calculus2 - mecs.jp

Calculus2

Hisashi Yokota

Contents

4 Partial Differentiation 34.1 Function of Several Variavles . . . . . . . . . . . . . . . . . . . . 34.2 Limit of Functions of Two Variables . . . . . . . . . . . . . . . . 74.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . 104.4 Total Differential and Tangent Plane . . . . . . . . . . . . . . . . 13

4.4.1 Tangent Plane and Normal Vector . . . . . . . . . . . . . 164.5 Partial Differentiation of Composite Functions . . . . . . . . . . 184.6 Higher Order Partial Derivatives . . . . . . . . . . . . . . . . . . 214.7 Extreme Values of Function of Two Variables . . . . . . . . . . . 24

4.7.1 Taylor Theorem for Two Variables . . . . . . . . . . . . . 254.8 Conditional Extrema . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.8.1 Implicit Functions . . . . . . . . . . . . . . . . . . . . . . 294.9 Lagrange Multiplier . . . . . . . . . . . . . . . . . . . . . . . . . 33

5 Multiple Integrals 375.1 Double Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 375.2 Repeated Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . 395.3 Change of Variables . . . . . . . . . . . . . . . . . . . . . . . . . 445.4 Improper Double Integrals . . . . . . . . . . . . . . . . . . . . . . 495.5 Application of Double Integrals . . . . . . . . . . . . . . . . . . . 53

5.5.1 Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535.5.2 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . 545.5.3 Volume of Solid . . . . . . . . . . . . . . . . . . . . . . . . 56

5.6 Triple Integrala . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

A ANSWERS 71

B ANSWERS B 81

1

Chapter 4

Partial Differentiation

4.1 Function of Several Variavles

Function of Two Variables Let D be a subset of R2. Then for each point (x, y) in D, there is a rule fsuch that there exists a unique z corresponding to (x, y). We call this rulef a function of two variables and denote by z = f(x, y).

NOTE Let D be a domian. x and y are called independent variable and zis called dependent variable. A function z = f(x, y) is called a function oftwo variables of x, y. The domain of f is the set of variables of x, y for whichf(x, y) is also a real.

Domain The domain of z =f(x, y)is given by D(f) =(x, y) ∈ R2 : f(x, y) ∈R.

Range The range of z = f(x, y)is given by R(f) = z ∈R : z = f(x, y), (x, y) ∈D(f).

Example 4.1 Find the domain of the following function

z = f(x, y) = 1− (x2 + y2)

Example4-1

Note that if x is a real num-ber, then x2 is also a real num-ber. Similarly, if y is a realnumber, then y2 is a real num-ber. Furthermore, sum, differ-ence, product, and quotient ofnon-zero real numbers are realnumbers, Thus the domain ofz = f(x, y) is xy-plane.

SOLUTION The domain of f(x, y) is the set of real numbers (x, y) such that1− (x2 + y2)is a real number. Thus

D(f) = (x, y) ∈ R2 : 1− (x2 + y2) ∈ R= (x, y) ∈ R2 = R2

Exercise 4.1 Find the domain of the following function.

z = f(x, y) =√1− (x2 + y2)

SOLUTION Note that√

1− (x2 + y2) is real if and only if 1 − (x2 + y2) ≥ 0.Thus

D(f) = (x, y) ∈ R2 :√

1− (x2 + y2) ∈ R= (x, y) ∈ R2 : x2 + y2 ≤ 1

Graph of Functions

For the function z = f(x, y), the set of points (x, y, z) such that (x, y, z) :z = f(x, y) is called graph. Thus the graph of a function of two variablesis a surface.

3

4 CHAPTER 4. PARTIAL DIFFERENTIATION

NOTE A surface is a collection of points. But it is not easy to draw a surfaceby plotting points. Then to draw the surface of a function of z = f(x, y), weuse the following techniques. If we look at the surface of a function from thedirection of x-axis, then we can only see the curve on yz-plane. If we look at thesurface of a function from the direction of y-axis, then we can only see the curveon xz-plane. From these observation, we can draw the surface of a function bydrawing the curve of a function z = f(0, y) on the yz-plane and the curve of afunction z = f(x, 0) on xz-plane. Finally, we let z = c and draw the curve onthe plane parallel to the xy-plane.

Graph of Function A graph of a function ortwo variables is given byG(f) = (x, y, z) : z =f(x, y), (x, y) ∈ D(f).

Contour A line of intersection of a plane z = c and z = f(x, y) is called contour orlevel curve.

NOTE Let (x, y) be a position of object on the xy-plane and z = f(x, y) be theatomospheric pressure at the point. Then z = c is the plane whose atomosphricpressure is c. Thus f(x, y) = c is the level curve of the atomospheric pressure c.

Example 4.2 Sketch the graph of the following function.

z = f(x, y) = 1− (x2 + y2)

Example4-2

SOLUTION Let x = 0 to get z = 1− y2. Then we have a parabola on yz-plane.We next let y = 0 to get z = 1 − x2. Then we have a parabola on xz-plane.Finally let z = c to get x2 + y2 = 1 − c. Thus we have a circle with the radius√1− c on z = c

Exercise 4.2 Sketch the graph of the following functions.1. z = f(x, y) =

√1− (x2 + y2) 2. z = f(x, y) = x2 − y2.

Exercise4-2-1

Exercise4-2-2

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SOLUTION 1. We can sketch the graph of z = f(x, y) =√1− (x2 + y2) by

squaring both sides to get z2 = 1 − (x2 + y2). Then htis is a sphere with theradius 1. x2 + y2 + z2 = 1. Since z ≥ 0, we have upper semisphere. 2. Let x = 0. Then z = f(0, y) = −y2. Thus we have a concave up parabolaon yz-plane. Let y = 0. Then z = f(x, 0) = −x2. Thus we have concave downparabola on xz-plane. Finally let z = c. Then we have hyperbola x2 − y2 = c.To sketch this, imagine the saddle on a horse back

4.1. FUNCTION OF SEVERAL VARIAVLES 5

Quadratic Surface We classify the surface represented by the following quadratic equation.

Ax2 +By2 + Cz2 +Dxy + Exz + Fyz +Gx+Hy + Iz + J = 0.

By suitable transformation, non degenerated equation can be clssified into9 surfaces.

1. Ellipsoidx2

a2+y2

b2+z2

c2= 1

2. hyperboloid of one sheetx2

a2+y2

b2− z2

c2= 1

3. hyperboloid of two sheetsx2

a2+y2

b2− z2

c2= −1

4. quadratic conex2

a2+y2

b2= z2

5. elliptic paraboloidx2

a2+y2

b2= z

6. hyperbolic paraboloidx2

a2− y2

b2= z

7. parabolic cylinder x2 = 4cy

8. elliptic cylinderx2

a2+y2

b2= 1

9. hyperboli cylinderx2

a2− y2

b2= 1

Degenerated Quadratic Degenerate quadraticequation is a equationwithout any solution oronly one solution. For ex-ample, 1+x2+y2+z2 = 0or x2 + y2 + z2 = 0.

Example 4.3 Classify the following surfaces.1. x2 + 4y2 − 16z2 = 0 2. x2 + 4y2 + 16z2 − 12 = 0

SOLUTION 1. For x = 0, we have 4y2 = 16z2. Then y2

4 = z2, z = ±y2 lines.

Also for y = 0, we have z = ±x4 lines. Thus, it is a quadratic cone

Example4-3-1.

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1

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2

2. For x = 0, we have 4y2 + 16z2 − 12 = 0. Then y2

3 + z234

= 1, ellipse. Also

for y = 0, we have x2 + 16z2 = 12. Thus x2

12 + y2

34

= 1 ellipse. Thus , it is a

ellipsoid

Example4-3-2.

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2

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1-0.50

0.5

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2

Exercise 4.3 Classify the following surfaces.1. x− 4y2 = 0 2. x2 − 4y2 − 2z = 0

SOLUTION Since x− 4y2 = 0, the value of z can be arbitrary. Thus we have aparabola for all z. Thus it is a parabolic cylinder Exercise4-3-1.

