vector field F is irrotational and conservative, a potential function f satisfies Laplace’s

equation. Such functions are said to be harmonic. The stream function, i.e., a potential g for

−N i+Mj also satisfies Laplace’s equation.

Sufficient Conditions for the Existence of a Potential Function

In Section 14.3 we saw that the integral of a gradient field is independent of path, and that a

necessary condition for F (x, y) = M (x, y) i + N (x, y) j to be a gradient field in a region D is

that∂M

∂y(x, y) =

∂N

∂x(x, y)

for each (x, y) ∈ D. Green’s Theorem enables us to discuss the sufficiency of this condition for

a vector field to be conservative:


Theorem 2 Assume that D is a simply connected region, F (x, y) =M (x, y) i+N (x, y) j and

∂M

∂y(x, y) =

∂N

∂x(x, y)

for each (x, y) ∈ D, and that these partial derivatives are continuous on D. Then the vectorfield F is conservative.

Proof

Step 1. To begin with, let us note that the line integral of a vector field is independent

of path if its integral around any simple closed curve is 0. We will not give a proof of

this statement under general conditions. For example, if C1 and C2 join P1 to P2 as in Figure

8, then C1 − C2 is a simple closed curve.

x

y

P1 P2

C2

C1

Figure 8

We have ZC1−C2

F · dσ = 0,

so that ZC1

F · dσ −ZC2

F · dσ = 0⇒ZC1

F · dσ =ZC2

F · dσ.

Step 2. If F (x, y) =M (x, y) i+N (x, y) j and

∂M

∂y(x, y) =

∂N

∂x(x, y)

for each (x, y) ∈ D, the line integrals of F are independent of path. By Step 1, it is

sufficient to show that ZC

F · dσ = 0

if C is a simple closed curve in D,. The interior of C is entirely in D since D is assumed to be

simply connected. By Green’s Theorem,ZC

F · dσ =ZC

Mdx+Ndy =

Z ZInt(C)

µ∂N

∂x− ∂M

∂y

¶dxdy = 0.

Step 3. Let’s fix a point P0 = (x0, y0) in D and define f (P ) = f (x, y) asZC(P0,P1)

F · dσ,

where C (P0, P1) is an arbitrary curve in D that joins P0 to P . The definition of f (x, y) makessense since the line integrals of F are independent of path, by Step 2.


Claim:

F (x, y) =∇f (x, y)for each (x, y) ∈ D.

x

y

P0

P x , yx x, y

Figure 9

Let |∆x| be small enough so that any point of the form (x+∆x, y) ∈ D. By independence ofpath, we can set

f (x+∆x, y)− f (x, y) =ZC(P0,P )+C∆x

F · dσ −ZC(P0,P )

F · d σ =ZC∆x

F · σ

where C∆x is the line segment from P = (x, y) to (x+∆x, y). We can parametrize C∆x byσ (t) = (x+ t∆x,y), where 0 ≤ t ≤ 1. Then,Z

C∆x

F · dσ =ZC∆x

Mdx+Ndy =

Z 1

0

µM (x+ t∆x, y)

dx

dt+N (x+ t∆x)

dy

dt

¶dt

=

Z 1

0

M (x+ t∆x, y)∆xdt

= ∆x

Z 1

0

M (x+ t∆x, y) dt

Therefore,

lim∆x→0

f (x+∆x, y)− f (x, y)∆x

= lim∆x→0

Z 1

0

M (x+ t∆x, y) dt

= lim∆x→0

µZ 1

0

M (x+ t∆x, y)−M (x, y)

¶+M (x, y)

= lim∆x→0

µZ 1

0

M (x+ t∆x, y)−Z 1

0

M (x, y) dt

¶+M (x, y)

= lim∆x→0

Z 1

0

(M (x+ t∆x)−M (x, y)) dt+M (x, y)

=M (x, y) ,

since

lim∆x→0

Z 1

0

(M (x+ t∆x)−M (x, y)) dt =

Z 1

0

lim∆x→0

(M (x+ t∆x)−M (x, y)) dt = 0,

by the continuity of M at (x, y).


Thus, we have shown that

∂f

∂x(x, y) = lim

∆x→0f (x+∆x, y)− f (x, y)

∆x=M (x, y) .

Similarly, by considering the vertical line that connects (x, y) to (x, y +∆y), we can show that

∂f

∂y(x, y) = N (x, y) .

Therefore„

F (x, y) =M (x, y) i+N (x, y) j =∂f

∂x(x, y) i+

∂f

∂y(x, y) j =∇f (x, y) ,

as claimed. ¥

Remark 1 The assumption that D is simply connected is essential. If

F(x, y) = − y

x2 + y2i+

x

x2 + y2j,

as in Example 2, we have∂

∂y

µ− y

x2 + y2

¶=

∂

∂x

µx

x2 + y2

¶for each (x, y) 6= (0, 0). But F is not conservative in D =

©(x, y) ∈ R2: (x, y) 6= (0, 0)ª, since

we showed that ZC

F·dσ = 2π

if C is any simple closed curve that contains the origin in its interior (if F were conservative

in D such a line integral would have been 0). There is no contradiction here, since F does not

satisfy the condition∂

∂y

µ− y

x2 + y2

¶=

∂

∂x

µx

x2 + y2

¶at the origin. D has a hole at the origin, so that it is not simply connected. ♦

Problems

In problems 1-4

a) Use Green’s Theorem to express the given line integral as a double integral,

b) Evaluate the double integral.

1. ZC

y3dx− x3dy,

where C is the positively oriented boundary of the disk bounded by the circle x2 + y2 = 4.

2. ZC

xy2dx+ 2x2ydy,

where C is the positively oriented boundary of the triangular region with vertices (0, 0) , (2, 0)and (2, 4).

3. ZC

cos (y) dx+ x2 sin (y) dy,


where C is the positively oriented boundary of the rectangular region with vertices (0, 0) , (5, 0) , (5,π)and (0,π).

4. ZC

xe−2xdx+¡x4 + 2x2y2

¢dy,

where C is the positively oriented boundary of the annular region bounded by the circles x2+y2 =1 an x2 + y2 = 4.


a) Use Green’s Theorem to express ZC

F · dσ

as a double integral,

b) Evaluate the double integral.

5.

F (x, y) = y2 cos (x) i+¡x2 + 2y sin (x)

¢j

and C = C1+C2+C3, where C1 is the line segment from (0, 0) to (2, 6), C2 is the line segmentfrom (2, 6) to (2, 0), and C3 is the line segment from (2, 0) to (0, 0) (pay attention to orientation).

6.

F (x, y) =¡ex + x2y

¢i+

¡ey − xy2¢ j

and C is the positively oriented boundary of the disk bounded by the circle x2 + y2 = 25.

7. Let

F(x, y) =x

x2 + y2i+

y

x2 + y2j

Assume that C1 is a positively oriented simple closed curve such that the circle of radius 1centered at the origin is contained in its interior. CalculateZ

C1

F · dσ

Caution: Green’s Theorem is not applicable in the interior of C1 since F is not defined at the

origin.