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0.50.75

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2. Since x2 − 4y2 = 2z, for x = 0, we have z = −2y2 a parabola. for y = 0,

we have z = x2

2 a parabola. Also for z = c, we have x2 − 4y2 = c a hyperbola.Thus it is a hyperbolic paraboloid

Exercise A

1. Find the domain and the region of the following functions．

(a) f(x, y) =√xy (b) f(x, y) =

1

x+ y(c) f(x, y) =

1

x2 + y2

(d) f(x, y) =x2

x2 + y2(e) f(x, y) = log(1− xy) (f) f(x, y, z) =

z

x2 − y2


2. Classify the following surfaces．

(a) x2 + 4y2 − 16z2 = 0 (b) x2 + 4y2 + 16z2 − 12 = 0 (c) x− 4y2 = 0

(d) x2 − 4y2 − 2z = 0 (e) 2x2 + 4y2 − 1 = 0 (f) x2 + 4y2 − 4z = 0

(g) 2x2 − 4y2 − 6 = 0 (h) x2 + y2 − 2z2 − 10 = 0 (i) x2 + y2 − 2z2 + 10 = 0

Exercise B

1. Find the domain of the region of the following functions and draw the do-main．

(a) f(x, y) = x2 − y2 (b) f(x, y) =x2

x2 + y2(c) f(x, y) = log (1− xy)

4.2. LIMIT OF FUNCTIONS OF TWO VARIABLES 7

4.2 Limit of Functions of Two Variables

Limit by Intuition For a function z = f(x, y), if (x, y) → (x0, y0), then f(x, y) → l. Then wecall l a limit of a function f(x, y) as (x, y) approaches (x0, y0) and denote

lim(x,y)→(x0,y0)

f(x, y) = l.

Understanding

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4

Unlike the function of sin-gle variable, there are in-finitely many approaches for(x, y) → (x0, y0). Thus in or-der to have the limit, for all(x, y) → (x0, y0), we must havef(x, y) → l.

Limit by ε− δ For every ε > 0, there exists δ > 0 such that whenever 0 <√

(x− x0)2 + (y − y0)2 < δ, then |f(x, y)− l| < ε. Then

lim(x,y)→(x0,y0)

f(x, y) = l NOTE Note that

√(x− x0)2 + (y − y0)2 < δ means that there are points (x, y)

in the circle of the center (x0, y0) and the radius less that δ. The relation|f(x, y)− l| < ε implies that the value of f(x, y) and l is very close to each other.

Example 4.4 Find the following limit.

lim(x,y)→(0,0)

x2y

x2 + y2

Example4.4

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SOLUTION As (x, y) → (0, 0), x2 + y2 → 0, x2y → 0. Now we must find outwhich one gets close to 0 faster. To do this, we compare the least degree in xand y of the numerator and the denominator.

Since (x, y) → (0, 0), larger thedegree, faster to converge.

Note that the least degree in x and y in the numerator is 3 and the leastdegree in x and y in the denominator is 2. Then we have a better chance ofconvergence. Thus we show that every approach to (0, 0), we have a uniquenumber to converge. Let

x = r cos θ, y = r sin θ

Then

0 ≤ | x2y

x2 + y2| = |r

3 cos2 θ sin θ

r2| ≤ |r|

Note that lim(x,y)→(0,0) 0 = 0 and limr→0 |r| = 0. Thus by squeezing theorem,we have

lim(x,y)→(0,0)

| x2y

x2 + y2| = 0.

and

lim(x,y)→(0,0)

x2y

x2 + y2= 0

Existence of Limit To show (x, y) → (0, 0), itis enough to show r → 0.


Exercise 4.4 Find the following limit.

lim(x,y)→(0,0)

xy

x2 + y2 Exercise4-4

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Non existence To show the limit doesnot exist, it is enough toshow that two different ap-proaches give a differentlimit. Nice way to do thisis to set y = mxk andchoose k so that the leastdegree of the numeratorand the least degree of thedenominator are the same.

SOLUTION Let y = mxk. Then we havemxk+1

x2 +m2x2k. Thus we let k = 1 and

y = mx.xy

x2 + y2=

mx2

x2(1 +m2)=

m

1 +m2(x = 0)

Thus

lim(x,y)→(0,0)

xy

x2 + y2=

m

1 +m2

Now note that this value depends on the value of m. If m = 0, then the limit is0 and if m = 1, then the limit is 1

2 . Thus as (x, y) → (0, 0), the limit does notexist

Continuous Functions If

lim(x,y)→(x0,y0)

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

then we say f((x, y)) is continuous at (x, y) = (x0, y0). If f(x, y) is contin-uous at all points of D ⊂ R2, then we say f(x, y) is continuous on D.

NOTE Note that f(x, y) is continuous at (x0, y0) if and only if1. lim(x,y)→(x0,y0) f(x, y) exists2. f(x0, y0) is defined3. lim(x,y)→(x0,y0) f(x, y) = f(x0, y0)

Understanding A sum of continuous func-tions is continuous, Adifference of continuousfunctions is continuous.A product of continu-ous functions is continu-ous, A quotient of continu-ous functions is continuousprovided the denominatoris not 0.

Properties of Continuous Functions Theorem 4.1 [If

lim(x,y)→(x0,y0)

f(x, y) = f(x0, y0), lim(x,y)→(x0,y0)

g(x, y) = g(x0, y0),

then

1. lim(x,y)→(x0,y0)

(f(x, y)± g(x, y)

)= f(x0, y0)± g(x0, y0)

2. lim(x,y)→(x0,y0)

kf(x, y) = kf(x0, y0) providek is constant

3. lim(x,y)→(x0,y0)

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

4. lim(x,y)→(x0,y0)

f(x, y)

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

g(x0, y0)(g(x0, y0) = 0)

Example 4.5 Determine the following function is continuous at (0, 0).

f(x, y) =

xy√x2+y2

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

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

Example4-5

Enough to showlim(x,y)→(0,0) f(x, y) = f(0, 0) .

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SOLUTION Note that f(x, y) is continuous except for the denominator is0. Thus we only need to check continuity at (0, 0). The least degree of the

4.2. LIMIT OF FUNCTIONS OF TWO VARIABLES 9

numerator is 2 and the least degree of the denominator is 1. Thus we letx = r cos θ, y = r sin θ. Then

0 ≤ | xy√x2 + y2

| = |r2 cos θ sin θ

r| ≤ |r cos θ sin θ| ≤ |r| −→ 0 (r → 0)

Therefore, lim(x,y)→(0,0) f(x, y) = 0. Since f(0, 0) = 0, f(x, y) is continuous at(0, 0)

Exercise 4.5 Determine the following function is continuous at (0, 0).

f(x, y) =

x2yx4+y2 , (x, y) = (0, 0)

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

Exercise4-5

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SOLUTION Note that f(x, y) is continuous except for the denominator is 0.Thus we only need to check continuity at (0, 0). The least degree of the numer-ator is 3 and the least degree of x in the denominator is 4. Then we let y = mx2

and

x2y

x4 + y2=

mx4

(1 +m2)x4=

m

1 +m2(x = 0)

Thus lim(x,y)→(0,0)x2yx4+y2 = n

1+m2 =

0 m = 012 m = 1

. Thus, f(x, y) is not contin-

uous at (0, 0) Exercise A

1. Find the limit of the following functions．

(a) lim(x,y)→(1,1)

x− y + 1

x+ y − 1(b) lim

(x,y)→(0,0)

2x− 3y

x+ y(c) lim

(x,y)→(0,0)

x2 − y2

x+ y

2. Determine the following functions are continuous at (0, 0)．

(a) f(x, y) =xy

x2 + y2 + 1(b) f(x, y) =

x2

x2+y2 , (x, y) = (0, 0)

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

(c) f(x, y) =

x2yx4+y2 , (x, y) = (0, 0)

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

Exercise B

2. Find the limit of the following functions as (x, y) → (0, 0)．

(a)

√xy

x2 + y2(b)

xy

x2 + y2 + y4(c)

xy

x2 + y2 + y

3. Determine whether the following functions are continuous at (0, 0) or not．

(a) f(x, y) =

x2yx2+y2 , (x, y) = (0, 0)

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

(b) f(x, y) =

x2−y2x2+y2 , (x, y) = (0, 0)

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

(c) f(x, y) =

xy log(x2 + y2), (x, y) = (0, 0)

−1, (x, y) = (0, 0)


4.3 Partial Derivatives

Figure 4.1: z = f(x, y)

Understanding A partial derivative off(x, y) at (x0, y0) with re-spect to x is a curve showsup by slicing the surfacegiven by z = f(x, y) by theplane y = y0. Thus differ-entiate z = f(x, y0) by xgives the answer.

Partial Differential Coefficient Consider the limit of a function z = f(x) = f(x, y) at point x0 = (x0, y0).Let y = y0 be a constant and a function f(x, y0) of x is differentiable atx = x0. Then

fx(x0, y0) = limh→0

f(x0 + h, y0)− f(x0, y0)

h

is called coefficient of partial derivative with respect to x. Similarly,Keep x = x0 a constant and a function f(x0, y) of y is differentiable aty = y0. Then

fy(x0, y0) = limk→0

f(x0, y0 + k)− f(x0, y0)

k

is called partial differential coefficient with respect to y. NOTE If fx(x0, y0) exists, then f(x, y) is called partially differentiable withrespect x at (x0, y0). If f(x, y) is differentiable at esch y in D, then f(x, y) iscalled partially differentiable on D with respect y.

Partially Differentiable f(x, y) is differentiablewith respect to x at(x0, y0) if and only iff(x, y0) is differentiable atx = x0. Similarly, f(x, y)is differentiable withrespect to y at (x0, y0)if and only if f(x0, y) isdifferentiable at y = y0.

Example 4.6 Find the partial differential coefficient of the following functionat (1, 1).

f(x, y) = x2 + y2

Example4-6

f(x, 1) is a curve given by slic-ing the surface z = f(x, y)with the plane y = 1. Thenfx(1, 1) is a derivative of thecurve z = f(x, 1) with respectto x at x = 1.

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SOLUTION To slice the surface f(x, y) = x2 + y2 by the plane y = 1, it isenough to consider f(x, 1) = x2 + 1. Now differentiate f(x, 1) by x. Then wehave fx(x, 1) = 2x. Thus, fx(1, 1) = 2

Exercise 4.6 Find the partial differential coefficient of the following functionat (1, 1).

f(x, y) =√xy

SOLUTION Let x = 1. Then slice the surface f(x, y) =√xy by the plane x = 1

to get the curve f(1, y) =√y. Now differentiate f(1, y) with respect to y to

obtain fy(1, y) =1

2√y . Thus, fy(1, 1) =

12

4.3. PARTIAL DERIVATIVES 11

Partial Derivatives A partial derivative of f(x, y) with respect to x is the derivative of f(x, y)by keeping y as constant. More precisely,

∂f(x, y)

∂x= fx(x, y) = lim

h→0

f(x+ h, y)− f(x, y)

h

Similarly for a partial derivative of f(x, y) with respect to y is

∂f(x, y)

∂y= fy(x, y) = lim

k→0

f(x, y + k)− f(x, y)

k

Finding a partial derivative, we say partial differentiation

Exercise4-6fy(1, 1) is the derivative f(y, 1)with respect to y at y = 1.

01

2

3 0

12

30

1

2

3

01

2

3

Understanding To find the partial deriva-tive of f(x, y) with respectto x, fix y and differentiatef(x, y) by x. ∂ is read del,dee, partial.

Example 4.7 Find the partial derivative of the followings.1. f(x, y) = x2 + 2xy + 3y3 2. g(x, y) = tan−1( yx )

Example4-7-1

Note that to find the partialderivative of f(x, y) with re-spect to y, we treat x as con-stant. Thus ∂(x2)/∂y = 0.

SOLUTION 1. To find fx(x, y), treat y as constant and differentiate f(x, y)with respect ot x.

fx(x, y) =∂(x2)

∂x+∂(2xy)

∂x+∂(3y3)

∂x= 2x+ 2y

Also, treat x as constant and differentiate with respect to y

fy(x, y) =∂(x2)

∂y+∂(2xy)

∂y+∂(3y3)

∂y

= 2x+ 9y2

Review 1. (tα)′ = αtα−1

2. (et)′ = et

3. (log |t|)′ = 1t

4. (sin t)′ = cos t5. (cos t)′ = − sin t6. (tan t)′ = 1

cos2 t

7. (sin−1 t)′ = 1√1−t2

8. (tan−1 t)′ = 11+t2

2. Let t = yx . Then

gx(x, y) =∂(tan−1 t)

∂x=∂(tan−1 t)

∂t· ∂t∂x

=1

1 + t2· (− y

x2) =

1

1 + ( yx )2· −yx2

=−y

x2 + y2

gy(x, y) =∂(tan−1 t)

∂y=∂(tan−1 t))

∂t· ∂t∂y

=1

1 + t2· ( 1x) =

1

1 + ( yx )2· xx2

=x

x2 + y2

Exercise 4.7 Find the partial derivatives of the followings.1. z =

√(x2 + y2) 2. z = log (x2 + y2)

SOLUTION 1. Let t = x2 + y2. Then

∂z

∂x=

∂(√t)

∂t· ∂t∂x

=1

2√t· 2x =

x√x2 + y2

∂z

∂y=

∂(√t)

∂t· ∂t∂y

=1

2√t· 2y =

y√x2 + y2

Exercise4-7-1.

(√t)′ = (t

12 )′ = 1

2 t− 1

2 = 12√t

( 1x )′ = (x−1)′ = −x−2 = − 1

x2

2. Let t = x2 + y2. Then

∂z

∂x=

∂(log t)

∂t· ∂t∂x

=1

t· 2x =

2x

x2 + y2

∂z

∂y=

∂(log t)

∂t· ∂t∂y

=1

t· 2y =

2y

x2 + y2

Exercise4-7-2.(log t)′ = 1

t


Example 4.8 Find the fx(0, 0), fy(0, 0). Then determine the following func-tion is partially differentiable with respect to x, y or not.

f(x, y) =

xyx2+y2 (x, y) = (0, 0)

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

Example4-8

fx(x, y) = y(x2+y2)−xy(2x)(x2+y2)2 =

y3−x2y(x2+y2)2 . Now substitute (0, 0)

to get fx(0, 0) =00 . Then con-

clude that f(x, y) is not par-tially differentiable is not right.The reason is that fx(x, y) isnot continuous at (0, 0). Thussubstituting (0, 0) is not al-lowed.

SOLUTION To find fx(0, 0), substitute y = 0 and find f(x, 0). Then

f(x, 0) =x · 0

x2 + 02=

0

x2= 0

Then fx(x, 0) = 0. Thus, fx(0, 0) = 0.Since f(0, y) = 0·y

02+y2 = 0y2 = 0, we have fy(0, y) = 0. Thus fy(0, 0) = 0.

Therefore, f(x, y) is partially differentiable with respect to x, y at (0, 0)

Exercise 4.8 Find the fx(0, 0), fy(0, 0). Then determine the following functionis partially differentiable with respect to x, y or not.

f(x, y) = log (1 + xy + y2)

Exercise4-8To find the function f(x, y) isdifferentiable with respect to xat (0, 0), it is enough to checkf(x, 0) is differentiable with re-spect to x

SOLUTION Since f(x, 0) = log(1+0+0) = log 1 = 0, fx(x, 0) = 0 and fx(0, 0) =0.