In problems 8 and 9, make use of the appropriate form of Green’s Theorem to calculateZC

F · nds,

where n is the unit normal that points towards the exterior of C:

8.

F (x, y) = x2y3i− xyjand C is the positively oriented boundary of the triangular region with vertices (0, 0) , (1, 0)and (1, 2).

9.

F (x, y) = x3i+ y3j

and C is the positively oriented boundary of the disk bounded by the circle x2 + y2 = 4.


14.7 Stokes’ Theorem

In this section we will discuss Stokes’ Theorem which is a generalization of Green’s Theorem to

R3.

The Meaning and Plausibility of Stokes’ Theorem

Assume that M is a smooth orientable surface that is parametrized by the function Φ : D ⊂R2 → R3. Thus, N = ∂uΦ × ∂vΦ is continuous on D, and let’s assume that N specifies the

positive orientation of M . Also assume that the boundary curve ∂M is parametrized by a

function φ : J ⊂ R → R3. Informally, the boundary of M is said to be positively oriented if

a person who walks along the boundary in the direction of the tangent determined by φ and

whose head is in the direction of the normal N to the surface sees M to his/her left.

Recall that we can express Green’s Theorem as follows:ZC

F · dσ =ZC

F ·Tds =Z Z

D

(∇×F) · kdS

if C is the positively oriented boundary of the simply connected planar region D. Stokes’

Theorem generalizes Green’s Theorem to surfaces than span a curve in space:

Theorem 1 Assume thatM is a smooth orientable surface, ∂M is its positively oriented bound-

ary and F is a smooth vector field (i.e., the components of F has continuous partial derivatives).

Then,

Z∂M

F · dσ =Z∂M

F ·Tds =Z Z

M

(∇×F) · ndS

You can find a discussion of the plausibility of Stokes’ Theorem at the end of this section.

Figure 1

Example 1 Let M be the image of z = g (x, y) = 4− x2 − y2, where x2 + y2 ≤ 4, and let

F (x, y, z) = 2zi+ xj+ y2k

Use Stokes’ Theorem in order to evaluateZ ZM

(∇×F) · ndS,

where n denotes the unit normal that points upwards.

14.7. STOKES’ THEOREM 275

Figure 2

Solution

The circle x2 + y2 = 4 is the boundary of M and can be parametrized by σ (t) = 2 cos (t) i +2 sin (t) j+0k, 0 ≤ t ≤ 2π (note that σ provides the boundary ∂M with the required orientation).

By Stokes’ Theorem,Z ZM

(∇×F) · ndS =Z∂M

F · dσ

=

Z∂M

¡2zi+ xj+ y2k

¢ · dr=

Z 2π

0

¡(0) i+ 2 cos (t) j+4 sin2 (t)k

¢ · (−2 sin (t) i+ 2 cos (t) j) dt=

Z 2π

0

4 cos2 (t) dt = 4π

¤Stokes’ Theorem leads to a nice interpretation of the curl of a vector field:

Proposition 1 Assume that Π is a plane that contains the point (x0, y0, z0) and let n denotea unit vector that is orthogonal to Π. Let Dr be the disk of radius r that is in the plane Πand centered at (x0, y0, z0). Let ∂Dr be the boundary of Dr that has positive orientation withrespect to the orientation of the plane as determined by n (as in Figure 3).Then

limr→0

1

πr2

Z∂Dr

F · dσ = limr→0

1

πr2

Z∂Dr

F ·Tds = (∇×F) (x0, y0, z0) · n.

Figure 3


Note that the integralR∂Dr

F ·Tds is the circulation of the vector field around ∂Dr. Therefore,

Proposition 1 says that the limiting case of the circulation of F as the disk shrinks to the point

(x0, y0, z0) is the component of the curl of F that is normal to the plane of the disks. Thus, wecan interpret (∇×F) (x0, y0, z0) · n as the circulation of the restriction of F to that plane atthe point (x0, y, z).

A plausibility argument for Proposition1:

By Stokes Theorem, Z∂Dr

F · dσ =Z Z

Dr

∇× F · n dS.

If r is small,Z ZDr

∇×F · n dS ∼= (∇×F (x0, y0, z0) · n) (area of Dr) = (∇×F (x0, y0, z0) · n)¡πr2

¢.

Thus,

1

πr2

Z∂Dr

F·dσ = 1

πr2

Z ZDr

∇×F·n dS ∼= 1

πr2(∇×F (x0, y0, z0) · n)

¡πr2

¢=∇× F (x0, y0, z0)·n.

This supports to the assertion that

limr→0

1

πr2

Z∂Dr

F · d σ =∇×F (x0, y0, z0) · n.

¥

Remark In Section 14.1 we noted that the condition∇×F = 0 is necessary for the field to beconservative. Stokes Theorem can be used to show that the condition is also sufficient

for F to be conservative in certain types of regions. For example, convex regions are

those that have the property that any two points in the region can be connected to each other

by a line segment that stays in the region. Interiors of spheres and ellipsoids are examples of

convex regions. The condition is sufficient for F to be conservative in a convex region. ♦

The proof of the above assertion is left to a post-calculus course in vector analysis.

A Plausibility Argument for Stokes’ Theorem (Optional)

Let’s express Stokes’ Theorem in terms of the components of the vector field F and in the

differential form notations for the line and surface integrals. If

F (x, y, z) =M (x, y, z) i+N (x, y, z) j + P (x, y, z) ,

then,

∇× F =µ∂P

∂y− ∂N

∂z

¶i−

µ∂P

∂x− ∂M

∂z

¶j+

µ∂N

∂x− ∂M

∂y

¶k,Z

∂M

F · dσ =Z∂M

Mdx+Ndy + Pdz,

andZ ZM

(∇×F) · ndS =Z Z

M

µ∂P

∂y− ∂N

∂z

¶dydz +

µ∂M

∂z− ∂P

∂x

¶dzdx+

µ∂N

∂x− ∂M

∂y

¶dxdy.

14.7. STOKES’ THEOREM 277

Therefore, Stokes’ Theorem takes the following form:Z∂M

Mdx+Ndy+Pdz =

Z ZM

µ∂P

∂y− ∂N

∂z

¶dydz+

µ∂M

∂z− ∂P

∂x

¶dzdx+

µ∂N

∂x− ∂M

∂y

¶dxdy.

It is enough to show that Z∂M

Mdx =

Z ZM

∂M

∂zdzdx− ∂M

∂ydxdy

(for F =M i). The cases F = N j and F = Pk are similar.

We will assume that Φ : D ⊂ R2 → R3 parametrizes the surface M and that the restriction of

Φ to the boundary of D, to be denoted by ∂D, parametrizes ∂M with the positive orientation.

Let’s set

Φ (u, v) = (x (u, v) , y (u, v) , z (u, v))

and

F (x, y, z) =M (x, y, z) i+N (x, y, z) j + P (x, y, z) k.

We have Z∂M

Mdx =

Z∂D

M

µ∂x

∂udu+

∂x

∂vdv

¶,

and Z ZM

∂M

∂zdzdx− ∂M

∂ydxdy =

Z ZD

µ∂M

∂z

∂ (z, x)

∂ (u, v)− ∂M

∂y

∂ (x, y)

∂ (u, v)

¶dudv.