Note that f(0, y) = log(1 + 0 + y2) = log(1 + y2). Thus

fy(0, y) =2y

1 + y2.

which implies that fy(0, 0) =01 = 0. Thus f(x, y) is partially differentiable with

respect to x and y at (0, 0) Exercise

1. Find the partial derivative of.the following functions.

(a) f(x, y) = 3x2 − xy + y (b) f(x, y) = x2e−y (c) z =√(x2 + y2)

(d) z = x sin y (e) z =x− y

x+ y

2. Find all 2nd partial derivatives of the following functions.．

(a) f(x, y) = ax2 + 2bxy + cy2 (b) f(x, y, z) = (x+ y2 + z3)2

(c) f(x, y) = sin(3x− 2y) (d) f(x, y) = xe2y

Exercise B

1. Find the partial derivative of the following functions.

(a) z = x3 + xy2 + y3 (b) z = ex sin y (c) z = log (x2 + y2)

2. Find all 2nd partial derivatives of the following functions.

(a) z = x3y + xy2 (b) z = xy2exy

(c) z = tan−1 (x2 + y2)

3. Determine the following functions are partially differentiable at the origin.．

(a) f(x, y) =

y3−x2yx2+y2 , (x, y) = (0, 0)

0, (x, y) = (0, 0)(b) f(x, y) = log (1 + xy + y2)

4.4. TOTAL DIFFERENTIAL AND TANGENT PLANE 13

4.4 Total Differential and Tangent Plane

Total Differential A function z = f(x, y) is defined on the region D. Let ∆x,∆y be theincrement of x and y. Then f(x+∆x, y+∆y)−f(x, y) is called an incrementof z and denoted by ∆z.If f(x, y) is partially differentiable, then

lim∆x→0

f(x+∆x, y +∆y)− f(x, y +∆y)

∆x= fx(x, y).

lim∆y→0

f(x, y +∆y)− f(x, y)

∆y= fy(x, y)

Thus∆z = fx(x, y)∆x+ fy(x, y)∆y + ε(x, y).

Now if lim(∆x,∆y)→(0,0)ε(x,y)√

(∆x)2+(∆y)2= 0, then we say a function z =

f(x, y) is totally diferentiable at (x, y) NOTE Note that we can express ∆z as

∆z = f(x+∆x, y +∆y)− f(x, y +∆y) + f(x, y +∆y)− f(x, y)

Note that f(x, y + ∆y) − f(x, y) is an increment of f as y moved to y + ∆y.Then we can write

f(x, y +∆y)− f(x, y) =f(x, y +∆y)− f(x, y)

∆y∆y

Similarly, f(x + ∆x, y) − f(x, y) is an increment of f as x moved to x + ∆x.Then we can write

f(x+∆x, y)− f(x, y) =f(x+∆x, y +∆y)− f(x, y +∆y)

∆x∆x

Putting together,

∆z =f(x+∆x, y +∆y)− f(x, y +∆y)

∆x∆x+

f(x, y +∆y)− f(x, y)

∆y∆y.

Total Differential The differential of x is denoted by dx . Then for z = f(x, y) is totallydifferentiable at (x, y), Let ∆x→ 0,∆y → 0. Then∆z = fx(x, y)∆x+ fy∆y + ε(x, y), where ∆x,∆y,∆z can be expressed asdx, dy, dz. Thus we write

dz = fx(x, y)dx+ fy(x, y)dy

dz is called taltal differential of z. Example 4.9 Find the total differential of the following functions.1. z = x2exy 2. f(x, y) = log(x2 + y2) 3. z = sinxy

14 CHAPTER 4. PARTIAL DIFFERENTIATION Example4-9

(et)′ = et

(log t)′ = 1t

(sin t)′ = cos t

SOLUTION 1. dz = zxdx+ zydy = (2xexy + x2yexy)dx+ x2 · xexydy= 2xexy(1 + x)dx+ x3exydy

2. df = fxdx+ fydy =2x

x2 + y2dx+

2y

x2 + y2dy

3. dz = zxdx+ zydy = y cosxydx+ x cosxydy

Exercise 4.9 Approximate the following number by using total differential.

√27

3√1021 Exercise4-9

Since√25 = 5, 3

√100 = 3, we

approximate using these val-ues. Increment of z = f(x, y)∆z can be approxiamted withthe total differential of z thatis dz.

SOLUTION Consider the function f(x, y) =√x 3√y = x

12 y

13 . Let x = 25, y =

1000. Then f(25, 1000) = 50. Now let ∆x = 2,∆y = 21. Then f(27, 1021) =f(25+∆x, 1000+∆y). Note that ∆f = f(25+∆x, 1000+∆y)−f(25, 1000) ≈ df .Thus

f(25 + ∆x, 1000 + ∆y) ≈ df + f(25, 1000)

Here, df = fx∆x + fy∆y = 12x

− 12 y

13∆x + 1

3x12 y−

23∆y. Thus for x = 25, y =

1000,∆x = 2,∆y = 21, we have df = ( 12 · 15 · 10)2 + ( 13 · 5 · 1

100 )21 = 2.35. From

this,√27

3√1021 ≈

√25

3√1000 + 2.35 = 52.35

Total Differentiability

Use the contrapositive of The-orem4.2, if f(x, y) is not con-tinuous at (x0, y0), then f(x, y)is not totally differentiable at(x0.y0).

Necessary Condition Theorem 4.2 If f(x, y) is totally differentiable at (x0, y0), then f(x, y) iscontinuous at (x0, y0).

NOTE Suppose that f(x, y) is totally differentiable at (x0, y0), then

f(x0 +∆x, y0 +∆y)− f(x0, y0) = fx(x0, y0)∆x+ fy(x0, y0)∆y

+ ε(x, y)

where lim(∆x,∆y)→(0,0)ε(∆x,∆y)√(∆x)2+(∆y)2

= 0. Now we show that f(x, y) is continu-

ous at (x0, y0). In other words, we show

lim(x,y)→(x0,y0)

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

Let x−x0 = ∆x, y−y0 = ∆y. Then (x, y) → (x0, y0) is equivalent to (∆x,∆y) →(0, 0). Thus,

lim(∆x,∆y)→(0,0)

[f(x0 +∆x, y0 +∆y)− f(x0, y0)] = 0

and f(x, y) is continuous at (x0, y0).


Sufficient Condition Theorem 4.3 Suppose that f(x, y) is partially differentiable with respectto x, y at (x0, y0) and fx, fy are continuous at (x0, y0). Then f(x, y) istotally differentiable at (x0, y0).

Checkf(x0 + ∆x, y0 + ∆y) −f(x0, y0) = fx(x0, y0)∆x +fy(x0, y0)∆y + ε(x, y)・・ε(x, y) → (0, 0)

Example 4.10 Determine the following function is totally differentiable ornot.

f(x, y) =

xyx2+y2 (x, y) = (0, 0)

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

Example4-10

-20

2-2

0

2-0.5

-0.250

0.250.5

-20

2

SOLUTION∆f = f(0 + ∆x, 0 + ∆y)− f(0, 0) =∆x∆y

(∆x)2 + (∆y)2. Then ∆f = fx(0, 0)∆x+ fy(0, 0)∆y + ε(x, y) = ε(x, y).

Then

CheckBy Example4.8, fx(0, 0) =0, fy(0, 0) = 0

ε(x, y) =∆x∆y

(∆x)2 + (∆y)2.

f(x, y) is totally differentiable at (0, 0) only if

lim(∆x,∆y)→(0,0)

ε(x, y)√(∆x)2 + (∆y)2

= 0

Let y = m(∆x)2. Then

lim(∆x,∆y)→(0,0)

ε(x, y)√(∆x)2 + (∆y)2

= lim(∆x,∆y)→(0,0)

∆x∆y((∆x)2 + (∆y)2

)3/2= lim

∆x→0

m

(1 +m2(∆x)2)3/2= m.

Then this limit depends on the value of m. Thus it is not totally differentiable

Checkf(x, y) is totally differentiableat (x0, y0) ⇒ f(x, y) is con-tinuous at (x0, y0). Now takethe contrapositive to this state-ment. Then f(x, y) is not con-tinuous at (x0, y0) ⇒ f(x, y)is not totally differentiable at(x0, y0).Alernative Solution By Example4.8, f(x, y) is partially differentiable at (0, 0).

But by Example4.4, f(x, y) is not continuous at (0, 0). Thus by Theorem4.2,f(x, y) is not totally diffrentiable at (0, 0)

Exercise 4.10 Determine whether the following function is totally diffrentiableat (0, 0).

f(x, y) = log(1 + xy + y2)

SOLUTION fx(x, y) =y

1 + xy + y2and fy(x, y) =

x+ 2y

1 + xy + y2. Then fx(x, y), fy(x, y)

is continuous at (0, 0). Thus by Theorem4.3, f(x, y) is totally differentiable at(0, 0).Altenative Solution ∆f = f(0+∆x, 0+∆y)− f(0, 0) = log(1+∆x∆y+(∆y)2)

and fx(x, y) =y

1 + xy + y2, fy(x, y) =

x+ 2y

1 + xy + y2. Then

∆f = fx(0, 0)∆x+ fy(0, 0)∆y + ε(x, y) = ε(x, y)

ε(x, y) = log(1 + ∆x∆y + (∆y)2). Thus,

lim(∆x,∆y)→(0,0)

ε√(∆x)2 + (∆y)2

= lim(∆x,∆y)→(0,0)

log(1 + ∆x∆y + (∆y)2)√(∆x)2 + (∆y)2

Now let ∆x = r cos θ,∆y = r sin θ. Then

log(1 + ∆x∆y + (∆y)2)√(∆x)2 + (∆y)2

=log(1 + r2 cos θ sin θ + r2 sin2 θ)√

(r cos θ)2 + (r sin θ)2

=log(1 + r2(cos θ sin θ + sin2 θ)

)r


Let cos θ sin θ + sin2 θ = k(θ). Then

limr→0

log(1 + kr2)

r

∗= lim

r→0

2kr

1 + kr2= 0

Thus f(x, y) is totally differentiable at (0, 0) Normal Vector

A vector orthogo-nal(perpendiculr) tothe plane is called nor-mal vector. Thus thevector such as n =(fx(x0, y0), fy(x0, y0),−1)is a normal vector.

4.4.1 Tangent Plane and Normal Vector

Tangent Plane If z = f(x, y) is totally differentiable at (x0, y0), then

f(x, y) = f(x0, y0) + fx(x0, y0)∆x+ fy(x0, y0)∆y + ε(∆x,∆y)

where (∆x,∆y) = (x− x0, y − y0). Note that

z = f(x0, y0) + fx(x0, y0)∆x+ fy(x0, y0)∆y

represents the equation of the plane goes through a point (x0, y0, f(x0, y0)).Note also that these two equations behave almost the same when (x, y) isclose to (x0, y0). Thus the equation above is called tangent plane of thefunction z = f(x, y) at (x0, y0, f(x0, y0)).

NOTE Given the point (x0, y0, f(x0, y0)) and take a point (x, y, z). Then we canform a vector (x− x0, y − y0, z − f(x0, y0). Now ((x− x0, y − y0, z − f(x0, y0) ·(fx(x0, y0), fy(x0, y0),−1) = fx(x− x0) + fy(x0, y0)− (z − f(x0, y0)) = 0 Thus(fx(x0, y0), fy(x0, y0),−1) is orthogonal to the tangent plane.

Normal Vector A vector A and a vectorB are orthogonal. Thentheir inner product is A ·B = 0. Example 4.11 Find the tangent plane and the normal line of the following

function at (1, 1, 6).z = f(x, y) = 3x+ xy + 2y

SOLUTION Since fx(x, y) = 3+y, fy(x, y) = x+2, the tangent plane of f(x, y)at (1, 1, 6) is given by the following.

z = f(1, 1) + fx(1, 1)(x− 1) + fy(1, 1)(y − 1) = 6 + 4(x− 1) + 3(y − 1).

Now let (x, y, z) be an arbitray point on the normal line. Then the vectorconnecting (x, y, z) and (1, 1, 6) is given by (x, y, z) − (1, 1, 6) and this vector(x− 1, y− 1, z− 6) is on the normal line with the normal vector (4, 3,−1), Thusthe normal line is

(x− 1, y − 1, z − 6) = t(4, 3,−1) or t =x− 1

4=y − 1

3=z − 6

−1

CheckLet (x0, y0, z0) be a initial pointand a point (x, y, z) be a termi-nal point. Then the positionvector is the vector with theinitial vector (0, 0, 0). Thus theposition vector is (x − x0, y −y0, z − z0).

Parallel When a vector A and avector B are parallel, weexpress A = tB providedby t real. Exercise 4.11 Find the tangent plane and the normal line of the following

function at (−2, 3, 2).

z =x2

4+y2

9 Exercise4-11

-20

2

-20

2

-5

0

5

-20

2

SOLUTION Since fx(x, y) =x

2, fy(x, y) =

2y

9, the tangent plane of f(x, y) at

(−2, 3, 2) is given by

z = f(−2, 3) + fx(−2, 3)(x+ 2) + fy(−2, 3)(y − 3)

= 2− (x+ 2) +2

3(y − 3)

= −x+2y

3− 2.


Now let (x, y, z) be an arbitray point on the normal line. Then the vectorconnecting (x, y, z) and (−2, 3, 2) is given by (x, y, z)− (−2, 3, 2) and this vector(x + 2, y − 3, z − 2) is on the normal line with the normal vector (−1, 23 ,−1),Thus the normal line is

(x+ 2, y − 3, z − 2) = t(−1,2

3,−1)

orx+ 2

−1=

3(y − 3)

2=z − 2

−1

Exercise A

1. Find the gradient and total differential of the following functions.

(a) f(x, y) = x3 + y2 (b) f(x, y) = 3x2 − xy + y (c) z = x2y−2

(d) z = x2y (e) z = ex cos y

2. Answer the following questions．

(a) Find the equation of the tangent plane to the surface whose normalvector is (3, 2,−1). Find the equation of the normal line through the point(1, 1, 1).

(b) Find the equation of the tangent plane to the surface z = xy at thepoint (2, 1, 2). Find the equation of the normal line through the point(2, 1, 2).

(c) Find the equation of the tangent plane to the surface z = x2+xy+2y2

at the point (1, 1, 4). Find the equation of the normal line through thepoint (1, 1, 4).

Exercise B

1. Find the gradient and total differential of the following functions．Find theequation of the tangent plane to the surface at the point corresponds to(1, 1). Find the equation of the normal line at the point corresponds to(1, 1).

(a) f(x, y) = x3y4 (b) f(x, y) = x3y + x2y4 (c) z = x2ye2x (d)z = cosxy

2. Approximate the following value by using total diferential.．

(a)√125

4√17 (b) sin (

6π

7) cos (

π

3)


4.5 Partial Differentiation of Composite Func-tions

∂Symbol d is used for the sign ofderivative of a function ofsingle variable. ∂ is usedfor the sign of derivativeof a function of more thanone variables.

Partial Derivative of Composite Functions Theorem 4.4 1. Suppose that z = f(x, y) is totally differentiable andx = x(t), y = y(t) are differentiable. Then a compostite function z =f((x(t), y(t)) is differentiable and the following is true.

dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt.

2. Suppose that z = f(x, y) is totally differentiable and x = x(r, s), y =y(r, s) are differentiable. Then a composite function z = f(x(r, s), y(r, s))is also differentiable and the following is true.

∂z

∂r=∂z

∂x

∂x

∂r+∂z

∂y

∂y

∂r,∂z

∂s=∂z

∂x

∂x

∂s+∂z

∂y

∂y

∂s NOTE 1. Let t be a time, x be a number of planktons, y be a water temparature,z be a number of minnows. Then dz

dt is an instanteneous change of a numberof minnows with respect to a time. But a number of minnows and a number ofplanktons are influenced by the temparature of the water. ∂z

∂x and ∂z∂y . Thus the

instanteneous chage of the number of planktons with respect to the temparatureis dx

dt and instance change of the water temparature with respect to the time isdydt . Thus, the instanteneous change of the number of minnows with respect tothe time is

dz

dt=∂z

∂x

dx

dt+∂z

∂y

dy

dt

Theorem4.4 2. Express by the tree diagram. 4.2Tree Diagram

z is a function of x and y.Then draw a line from z tox, and z to y. Now x andy are functions of r ands. Then draw lines fromx and y to r and s.

Figure 4.2: Composite Functions

Proof 2.

∂z

∂r= lim

∆r→0

∆z

∆r= lim

∆r→0∂z∂x

∆x

∆r+∂z

∂y

∆y

∆r+ ε1

∆x

∆r+ ε2

∆y

∆r

=∂z

∂x

∂x

∂r+∂z

∂y

∂y

∂r

Example 4.12 Finddz

dt1. z = exy

2

, x = t cos t, y = t sin t 2. z = f(t2, et)

4.5. PARTIAL DIFFERENTIATION OF COMPOSITE FUNCTIONS 19

SOLUTION 1.

dz

dt=

∂z

∂x

dx

dt+∂z

∂y

dy

dt

= y2exy2

(cos t− t sin t) + 2xyexy2

(sin t+ t cos t)

Example4-12-1.

z is a function of x and y, x andy are functions of t. Then drawa line from z to x, y and linesfrom x to t, from y to t.