Therefore„ we need to show thatZ∂D

M

µ∂x

∂udu+

∂x

∂vdv

¶=

Z ZD

µ∂M

∂z

∂ (z, x)

∂ (u, v)− ∂M

∂y

∂ (x, y)

∂ (u, v)

¶dudv.

By Green’s Theorem,Z∂D

M

µ∂x

∂udu+

∂x

∂vdv

¶=

Z ZD

µ∂

∂u

µM

∂x

∂v

¶− ∂

∂v

µM

∂x

∂u

¶¶dudv

=

Z ZD

µµ∂M

∂u

∂x

∂v+M

∂2x

∂u∂v

¶−µ∂M

∂v

∂x

∂u+M

∂2x

∂v∂u

¶¶dudv

=

Z ZD

µ∂M

∂u

∂x

∂v− ∂M

∂v

∂x

∂u

¶dudv

=

Z ZD

µµ∂M

∂x

∂x

∂u+

∂M

∂y

∂y

∂u+

∂M

∂z

∂z

∂u

¶∂x

∂v−µ∂M

∂x

∂x

∂v+

∂M

∂y

∂y

∂v+

∂M

∂z

∂z

∂u

¶∂x

∂u

¶dudv

=

Z ZD

µ∂M

∂x

µ∂x

∂u

∂x

∂v− ∂x

∂v

∂x

∂u

¶+

∂M

∂y

µ∂y

∂u

∂x

∂v− ∂y

∂v

∂x

∂u

¶+

∂M

∂z

µ∂z

∂u

∂x

∂v− ∂z

∂u

∂x

∂u

¶¶dudv

=

Z ZD

µ∂M

∂z

∂ (z, x)

∂ (u, v)− ∂M

∂y

∂ (x, y)

∂ (u, v)

¶dudv.

¥


Problems

In problems 1 and 2 use Stokes’ Theorem to expressZ ZM

(∇×F) · ndS

as a line integral on C. Evaluate the line integral:

1.

F (x, y, z) = yi+ 2xj+ zk,

M is the graph of

z =p16− x2 − y2 where x2 + y2 ≤ 16.

(M is a hemisphere of radius 4 centered at the origin). C is the circle of radius 4 in the xy-plane

centered at the origin. M is oriented so that the normal points upward and C is oriented so

that it is traversed counterclockwise.

2.

F (x, y, z) = yi− xj+ z2k,M is the portion of the sphere

x2 + y2 + (z − 1)2 = 2above the xy-plane. C is the circle of radius 1 in the xy-plane centered at the origin. M is

oriented so that the normal points upward and C is oriented so that it is traversed counterclock-

wise.

In problems 3 and 4 use Stokes Theorem to express the line integralZC

F · dσ

as a surface integral on M . Evaluate the surface integral:

3.

F (x, y, z) = zi+ x3j+ y2k

M is the part of the plane z = 4 within the cylinder x2+y2 = 9 and the normal points upward.Cis the intersection of the plane and the cylinder. The orientation of C is counterclockwise (a

person whose head is in the direction of the normal to the plane and traverses C sees the plane

on his/her left).

4.

F (x, y, z) = zi− xj+ 2yk.M is the part of the plane z = y + 2 that is inside the cylinder x2 + y2 = 4 and the normalpoints upward. C is the intersection of the plane and the cylinder. The orientation of C

is counterclockwise (a person whose head is in the direction of the normal to the plane and

traverses C sees the plane on his/her left).

14.8 Gauss’ Theorem

The Meaning and Plausibility of Gauss’ Theorem

Gauss’ Theorem establishes an important link between the flux of a vector field across a closed

surface such a sphere and the divergence of the vector field in the interior of the surface:

14.8. GAUSS’ THEOREM 279

Theorem 1 Assume that W is a region in R3 that is bounded by the smooth orientable closedsurface ∂W . If n is the unit normal that points towards the exterior of W then

Z Z∂W

F · ndS =Z Z Z

W

∇ · FdV

Figure 1

Gauss’ Theorem is also referred to as the Divergence Theorem since the statement involves

the divergence of the vector field that is being discussed. You can find a plausibility argument

for Gauss’ Theorem at the end of this section.

Example 1 Let D be the interior of the sphere x2 + y2 + z2 = 4 and let

F (x, y, z) = 2x3i+ 2y3j+ 2z3k.

Use the Divergence Theorem in order to determineZ Z∂D

F · ndS,

where n is the unit normal to ∂D that points towards the exterior of D.

Solution

We have

(∇ · F) (x, y, z) = ∂

∂x

¡2x3

¢+

∂

∂y

¡2y3¢+

∂

∂z

¡2z3¢= 6

¡x2 + y2 + z2

¢.

Therefore, Z Z∂D

F · ndS =Z Z Z

D

(∇ · F) dxdydz

=

Z Z ZD

6¡x2 + y2 + z2

¢dxdydz

= 6

Z ρ=2

ρ=0

Z π

φ=0

Z θ=2π

θ=0

ρ2ρ2 sin (φ) dθdφdρ

= 6

Z ρ=2

ρ=0

Z π

φ=0

Z θ=2π

θ=0

ρ4 sin (φ) dθdφdρ

=768π

5


¤

Gauss’ Theorem leads to an interpretation of divergence as outward flux per unit volume at a

point:

Proposition 1 Let Dr be the ball of radius r that is centered at (x0, y0, z0) and let ∂Dr bethe sphere that bounds Dr. If V (Dr) denotes the volume of Dr then

limr→0

1

V (Dr)

Z Z∂Dr

F · ndS = (∇ · F) (x0, y0, z0)

(assuming that the divergence of F is continuous).

A plausibility argument for the Proposition:

By Gauss’ Theorem Z Z∂Dr

F · ndS =Z Z Z

Dr

∇ · FdV.

If r is small, Z Z ZDr

∇ · FdV ∼= (∇ · F) (x0, y0, z0)V (Dr)

so that1

V (Dr)

Z Z∂Dr

F · ndS ∼= (∇ · F) (x0, y0, z0) .

This makes it plausible that

limr→0

1

V (Dr)

Z Z∂Dr

F · ndS = (∇ · F) (x0, y0, z0)

¥Thus, we can interpret (∇ · F) (x0, y0, z0) as the flux per unit volume at the point (x0, y0, z0).

Example 2 (Another law by Gauss) Let W be a region in R3 and assume that the origin isnot on the boundary of W (∂W ). If r (x, y, x) = xi+ yj+ zk is the position vector of the point(x, y, z) then Z Z

∂W

r · n||r||3 dS = 4π

if the origin is in W . Otherwise the integral is 0.

Indeed, if the origin is not in W ,Z Z∂W

r · n||r||3 dS =

Z Z ZW

∇ ·Ã

r

||r||3!dV

by Gauss’ Theorem. Now,

∇ ·Ã

r

||r||3!

= ∇ ·Ã

1

(x2 + y2 + z2)3/2

(xi+ yj+ zk)

!

=∂

∂x

Ãx

(x2 + y2 + z2)3/2

!+

∂

∂y

Ãy

(x2 + y2 + z2)3/2

!+

∂

∂z

Ãz

(x2 + y2 + z2)3/2

!.


We have

∂

∂x

Ãx

(x2 + y2 + z2)3/2

!=

¡x2 + y2 + z2

¢3/2 − x (3/2) ¡x2 + y2 + z2¢1/2 (2x)(x2 + y2 + z2)

3

=

¡x2 + y2 + z2

¢1/2 ¡¡x2 + y2 + z2

¢− 3x2¢(x2 + y2 + z2)

3

=

¡−2x2 + y2 + z2¢(x2 + y2 + z2)5/2

.

Thus,

∇ ·Ã

r

||r||3!=

¡−2x2 + y2 + z2¢(x2 + y2 + z2)

5/2+

¡−2y2 + x2 + z2¢(x2 + y2 + z2)

5/2+

¡−2z2 + x2 + y2¢(x2 + y2 + z2)

5/2= 0.

Therefore, Z Z∂W

r · n||r||3 dS =

Z Z ZW

∇ ·Ã

r

||r||3!dV = 0.

If (0, 0, 0) is in W , we apply the divergence theorem to the region W 0 between ∂W and ∂Ba,

where B is the ball of sufficiently small radius a centered at (0, 0, 0):Z Z∂W

r · n||r||3 dS −

Z Z∂Ba

r · n||r||3 dS =

Z Z ZW 0∇ ·

Ãr

||r||3!dV = 0.

The (−) sign in front of the integral on the integral on the inner boundary ∂B is due to the factthat −n points towards the interior of Ba which is the exterior of W 0. Thus,Z Z

∂W

r · n||r||3 dS =

Z Z∂Ba

r

||r||3 ·ndS =Z Z

∂Ba

r

||r||3 ·r

||r||dS =Z Z

∂Ba

||r||2||r||4 dS =

1

a2

Z Z∂Ba

dS =1

a2

¡4πa2

¢= 4π.

¤

Example 3 (The Continuity Equation) Let ρ(x, y, z, t) be the density of the matter (forexample water) at a point (x, y, z) and time t. Let v (x, y, z, t) be the corresponding velocity.The amount of material in a region W at the instant t isZ Z Z

W

ρ (x, y, z, t) dV.

This changes at the rate

d

dt

Z Z ZW

ρ (x, y, z, t) dV =

Z Z ZW

∂ρ

∂tdV.

The rate at which the material flows through the boundary ∂W isZ Z∂W

ρv · ndS

If material is conserved, the amount of matter in W decreases at the rate at which matter flows

outward across ∂W . Thus Z Z ZW

∂ρ

∂tdV = −

Z Z∂W

ρv · ndS


By Gauss’ Theorem, Z Z Z∂W

ρv · ndS =Z Z Z

W

∇ · (ρv) dV

Therefore, Z Z ZW

∂ρ

∂tdV = −

Z Z ZW

∇ · (ρv) dV

for every W . This implies that∂ρ

∂t+∇ · (ρv) = 0

for each (x, y, z, t) (this can be justified by considering the equality of the integrals for a sequenceof regions that shrink towards a given point (x, y, z) at a given time t). This is the continuityequation.of continuum mechanics.

For example, if the density ρ is constant we have

∇ · v = 0.Thus, if the fluid is incompressible, the divergence of the velocity field is 0. ¤

Example 4 (Electrostatics) Gauss’ law for electrostatics can be expressed in the following

form: Z Z∂W

F · ndS = 1

∈0

Z Z ZW

ρdV.

Here ∈0is a universal constant and ρ denotes the charge density. Thus, the flux across a

closed surface is equal to the total charge in the interior. By Gauss’ Theorem,Z Z∂W

F · ndS =Z Z Z

W

∇ · FdV.

Therefore, Z Z ZW

∇ · FdV = 1

∈0

Z Z ZW

ρdV

for every W . Thus we must have

∇ · F (x, y, z) = 1

∈0 ρ (x, y, z)

at each (x, y, z). If F is conservative and −Φ is a potential so thatF = −∇Φ,

then

∇ · (∇Φ) = − ρ

∈0We have

∇ · (∇Φ) = ∆Φ,where

∆Φ =∂2Φ

∂x2+

∂2Φ

∂y2+

∂2Φ

∂z2.

Thus

∆Φ =− ρ

∈0at each point (x, y, z). This is Poisson’s Equation for −ρ/ ∈0 . ¤


A plausibility argument for Gauss’ Theorem (Optional)

In terms of the components of F =M i+N j+ Pk, and the differential form notation,Z Z∂W

F · ndS =Z Z

∂W

Mdydz +Ndzdx+ Pdxdy.

The integral over W isZ Z ZW

∇ · FdV =Z Z Z

W

µ∂M

∂x+

∂N

∂y+

∂P

∂z

¶dxdydz.

Therefore, Gauss’ Theorem reads as follows in the differential form notation:Z Z∂W

Mdydz +Ndzdx+ Pdxdy =

Z Z ZW

µ∂M

∂x+

∂N

∂y+

∂P

∂z

¶dxdydz

For example, let us confirm thatZ Z∂W

Pdxdy =

Z Z ZW

∂P

∂zdxdydz

if W is xy-simple, i.e., W can be expressed as

(x, y, z) : ϕ1 (x, y) ≤ z ≤ ϕ2 (x, y) „ for each (x, y) ∈ D,

where D is a region in the xy-plane. In this case,Z Z ZW

∂P

∂zdxdydz =

Z ZD

ÃZ z=ϕ2(x,y)

z=ϕ1(x,y)

∂P

∂z(x, y, z) dz

!dxdy

=

Z ZD

(P (x, y,ϕ2 (x, y))− P (x, y,ϕ1 (x, y))) dxdy,

by the Fundamental Theorem of Calculus. Note thatZ ZD

(P (x, y,ϕ2 (x, y))− P (x, y,ϕ1 (x, y))) dxdy

=

Z ZD

P (x, y,ϕ2 (x, y)) dxdy −Z Z

D

P (x, y,ϕ1 (x, y)) dxdy,

The first integral is the part of the surface integralZ Z∂W

Pdxdy

that corresponds to the upper part of the surface. Indeed, that part of ∂W is the graph G2 of

the function ϕ2 on the region D, the outward unit normal is k, andZ ZG2

Pdxdy =

Z ZG2

P i · kdS =Z ZD

P i · kdxdy =Z Z

D

P (x, y,ϕ2 (x, y)) dxdy.

Similarly, the integral

−Z Z

D

P (x, y,ϕ1 (x, y)) dxdy


is the part of the surface integral Z Z∂W

Pdxdy

that corresponds to the lower surface that is the graph G1 of the function ϕ1. There, the outward

unit normal is −k, andZ ZG1

Pdxdy =

Z ZG1

P i · (−k) dS = −Z Z

D

P i · kdxdy = −Z Z

D

P (x, y,ϕ2 (x, y)) dxdy.

The part of the integral Z Z∂W

Pdxdy

that corresponds to the lateral boundary S of W add up to 0, sinceZ ZS

Pdxdy =

Z ZS

P i · ndS,

where the unit normal n is orthogonal is parallel to the xy-plane. In particular i · n = 0. Thus,Z ZS

P i · ndS.