2. Let x = t2, y = et. Then z = f(x, y) and

dz

dt=

∂z

∂x

dx

dt+∂z

∂y

dy

dt= 2t

∂z

∂x+ et

∂z

∂y

Example4-12-2.

Since z = f(x, y) is not known,∂z∂x is the final form

Exercise 4.12 1. Find ∂z∂r ,

∂z∂s of the following function.

z = tan−1 y

x, x = r3 − 3rs2, y = 3r2s− s3

2. For z = f(x2y), Show x(∂z

∂x) = 2y(

∂z

∂y). Exercise4-12-1.

SOLUTION 1.

∂z

∂r=

∂z

∂x

∂x

∂r+∂z

∂y

∂y

∂r

=1

1 + (xy )2(− y

x2)(3r2 − 3s2) +

1

1 + ( yx )2(1

x)(6rs)

=−y

x2 + y2(3r2 − 3s2) +

x

x2 + y2(6rs)

∂z

∂s=

∂z

∂x

∂x

∂s+∂z

∂y

∂y

∂s=

1

1 + ( yx )2(− y

x2)(−6rs)

+1

1 + ( yx )2(1

x)(3r2 − 3s2) =

−yx2 + y2

(6rs) +x

x2 + y2(3r2 − 3s2) Exercise4-12-2.

2. Let x2y = u. Then z = f(u) and by the tree diagram,

∂z

∂x=dz

du

∂u

∂x= f ′(u)2xy,

∂z

∂y=dz

du

∂u

∂y= f ′(u)x2

Thus

x∂z

∂x= f ′(u)2x2y = 2y

∂z

∂y

Exercise A

1. Finddz

dt.

(a) z = x2 + 2y, x = 2t, y = t3 (b) z = x2 + y2, x = cos t, y = sin t

(c) z = x2 + xy + 2y2, x = cos t, y = sin t (d) z = x3y2, x = t2, y = t3

2. Find∂z

∂u,∂z

∂v．

(a) z = x2 + y2, x = u− 2v, y = 2u+ v

(b) z = x2 + xy + 2y2, x = u+ v, y = uv (c) f(x, y) = x2y2, x = uv, y = v2

Exercise B


1. Finddz

dtprovided f is in C(1).

(a) z = log (x2 + y2), x = t+1

t, y = t(t− 1) (b) z = f(t2, et)

(c) z = f(2t, 4t2) (d) z = x2 − 2y2, x = cos t, y = sin t

2. Find∂z

∂r,∂z

∂s．

(a) z = tan−1 y

x, x = r3 − 3rs2, y = 3r2s− s3

(b) z = logy

x, x = (r − 1)2 + s2, y = (r + 1)2 + s2

(c) z =√x2 + y2, x = r cos s, y = r sin s, (r > 0)

3. For z = f(x, y), x = r cos θ, y = r sin θ, show that

zr = zx cos θ + zy sin θ, zθ = r(−zx sin θ + zy cos θ).

4.6. HIGHER ORDER PARTIAL DERIVATIVES 21

4.6 Higher Order Partial Derivatives

The second Order PartialDerivatives Suppose that partial derivatives fx, fy are again partially differentiable withrespect to x, y. Then

∂

∂x(∂f

∂x) =

∂2f

∂x2= fxx,

∂

∂y(∂f

∂x) =

∂2f

∂y∂x= fxy

∂

∂x(∂f

∂y) =

∂2f

∂x∂y= fyx,

∂

∂y(∂f

∂y) =

∂2f

∂y2= fyy

We say fxx, fxy, fyx, fyy The second partial derivatives of f .

Evaluation ∂∂y (

∂f∂x ),

∂2f∂y∂x , To evaluate

fxy, first differentiate withrespect to x.

Schwartz’s Lemma If f(x, y) is the class C2 on D, then fxy = fyx.

Interchange the order If a function is the classC2・・, then it is possibleto interchange the order ofdifferentiation.

Example 4.13 Find the 2nd order derivatives of f(x, y) = log(x2 + y2).

SOLUTION fx(x, y) =2x

x2 + y2, fy(x, y) =

2y

x2 + y2,

fxx(x, y) =∂fx(x, y)

∂x=

2(x2 + y2)− 2x(2x)

(x2 + y2)2=

2y2 − 2x2

(x2 + y2)2,

fxy(x, y) =∂fx(x, y)

∂y=

−2x(2y)

(x2 + y2)2=

−4xy

(x2 + y2)2,

fyy(x, y) =∂fy(x, y)

∂y=

2(x2 + y2)− 2x(2y)

(x2 + y2)2=

2x2 − 2y2

(x2 + y2)2

Class Cn If f(x, y) has the nthderivatives on D and theyare continuous, thenwe sayf(x, y) is the class Cn.

Exercise 4.13 Show fxy(0, 0) = fyx(0, 0) for the following function.

f(x, y) =

(xy)x

2−y2x2+y2 (x, y) = (0, 0)

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

SOLUTION To show fxy(0, 0) = fyx(0, 0), we first evaluate these values.

fx(0, y) = limh→0

f(h, y)− f(0, y)

h= limh→0

(hy)h2 − y2

h(h2 + y2)= −y

Thus, fxy(0, y) = −1 and fxy(0, 0) = −1. Next

fy(x, 0) = limk→0

f(x, k)− f(x, 0)

k= limk→0

(xk)x2 − k2

k(x2 + k2)= x

Thus fyx(x, 0) = 1 aand fyx(0, 0) = 1.Therefore, fxy(0, 0) = −1 = fyx(0, 0) = 1 Example 4.14 If u = log(x2 + y2), show uxx + uyy = 0.

SOLUTION ux = 2xx2+y2 .

uxx =∂ux∂x

=2(x2 + y2)− 4x2

(x2 + y2)2=

2(−x2 + y2)

(x2 + y2)2.

Note uy = 2yx2+y2 . Then

uyy =∂uy∂y

=2(x2 + y2)− 4y2

(x2 + y2)2=

2(x2 − y2)

(x2 + y2)2.

Thus, uxx + uyy = 0


Exercise 4.14 Show the following function satisfies the laplace equation. uxx+uyy + uzz = 0

u =1√

x2 + y2 + z2

Laplace Equation uxx + uyy = 0 is calledtwo dimensionalLaplace equation.uxx + uyy + uzz = 0 iscalled three dimen-sional Laplace equa-tion and expressed by∆u. This represents thevelocity potential of theimcompressible fluid, thepotential the electrostaticfield, the steady statetemperature distributionof the heat conduction.

SOLUTION Let u = (x2 + y2 + z2)−12 . Then

ux = −1

2(x2 + y2 + z2)−

32 (2x) = −x(x2 + y2 + z2)−

32

uxx =∂ux∂x

= −(x2 + y2 + z2)−32 − x(−3

2)(x2 + y2 + z2)−

52 (2x)

= −(x2 + y2 + z2)−52

(x2 + y2 + z2 − 3x2

)= −(x2 + y2 + z2)−

52 (−2x2 + y2 + z2)

uy = −1

2(x2 + y2 + z2)−

32 (2y) = −y(x2 + y2 + z2)−

32

uyy =∂uy∂y

= −(x2 + y2 + z2)−32 − y(−3

2)(x2 + y2 + z2)−

52 (2y)

= −(x2 + y2 + z2)−52

(x2 + y2 + z2 − 3y2

)= −(x2 + y2 + z2)−

52 (x2 − 2y2 + z2)

uz = −1

2(x2 + y2 + z2)−

32 (2z) = −z(x2 + y2 + z2)−

32

uzz =∂uz∂z

= −(x2 + y2 + z2)−32 − z(−3

2)(x2 + y2 + z2)−

52 (2z)