Therefore, we confirmed the equalityZ Z ZW

∂P

∂zdxdydz =

Z Z∂W

Pdxdy.

Problems

In problems 1-4 evaluate the flux Z ZM

F · ndS,

where M is a closed surface that encloses the region W by using Gauss’ Theorem (n points

towards the exterior of W ):

1.

F (x, y, z) = xi+ yj+ zk,

M is the sphere of radius 4 centered at the origin, and W is the interior of M .

2.

F (x, y, z) = z3k

M is the sphere of radius 2 centered at the origin, and W is the interior of M .

3.

F (x, y, z) = xi+ yj+ zk,

and M is the surface of the cube

W = (x, y, z) : 0 ≤ x ≤ 4, 0 ≤ y ≤ 4, 0 ≤ z ≤ 4 .4.

F (x, y, z) = x2i

and M is the surface of the cube

W = (x, y, z) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 .

Appendix K

Answers to Some Problems

Answers of Some Problems of Section 11.1

1.

If x = c then2c+ y + 4z = 8⇒ y + 4z = 8− 2c.

These are lines.

If y = c then2x+ c+ 4z = 8⇒ 2x+ 4z = 8− c.

These are lines.

If z = c then2x+ y + 4c = 8⇒ 2x+ y = 8− 4c.

These are lines.

3.

If x = c thenz = 4c2 + 9y2

285

286 APPENDIX K. ANSWERS TO SOME PROBLEMS

These are parabolas.

If y = c thenz = 4x2 + 9c2

These are parabolas.

If z = c thenc = 4x2 + 9y2

These are ellipses if c > 0. If c = 0 the curve is reduced to a single point, the origin. The surfacedoes not have any points below the xy-plane corresponding to c < 0.

5.

If x = c thenc2 + 2y2 + 4z2 = 4⇒ 2y2 + 4z2 = 4− c2.

These are ellipses if −2 < c < 2. The curve is reduced to a single point if c = ±2. The surfacedoes not have any points corresponding to x < −2 or x > 2.If y = c then

x2 + 2c2 + 4z2 = 4⇒ x2 + 4z2 = 4− 2c2

These are ellipses if −√2 < c < √2. The curve is reduced to a single point if c = ±√2. Thesurface does not have any points corresponding to y >

√2 or y < −√2.

If z = c thenx2 + 2y2 + 4c2 = 4⇒ x2 + 2y2 = 4− 4c2

These are ellipses if −1 < c < 1. The curve is reduced to a single point if c = ±1. The surfacedoes not have any points corresponding to z > 1 or z < −1.7.

If x = c thenc2 − 9y2 − 4z2 = 1⇒ 9y2 + 4z2 = c2 − 1

These are ellipses if c < −1 or c > 1. The curve is reduced to a single point if c = ±1. Thesurface does not have any points corresponding to −1 < x < 1.

287

If y = c thenx2 − 9c2 − 4z2 = 1⇒ x2 − 4z2 = 1 + 9c2

These are hyperbolas.

If z = c thenx2 − 9y2 − 4c2 = 1⇒ x2 − 9y2 = 1 + 4c2


9.

If x = c theny2 − c2 + z2 = 1⇒ y2 + z2 = 1 + c2

These are circles.

If y = c thenc2 − x2 + z2 = 1⇒ −x2 + z2 = 1− c2


If z = c theny2 − x2 + c2 = 1⇒ y2 − x2 = 1− c2



1.

a) v = (2, 3)

b) Q2 = (4, 4)

1 2 3 4x

1

2

3

4

5

y

P1

P2

Q1

Q2

3.

a) v =(2, 5)

b) Q2 = (5, 7)


2 3 4 5x

-3

2

7

y

P1

P2Q1

Q2

5.

a) v+w = (5, 5)b)

2 3 5

1

4

5

v

wv+w

7.

a) v+w = (0, 3)b)

-2 -1 2

-1

3

4

v

w

v+w

9.

a) v+w = (1, 6)b)

-1 2

2

4

6

v

w

v+w

289

11.

a) v −w = (4, 2)

2-2 4

4

2 v

w

v- w

13. 2v− 3w = (12,−17)15.

a)

u =

µ3

5,4

5

¶.

b)

3

4

v

u

17.

a) v = 3i+ 2j, w = −2i+4jb) 2v − 3w = 12i− 8j19.

a) v = −2i+ 3j+ 6k, w = 4i− 2j+ kb) −v+ 4w =18i− 11j− 2k


1.

a)

||v|| = 1, ||w|| =√2, v ·w = 1.

b)

θ =π

4.

3.

a)

||v|| =√2, ||w|| =

√2, v ·w =

√3

b)

θ =π

6

5.

a)

||v|| =√2, ||w|| =

√2, v ·w = 1

b)

θ =π

3.

7.


a)

||v|| =√5, ||w|| =

√2, v ·w = −1

b)

cos (θ) = − 1√10.

c)

θ ∼= 1. 892 559.

a)

||v|| =√5, ||w|| =

√10, v ·w = −1

b)

cos (θ) = − 1√50.

c)

θ ∼= 1. 712 6911.

a)

u =

µ2√13,3√13

¶b) The direction cosines of v are

2√13and

3√13

13.

a)

u =

µ−35,4

5

¶

b) The direction cosines of v are

−35and

4

5

15.

a)

u =

µ6√40,2√40

¶b)

compwv =26√40

c)

Pwv =

µ39

10,13

10

¶d)

v2 =

µ− 910,27

10

¶17.

a)

u =

µ2√5,− 1√

5

¶b)

compwv =4√5

c)

Pwv =

µ8

5,−45

¶d)

v2 =

µ−7,−6

5

¶Answers of Some Problems of Section 11.4

1.

v ×w = 5k3.

v×w = −11i− 11j+ 11k5. 12

7. 7

9.

3x+ 2y + z = 13

11.

−x+ y + 2z = −813.

−5 + 2y + 5z = 9


1.

a)

σ0 (t) = i+ 8 sin (4t) cos (4t) j, σ0³π3

´= i+ 2

√3j, T

³π3

´=

1√13i+

2√3√13j

b)

291

L (u) =

µπ

3+ u,

3

4+ 2√3u

¶, −∞ < u ≤ +∞.

3.

a)

σ0 (t) =

Ã−3t2 + 3(t2 + 1)

2 ,6t

(t2 + 1)2

!, σ0 (1) =

µ0,3

2

¶, T (1) = j

b)

L (u) =

µ3

2,3

2+3

2u

¶, u ∈ R.

5.

v (t) = σ0 (t) = e−t (cos (t)− sin (t)) i− e−t (cos (t) + sin (t)) jv (π) =

¡−e−π, e−π¢ = −e−πi+ e−πjSpeed at t = π is

||v (π)|| =√2e−π.


1.

a (t) = −48 cos (4t) i− 48 sin (4t) j

a (π/12) = −24i− 24√3j

3.

a (t) = − 2t

(t2 + 1)2 j.

a (1) = −24j = −1

2j.

5.

T (π/6) =1

2

³−1,√3´.

N (π/6) = −12

³√3, 1´.