= −(x2 + y2 + z2)−52

(x2 + y2 + z2 − 3z2

)= −(x2 + y2 + z2)−

52 (x2 + y2 − 2z2)

Thus,

uxx + uyy + uzz = −(x2 + y2 + z2)−52 (2x2 + 2y2 + 2z2 − 2x2 − 2y2 − 2z2) = 0

Example 4.15 Show the following functions are harmonic.1. z = x2 − y2 2. z = x

x2+y2

SOLUTION 1. zx = 2x, zxx = 2, zy = −2y, zyy = −2. Thus, ∆z = zxx + zyy =

4.6. HIGHER ORDER PARTIAL DERIVATIVES 23

2− 2 = 0 2.

zx =x2 + y2 − x(2x)

(x2 + y2)2=

−x2 + y2

(x2 + y2)2,

zxx =−2x(x2 + y2)2 − (−x2 + y2)(2(x2 + y2)(2x))

(x2 + y2)4

=(x2 + y2)

(− 2x(x2 + y2)− (4x)(−x2 + y2)

)(x2 + y2)4

=−2x3 − 2xy2 + 4x3 − 4xy2

(x2 + y2)3=

2x3 − 6xy2

(x2 + y2)3,

zy =−x(2y)

(x2 + y2)2=

−2xy

(x2 + y2)2,

zyy =−2x(x2 + y2)2 − (−2xy)(2(x2 + y2)(2y))

(x2 + y2)4

=(x2 + y2)

(− 2x(x2 + y2) + 8xy2

)(x2 + y2)4

=−2x3 − 2xy2 + 8xy2

(x2 + y2)3=

−2x3 + 6xy2

(x2 + y2)3.

Thus,

∆z = zxx + zyy =2x3 − 6xy2

(x2 + y2)3+

−2x3 + 6xy2

(x2 + y2)3= 0

Harmonic Functions When we express uxx +uyy, uxx+uyy+uzz by ∆u.Then ∆ is called Lapla-ian and the class C2 func-tion u satisfying the equa-tion ∆u = 0 is called har-monic function. Example4-15-2.

zx = ∂∂x

(x

x2+y2

)=

∂x∂x (x2+y2)−x( ∂

∂x (x2+y2))

(x2+y2)2 =x2+y2−2x2

(x2+y2)2 = −x2+y2

(x2+y2)2

Exercise 4.15 Show the following functions are harmonic.1. z = ex sin y 2. z = tan−1 y

x

Exercise4-15-2.

zx = ∂∂x

(tan−1 y

x

)=

∂∂x

(tan−1 t) =

∂∂t

(tan−1 t) ∂t∂x = 1

1+t2 · ∂(yx )

∂x =1

1+t2 (−yx2 ) = 1

1+( yx )2 (−

yx2 ) =

yx2+y2

SOLUTION 1. zx = ex sin y, zxx = ex sin y, zy = ex cos y, zyy = −ex sin y.Thus, ∆z = zxx + zyy = ex sin y − ex sin y = 0

2. zx =1

1 + ( yx )2· −yx2

=−y

x2 + y2, zxx = − −y(2x)

(x2 + y2)2=

2xy

(x2 + y2)2

zy =1

1 + ( yx )2· 1x=

x

x2 + y2, zyy = − x(2y)

(x2 + y2)2= − 2xy

(x2 + y2)2. Thus,

∆z = zxx + zyy =2xy

(x2 + y2)2− 2xy

(x2 + y2)2= 0


4.7 Extreme Values of Function of Two Vari-ables

Extreme Values Definition 4.1 For all points (x, y) in the δ neighborhood of (x0, y0)

a

1. If f(x0, y0) ≤ f(x, y), then f(x, y) takes minimum at (x0, y0) andf(x0, y0) is called local minimum of f(x, y).2. If f(x0, y0) ≥ f(x, y), then f(x, y) takes maximum at (x0, y0) andf(x0, y0) is called local maximum of f(x, y).

aδ neighborhood of (x0, y0) is a circular region centered at (x0, y0) with the radius δ.

NOTE A locam minimum and a local maximum togrther are called extrema.

Understanding

-20

2

-20

2

0

5

10

15

-20

2

If the graph of function issmooth, then fx(x, y) =0, fy(x, y) = 0.

-20

2

-20

2

0

1

2

3

4

-20

2

If the graph has a sharpedge, then the function isnot differentiable at thesharp edge.

Necessary Condition Theorem 4.5 If f(x, y) has a extremum at (x0, y0), then .(fx(x0, y0), fy(x0, y0)) exists and (fx(x0, y0), fy(x0, y0)) = (0, 0), or(fx(x0, y0), fy(x0, y0)) does not exist.

Proof Since f(x, y0) takes a extreme value at x = x0, fx(x0, y0) = 0 or fx(x0, y0)does not exist. Similarly for f(x0, y). fy(x0, y0) = 0 or fy(x0, y0) does notexist

Example 4.16 Find the extrema of the following function

f(x, y) = x2 + xy + 3y2 + x+ y Example4-17

2x0 + y0 + 1 = 0 implies y0 =−2x0 − 1. substitute this inotx0 + 6y0 + 1 = 0 to get x0 +6(−2x0−1)+1 = −11x0−5 =0. Thus, x0 = − 5

11 .

SOLUTION If f(x, y) takes the extreme value at (x0, y0), then

fx(x0, y0) = 0 implies 2x0 + y0 + 1 = 0

fy(x0, y0) = 0 implies x0 + 6y0 + 1 = 0

Now solve for x0, y0. Then

x0 = − 5

11, y0 = − 1

11.

Thus f(x, y) may takes the extreme value at (− 511 ,−

111 ). Now we have to check

to see if this is a local maximum or minimum.owari

Exercise 4.16 Find the extrema of the following function.

f(x, y) = x2 − y2 Exercise4-17

-4-2

02

4-4

-2

0

2

4

-20

0

20

-4-2

02

4

fx(x, y) = 0, fy(x, y) = 0 is notsufficient condition for the ex-istence of a limit.

f(x, y) = x2 − y2. Find ∆at (0, 0). Then fxx(0, 0) = 2,fxy(0, 0) = 0, fyy(0, 0) = −2.Thus, ∆ = 2(−2)− 02 = −4 <0 and f(0, 0) is not extremevalue.

SOLUTION If f(x, y) takes the extreme value at (x0, y0), then

fx(x0, y0) = 0 ⇒ 2x0 = 0

fy(x0, y0) = 0 ⇒ 2y0 = 0

Thus a point (0, 0) is a critical point. Now as (x, y) approaches (0, 0) alongx-axis, we have f(x, y) > f(0, 0), along y-axis we have f(x, y) < f(0, 0). Thusf does not take the extreme value at (0, 0)

4.7. EXTREME VALUES OF FUNCTION OF TWO VARIABLES 25

Characterization Local Extrema Theorem 4.6 (Second Derivative Test) Let f(x, y) be the class C2 at(x0, y0) in the region D. If fx(x0, y0) = fy(x0, y0) = (0, 0), then denotefxx(x0, y0) = A, fxy(x0, y0) = B, fyy(x0, y0) = C,∆ = AC −B2.1. If ∆ > 0, A > 0, then f(x0, y0) is a local minimum.2. If ∆ > 0, A < 0, then f(x0, y0) is a local maximum.3. If ∆ < 0, then f(x0, y0) is a saddle point4. If ∆ = 0, then test is no conclusive. Check

For Q =A2

[(h+ Bk

A )2 + (AC−B2)k2

A2

],

(h + BkA )2 and k2

A2 are squaresand thus non negative.Therefore, the sign of Q isdetermined by the sign of Aand AC −B2.

4.7.1 Taylor Theorem for Two Variables

Consider approximating z = f(x, y) by a quadratic polynomial of x and y. Let

Aroximation Note that the total dif-ferential is an approxima-tion of the surface z =f(x, y) at (x0, y0) by thetangent plane. If we ap-proximate the surface bythe quadratic polynomial,we expect better approxi-mation.

f(x, y) = a0 + a1x+ a2y+ a3x2 + a4xy+ a5y

2 +R3, where R3 is an error term.Then find all 2nd order partial derivatives of z = f(x, y). If R3 is so small thatwe can neglect, then

f(x, y) = a0 + a1x+ a2y + a3x2 + a4xy + a5y

2

fx(x, y) = a1 + 2a3x+ a4y, fy(x, y) = a2 + a4x+ 2a5y

fxx(x, y) = a3, fxy(x, y) = a4, fyy(x, y) = a5.