7.

a)

s (t) =

Z t

0

q4 sin2 (τ) + 9 cos2 (τ)dτ .

b)

s (π) ∼= 7. 932 729.

κ (π/2) =3

4.

11. The tangential component of the acceleration is always 0. The normal component of the

acceleration is 8.



In the plots of the level curves for problems 1-5, the smaller values of f are indicated by the

darker color.

1.

a)

b)

3.

a)

b)

5.

293

a)

b)

7.

The level surfaces are spheres. In the pictures, the outer sphere is shown partially in order to

show the inner sphere.

9. The pictures show two level surfaces of f .



1.∂

∂x

¡4x2 + 9y2

¢= 8x,

∂

∂y

¡4x2 + 9y2

¢= 18y

3.∂

∂x

p2x2 + y2 =

2xp2x2 + y2

,∂

∂y

p2x2 + y2 =

yp2x2 + y2

5.∂

∂xe−x

2−y2 = −2xe−x2−y2 , ∂

∂ye−x

2−y2 = −2ye−x2−y2

7.∂

∂xln¡x2 + y2

¢=

2x

x2 + y2,

∂

∂yln¡x2 + y2

¢=

2y

x2 + y2

9.

∂

∂xarctan

Ãyp

x2 + y2

!= − xy

(x2 + 2y2)px2 + y2

,

∂

∂yarctan

Ãyp

x2 + y2

!=

x2

(x2 + 2y2)px2 + y2

11.

∂

∂x

Ã1p

x2 + y2 + z2

!= − x

(x2 + y2 + z2)3/2

Similarly,

∂

∂y

Ã1p

x2 + y2 + z2

!= − z

(x2 + y2 + z2)3/2

13.

∂

∂xarcsin

Ã1p

x2 + y2 + z2

!= − xp

x2 + y2 + z2 − 1 (x2 + y2 + z2)15.

a)

i+3√10k and j+

1√10k

b)

L1 (u) =

µ3 + u, 1,

√10 +

3√10u

¶,

295

and

L2 (u) =

µ3, 1 + u,

√10 +

1√10u

¶, u ∈ R.

17.

a)

i+ 2ek and j

b)

L1 (u) = (1 + u, 0, e+ 2eu) ,

and

L2 (u) = (1, u, e) , u ∈ R.

19.∂f

∂x(x, y) = 8x,

∂f

∂y(x, y) = 18y,

∂2f

∂x2(x, y) = 0

21.

∂f

∂x(x, y) = −2xe−x2+y2 , ∂2f

∂x2(x, y) = −2e−x2+y2 + 4x2e−x2+y2 , ∂2f

∂y∂x(x, y) = −4xye−x2+y2

23.∂f

∂z(x, y, z) = xyex−y+2z + 2xyzex−y+2z,

∂2f

∂x∂z(x, y, z) = ex−y+2z (y + xy + 2yz + 2xyz)

∂2f

∂y∂z(x, y, z) = ex−y+2z (x− xy + 2xz − 2xyz)

25.

fxy (x, y) = − 16xy

(x2 + 4y2)2 = fyx (x, y)


1.

a)

L (x, y) = 125 + 45 (x− 3) + 60(y − 4).b)

f (3.1, 3.9) ∼= L (3.1, 3.9) = 123.5c) According to a calculator, f (3.1, 3.9) ∼= 123. 652. The absolute error is

|f (3.1, 3.9)− 123.5| ∼= 0.152.

3.

a)

L (x, y) =π

6− 14

³x−√3´+

√3

4(y − 1) .


b)

f (1.8, 0.8) ∼= L (1.8, 0.8) ∼= 0.420 009c) According to a calculator f (1.8, 0.8) ∼= 0.418 224. The absolute error is

|f (1.8, 0.8)− L (1.8, 0.8)| ∼= 1.8× 10−3

5.

a)

L (x, y, z) =√14 +

1√14(x− 1) + 2√

14(y − 2) + 3√

14(z − 3) .

b)

f (0.9, 2.2, 2.9) ∼= L (0.9, 2.2, 2.9) ∼= 3. 741 66c). According to a calculator f (0.9, 2.2, 2.9) ∼= f (0.9, 2.2, 2.9). The absolute error is

|f (0.9, 2.2, 2.9)− L (0.9, 2.2, 2.9)| ∼= 8× 10−3

7.

a)

df =yz

2√xyz

dx+xz

2√xyz

dy +xy

2√xyz

dz.

b)

f (0.9, 2.9, 3.1)− f (1, 3, 3) ∼= df (1, 3, 3,−0.1,−0.1, 0.1) = −0.15Therefore,

f (0.9, 2.9, 3.1) ∼= f (1, 3, 3)− 0.15 = 3− 0.15 = 2. 85c) According to a calculator f (0.9, 2.9, 3.1) ∼= 2. 844 47. The absolute error is

|f (0.9, 2.9, 3.1)− 2. 85| ∼= 5.5× 10−3


1.

a)

d

dtf (x (t) , y (t)) =

dz

dt=

∂z

∂x

dx

dt+

∂z

∂y

dy

dt

=

Ãxp

x2 + y2

!cos (t) +

Ãyp

x2 + y2

!(−2 sin (t))

= − 3 sin (t) cos (t)qsin2 (t) + 4 cos2 (t)

b) We have

f (x (t) , y (t)) = f (sin (t) , 2 cos (t)) =

qsin2 (t) + 4 cos2 (t).

Therefore,

d

dtf (x (t) , y (t)) =

d

dt

qsin2 (t) + 4 cos2 (t) = − 3 sin (t) cos (t)q

sin2 (t) + 4 cos2 (t)

3.

297

a)

∂

∂uf (x (u, v)) =

∂z

∂u=dz

dx

∂x

∂u=

µ1

x

¶Ã− u

(u2 + v2)3/2

!= − u

u2 + v2

and

∂

∂vf (x (u, v)) =

∂z

∂v=dz

dx

∂x

∂v=

µ1

x

¶Ã− v

(u2 + v2)3/2

!= − v

u2 + v2.

b) We have

f (x (u, v)) = f

µ1√

u2 + v2

¶= ln

³pu2 + v2

´.

Therefore.∂

∂uf (x (u, v)) =

∂

∂uln

µ1√

u2 + v2

¶= − u

u2 + v2,

and∂

∂vf (x (u, v)) =

∂

∂vln

µ1√

u2 + v2

¶= − v

u2 + v2.

5.

a)

∂

∂uf (x (u, v) , y (u, v)) =

∂z

∂u=

dz

dx

∂x

∂u+dz

dy

∂y

∂u

=

µ− y

x2 + y2

¶cos (v) +

µx

x2 + y2

¶sin (v) = 0,

∂

∂vf (x (u, v) , y (u, v)) =

∂z

∂v=

dz

dx

∂x

∂v+dz

dy

∂y

∂v

=

µ− y

x2 + y2

¶(−u sin (v)) +

µx

x2 + y2

¶(u cos (v)) = 1.

b) We have

f (x (u, v) , y (u, v)) = f (u cos (v) , u sin (v)) = arcsin (sin (v)) = v.

Therefore,∂

∂uf (x (u, v) , y (u, v)) = 0 and

∂

∂vf (x (u, v) , y (u, v)) = 1.

9.

b)

u (x, 0) = sin (x) , u (x, 2) = cos (x) , u (x, 4) = − cos (x)c)

sin(x)

-10 -5 5 10

-1.0

-0.5

0.5

1.0

x

y

cos(x)

-10 -5 5 10

-1

1

x

y

-cos(x)

-10 -5 5 10

-1

1

x

y


11.

a)∂z

∂x= −x

z,∂z

∂y=y

z

b)

∂z

∂x

¯x=3,y=6,z=6

= −36= −1

2,∂z

∂y

¯x=3,y=3,z=6

=6

6= 1.

Therefore, the tangent plane is the graph of the equation

z = 6− 12(x− 3) + (y − 6) .

z

6

3

y6 x


1.

a)

∇f (x, y) = 8xi+ 18yjb)

Duf (3, 4) =24√5

5

3.

a)

∇f (x, y) = 2xex2−y2i− 2yex2−y2jb)

Duf (2, 1) = −√10e3

5.

a)

∇f (x, y, z) = 2xi− 2yj+ 4zk.b)

Duf (1,−1, 2) = 8√3

7.

a)

v = ∇f (2, 3) = − 2

133/2i− 3

133/2j

299

The corresponding rate of increase of f is

||v|| = 1

13

b)

w = −∇f (2, 3) = 2

133/2i+

3

133/2j

The corresponding rate of decrease of f is 1/13.

9.

a)

d

dtf (σ (t))

¯t=π/6

= ∇f³σ³π6

´´· dσdt

³π6

´=³2√3e2i− 2e2j

´·³−i+

√3j´= −4

√3e.

b)

Ddσ/dtf (σ (π/6)) = −2√3e2

11.

a) Let f (x, y) = 2x2 + 3y2.∇f (x, y) = 4xi+ 6yj.∇f (2, 3) = 8i+ 18j

is orthogonal to the curve f (x, y) = 35 at (2, 3).

b) The tangent line is the graph of the equation

8 (x− 2) + 18 (y − 3) = 0

13.

a) Let

f (x, y) = e25−x2−y2 .

∇f (3, 4) = −6i− 8jis orthogonal to the curve f (x, y) = 1 at (3, 4).

b) The tangent line is the graph of the equation

−6 (x− 3)− 8 (y − 4) = 0.15.

a) Let f (x, y, z) = x2 − y2 + z2.∇f (2, 2, 1) = 4i− 4j+ 2k

is orthogonal to the surface at (2, 2, 1).b) The plane that is tangent to the surface at (2, 2, 1) is the graph of the equation

4 (x− 2)− 4 (y − 2) + 2 (z − 1) = 0.17.

Let f (x, y, z) = x− sin (y) cos (z).∇f (1,π/2, 0) = i

is orthogonal to the surface at (1,π/2, 0).b) The plane that is tangent to the surface at (1,π/2, 0) is the graph of the equation

x− 1 = 0⇔ x = 1.



1.(−1/5,−3/5) is the only critical point. The function has a saddle point at (−1/5,−3/5) .3. Any point on the line y = −x is a critical point. The function attains its absolute maximumor minimum on the line y = −x.5. (−18/5,−11/15).is the only critical point. The function has a local (and absolute) minimumat (−18/5,−11/15).7. The critical points are (0, 0), (1,−1) and (−1, 1). The function has a saddle pioint at (0, 0),local maxima at (1,−1) and (−1, 1).


1. The maximum value of f on the circle x2 + y2 = 4 is f¡√2,√2¢= 2√2 and the minimum

value is f¡−√2,−√2¢ = −2√2.

3. The minimum value of f (x, y) subject to 4x2 + y2 = 8 is -−2, the maximum value is 2.

5. The minimum value of f in D is 0, and its maximum value in D is 8.


1.81

23.

1

3e12 − e4 − 1

3e6 + e2

5.8

3ln (2)

7.π

12


1. The integral is 32.

3.

2

4

x

y

The integral is 64/3.5.

1

1

x

y

The integral is1

2e− 1

7.

y

-2

01

2

0

1

x

-1

0

2z

4

6

-1 0 1

1

x

y

The integral is

76

359.

301

02

1

6

z

4

x

2

00

3

y

21

2

3

x

y

The integral is 6.

11.

a)

b) The integral is

√2

4


1.

a)

4x

4

y

b) The integral is 32.

3.

a)

3

-3

3

x

y

b) The integral is 9π.

5.

a)


1 2

1

2

x

y

b) The integral is

3π2

64

7.

a)

1 2 3

-1

1

x

y

b) The area of D is π/4.

9. The volume of D is4π

3123/2

11.

a)

b) The integral is 2/3.


1. The volume of the region is 81π.3. Z Z Z

D

x2dxdydz =1

3

5. Z Z ZD

zex+ydxdydz = 2 (e− 1)2 .

7. Z Z ZD

zdxdydz =1

2

Z ZR

dxdy =π

8.

303

9. Z Z ZD

xydxdxydz =3

28


1. ³√2,√2, 1´

3. Ã−32,3√3

2, 4

!5.

r =√2, θ =

3π

4, z = 4

7.

r =√2, θ = −π

4, z = 2

9. Z Z ZD

px2 + y2dxdydz = 384π

11. Z Z ZD

ezdxdydz = π¡e6 − 5− e¢

13. The volume of the region is4π

3

³8− 33/2

´15. Ã√

2

2,

√6

2,√2

!17. ³

−√6,√6, 2´

19.

ρ =√2, φ =

3π

4, θ = −π

221.

ρ = 2√2, φ =

π

6, θ =

3π

423. Z Z Z

D

zdxdydz =15

16π

25. Z Z ZD

x2dxdydz =1562

15π

27. Z Z ZD

dxdydz = 9³√3− 1

´π

29. The volume of the region D is

8√2

3π



1.

a)

∇ · F (x, y, z) = yzb)

∇× F (x, y, z) = −x2i+ 3xyi− xzk3.

a)

∇ · F (x, y, z) = 0.b)

∇×F (x, y, z) = (−x cos (xy) + x sin (xz)) i+ y cos (xy) j− z sin (xz)k5.

a)

∇ · F (x, y, z) = 2y + 2zb)

∇×F (x, y, z) = 0


1. ZC

y3ds =1

54

³1453/2 − 1

´3. Z

C

xeyds =e5

2− e2

5. ZC

(x− 2) e(y−3)(z−4)ds =√14

12

¡e6 − 1¢

7. ZC

¡x2i+ y2j

¢ · dσ = −π2

9. ZC

(cos (y) i+ sin (z) j+ xk) · dσ = 3

4

11. ZC

(−2xyi+ (y + 1) j) ·Tds = 28

3

13.

a) ZC

F·d σ =ZC

−xydx+ 1

x2 + 1dy.

b) ZC

−xydx+ 1

x2 + 1dy = ln

µ2

17

¶+255

4

15. ZC

3x2dx− 2y3dy = −32

17. ZC

− sin (x) dx+ cos (x) dy = −6.

305


1.

b)

f (x, y) = x2 − 3yx+ g (y) = x2 − 3yx+ 2y2 − 8y +Kis a potential for F (K is an arbitrary constant).

3.

b)

f (x, y) = ex sin (y) +K

is a is a potential for F (K is an arbitrary constant).

5.

a)

f (x, y) =1

2x2y2

is a potential for F.

b) ZC

F · dσ = 2.7.

a)

f (x, y, z) = xyz + z2

is a potential for F.

b) ZC

F · dσ = 779. Z

C

y2

1 + x2dx+ 2y arctan (x) dy =

ZC

df,

where

f (x, y) = y2 arctan (x) .ZC

y2

1 + x2dx+ 2y arctan (x) dy = π.


1.

a)

N³π6, 1´=3√3

2j+

3

2k

b)

3√3

2

Ãy − 3

√3

2

!+3

2

µz − 3

2

¶= 0

3.

a)

N³2,π

4

´= 8√2i−

³3√2 + 6

´j+ 12k

b)

8√2³x− 3

√2´−³3√2 + 6

´(y − 4) + 12

³z − 2

√2´= 0.


5.

a)

N³2,π

3

´= −√3

4i−34j+

1

2k

b)

−√3

4

µx− 1

2

¶− 34

Ãy −√3

2

!+1

2

³z −√3´= 0.

7.

a)

N³1,π

6

´= −e2i+ 1

2ej+

√3

2k

b)

−e2 (x− 1) + 12e

µy − 1

2e

¶+

√3

2

Ãz −√3

2e

!= 0.


1 The area of the surface is 48π.

3. The area of the surface is 8π.

5 The area of the surface is

2π

µ1

2e2p(e4 + 1) +

1

2arcsinh

¡e2¢− 1

2

√2− 1

2arcsinh (1)

¶7. Z Z

S

y2dS = 8π

9. Z ZS

xdS = −4π

13. ZS

xi+ yj+ zk

(x2 + y2 + z2)· dS = 12π

15. Z ZS

(xi+yj+ zk) · dS =16π


1.

a) ZC

y3dx− x3dy = −3Z Z

D

¡x2 + y2

¢dxdy

b)

−3Z Z

D

¡x2 + y2

¢dxdy = −24π

3.

307

a) Let D be the rectangular region with vertices (0, 0) , (5, 0) , (5,π) and (0,π). ThenZC

cos (y) dx+ x2 sin (y) dy =

Z ZD

(2x+ 1) sin (y) dxdy

b) Z ZD

(2x+ 1) sin (y) dxdy = 60

5.

a) Let D be the triangular region with vertices (0, 0) , (2, 6) and (2, 0).ZC

F · dσ = −Z Z

D

2xdxdy

b)

−Z Z

D

2xdxdy = −16

7. ZC1

F · dσ = 0.

9. ZC

F · nds = 24π


1.

16π

3.243

4π


1. 3

3. 192

Appendix L

Basic Differentiation and

Integration formulas

Basic Differentiation Formulas

1.d

dxxr = rxr−1

2.d

dxsin (x) = cos (x)

3.d

dxcos (x) = − sin(x)

4.d

dxsinh(x) = cosh(x)

5.d

dxcosh (x) = sinh(x)

6.d

dxtan(x) =

1

cos2(x)

7.d

dxax = ln (a) ax

8.d

dxloga (x) =

1

x ln (a)

9.d

dxarcsin (x) =

1√1− x2

10.d

dxarccos (x) = − 1√

1− x2

11.d

dxarctan(x) =

1

1 + x2

Basic Antidifferentiation Formulas

C denotes an arbitrary constant.

1.

Zxrdx =

1

r + 1xr+1 + C (r 6= −1)

2.

Z1

xdx = ln (|x|) + C

3.

Zsin (x) dx = − cos (x) + C

4.

Zcos (x) dx = sin (x) + C

5.Rsinh(x)dx = cosh(x) + C

6.Rcosh(x)dx = sinh(x) + C

7.

Zexdx = ex + C

8.

Zaxdx =

1

ln (a)ax + C (a > 0)

9.

Z1

1 + x2dx = arctan (x) + C

10.R 1√

1− x2 dx = arcsin (x) + C

309

Index

Absolute extrema, 122

Acceleration, 46

Arc length, 52

Binormal, 49

Cartesian Coordinates, 1

Chain rule, 85

Circulation of a vector field, 268

Conservative vector fields, 215

Continuity, 66

Continuity equation, 283

Curvature, 53

radius of curvature, 54

Cylindrical coordinates, 169

Differential, 80

Directional derivatives, 94

Distance traveled, 52

Divergence Theorem, 281

Double integrals, 133, 140

double integrals in polar coordinates, 146

Fubini’s Theorem, 136

Riemann sums, 134

Doubly connected region, 265

Flux across a curve, 270

Flux integrals, 249

differential form notation, 255

Gauss’ Theorem, 280

Gradient, 97

chain rule and the gradient, 101

gradient and level curves, 101

gradient and level surfaces, 104

Green’s Theorem, 261

Implicit differentiation, 90

Incompressible flow, 271

Irrotational flow, 271

Jacobian, 183

Lagrange multipliers, 124

Level curves, 62

Level surface, 64

Limit, 66

Line integral of a vector field, 200

Line integrals, 196

differential form notation, 204

fundamental theorem of line integrals, 212

independence of path, 213

line integrals of conservative fields, 212

Line integrals of scalar functions, 196

Linear approximations, 75

Mass, 154

mass density, 154

Maxima and minima, 107

discriminant, 112

Second derivative test, 112

Mobius band, 230

Moments, 155

center of mass, 155

Moving frame, 48

Normal distribution, 158

Orientation of a curve, 197

Parametrized curves, 35

derivative of a vector-valued function, 40

tangent line, 42

tangent vectors, 40

unit tangent, 42

Parametrized surfaces, 222

normal vectors, 227

orientation, 230

tangent planes, 227

Partial derivatives, 65

higher-order partial derivaives, 71

Planes

normal vactor, 31

Potential, 215

Potential function

existence of a potential, 271

Principal normal, 49

Probability, 157

310

INDEX 311

joint density function, 157

Projection, 23

Random variables, 157

Real-valued functions

graphs, 61

Real-valued functions of several variables, 61

Simple closed curve, 261

Simply connected region, 261

Spherical coordinates, 172

Stokes’ Theorem, 276

Surface area, 241

Surface integrals of scalar functions, 246

Surfaces of revolution, 226

Tangent plane

tangent plane to a graph, 75

Transformations, 182

Triangle Inequality, 19

Triple Integrals

change of variables, 181

triple integrals in Cartesian coordinates,

159

Triple integrals

triple integrals in cylindrical coordinates,

170

triple integrals in spherical coordinates, 174

Vectors, 6

addition, 8

angle between vectors, 20

component, 24

cross product, 27

direction cosines, 23

dot product, 18

length, 6

linear combination, 12

normalization, 12

orthogonal vectors, 22

orthogonality, 20

scalar multiplication, 8

scalar triple product, 30

standard basis vectors, 13

subtraction, 11

unit vector, 12

Velocity, 45, 46

Work, 200

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