Now put x = 0, y = 0. Then

f(0, 0) = a0, fx(0, 0) = a1, fy(0, 0) = a2

fxx(0, 0) = a3, fxy(0, 0) = a4, fyy(0, 0) = a5

Thus we can express all coefficients of f by the 2nd order partial derivatives.Therefore,

f(x, y) = a0 + a1x+ a2y + a3x2 + a4xy + a5y

2

= f(0, 0) + fx(0, 0)x+ fy(0, 0)y

+ fxx(0, 0)x2 + fxy(0, 0)xy + fyy(0, 0)y

2

δ neighborhood A δ neighborhoodof (x0, y0) is a setof (x, y) such that|(x− x0, y − y0)| < δ. Partial Differential Operator Let h, k be constants, Wedefine h ∂

∂x + k ∂∂y by

(h ∂∂x + k ∂

∂y )f(x0, y0) =

h∂f∂x (x0, y0) + k ∂f∂y (x0, y0).

Taylor Theorm for Two Variables Theorem 4.7 If z = f(x, y) is the class Cn in the δ neighborhood of(x0, y0), then for (x0 + h, y0 + k),

f(x0 + h, y0 + k) = f(x0, y0) +

(h∂

∂x+ k

∂

∂y

)f(x0, y0)

+1

2!

(h∂

∂x+ k

∂

∂y

)2

f(x0, y0) + · · ·

+1

(n− 1)!

(h∂

∂x+ k

∂

∂y

)n−1

f(x0, y0) +Rn

where,

Rn =1

n!

(h∂

∂x+ k

∂

∂y

)nf(x0 + θh, y0 + θk) (0 < θ < 1)

(h∂

∂x+ k

∂

∂y

)mf(x, y) =

(h∂

∂x+ k

∂

∂y

)(h∂

∂x+ k

∂

∂y

)m−1

f(x, y), m = 2, 3, ··


NOTE Let F (t) = f(x0 + ht, y0 + kt) (0 ≤ t ≤ 1). Then F (t) is the class Cn int. Thus by Maclausin theorem,

F (t) = F (0) +1

1!F ′(0)t+

1

2!F ′′(0)t2 + · · ·+ 1

(n− 1)!F (n−1)(0)tn−1 +Rn,

Rn =1

n!F (n)(θt)tn (0 < θ < 1).

Thus for t = 1,

F1. = f(x0 + h, y0 + k) = f(x0, y0) +

(h∂

∂x+ k

∂

∂y

)f(x0, y0)

+1

2!

(h∂

∂x+ k

∂

∂y

)2

f(x0, y0) + · · ·

+1

(n− 1)!

(h∂

∂x+ k

∂

∂y

)n−1

f(x0, y0) +Rn

Maclaurin Theoremf(x, y) = f(0, 0) +xfx(0, 0) + yfy(0, 0) +12 (x

2fxx(0, 0) + 2xyfxy(0, 0) +

y2fyy(0, 0)) + · · · + 1n! (x

∂∂x +

y ∂∂y )

n−1f(θx, θy) (0 < θ < 1)

Example 4.17 Given f(x, y) = ex cos y. Find the Taylor polynomial of 2nddegree at (0, 0).

Example4-16

By Maclaurin theorem,f(x, y) = f(0, 0) + xfx(0, 0) +yfy(0, 0) + 1

2 (x2fxx(0, 0) +

2xyfxy(0, 0) + y2fyy(0, 0)) +R3s. SOLUTION We first find all 2nd partial derivatives of f(x, y) = ex cos y. fx =

ex cos y, fy = ex(− sin y) = −ex sin y, fxx = ex cos y, fxy = −ex sin y, fyy =−ex cos y. Theorem4.7, let x0 = 0, y0 = 0, x = h, y = k. Then f(0, 0) =1, fx(0, 0) = 1, fy(0, 0) = 0, fxx(0, 0) = 1, fxy(0, 0) = 0, fyy(0, 0) = −1. Thus

f(x, y) = f(0, 0) + xfx(0, 0) + yfy(0, 0) +1

2!

x2fxx(0, 0)

+ 2xyfxy(0, 0) + y2fyy(0, 0)+R3

= 1 + x+1

2!

x2 − y2

+R3 Exercise4-16

Note that Taylor polynomialof 2nd degree of a functionat (π2 , 1) means expressing thefunction using (x− π

2 ) and (y−1).

Exercise 4.17 Find the 2nd degree Taylor polynomial of f(x, y) = sinxy at(π2 , 1).

SOLUTION fx = y cosxy, fy = x cosxy, fxx = −y2 sinxy, fxy = cosxy −xy sinxy, fyy = −x2 sinxy. Thus in Theorem4.7, let x = x0 + h = π

2 + h,y = y0 + k = 1 + k. Then

f(x, y) = f(π

2, 1) + (x− π

2)fx(

π

2, 1) + (y − 1)fy(

π

2, 1) +

1

2!

(x− π

2)2fxx(

π

2, 1)

+ 2(x− π

2)(y − 1)fxy(

π

2, 1) + (y − 1)2fyy(

π

2, 1)+R3

= − 1

2!

(x− π

2)2 − (x− π

2)(y − 1)− π2

4(y − 1)2

+R3

Proof By Taylor theorem, for (x0 + h, y0 + k) ∈ D, we have

f(x0 + h, y0 + k) = f(x0, y0) + hfx(x0, y0) + kfy(x0 + y0)

+1

2[h2fxx(x0 + θh, y0 + θk) + 2hkfxy(x0 + θh, y0 + θk)

+ k2fyy(x0 + θh, y0 + θk)], (0 < θ < 1)

Now let Q = f(x0 + h, y0 + k)− f(x0, y0). Then fx(x0, y0) = fy(x0, y0) = 0 and

Q =1

2[h2fxx(x0+θh, y0+θk)+2hkfxy(x0+θh, y0+θk)+k

2fyy(x0+θh, y0+θk)].

4.7. EXTREME VALUES OF FUNCTION OF TWO VARIABLES 27

Next let A = fxx(x0 + θh, y0 + θk), B = fxy(x0 + θh, y0 + θk), C = fyy(x0 +θh, y0 + θk). Then

Q =1

2

(Ah2 + 2Bhk + Ck2

)=A

2

[(h+

Bk

A)2 +

(AC −B2)k2

A2

].

Thus the sign of Q is determinde by the sign of A and the sign of AC −B2.1. If ∆ = AC − B2 > 0, A > 0, then since f(x, y) is the class C2 function,

for any h, k such that |h|, |k| is sufficiently small and never 0 simulteneously, wehave Q > 0. Thus, f(x0, y0) is a local minimum.

CheckQ = positive

[positive +

(positive)positivepositive

]. Thus

Q > 0.

2. If ∆ = AC − B2 > 0, A < 0, then since f(x, y) is the class C2 func,for any h, k such that |h|, |k| is sufficiently small and never 0 simulteneously, wehave Q < 0. Thus f(x0, y0) is a local maximum.

CheckQ = negative

[positive +

(positive)positivepositive

]. Thus

Q < 0.3. If ∆ = AC −B2 < 0 and A > 0, then C < 0 which gives a saddle point.Similarly, if ∆ < 0 and A < 0, then C > 0 which give a saddle point.

Example 4.18 Find the local extrema of the following function.

f(x, y) = x2 + xy + 3y2 + x+ y

SOLUTION In Example4.16, we found the critical point. Now we check to seewhether the function takes a local extremum at the critical point. Now by the2nd derivative test,

fxx(x, y) = 2, fxy(x, y) = 1, fyy(x, y) = 6,∆ = 11

Thus, f(− 511 ,−

111 ) = − 3

11 is a local minimum.

Exercise 4.18 Find the local extrema of th following function.

f(x, y) = x3 + y3 − 3xy CheckMultiply the equation 4.1 by y0and multiply the equation 4.2by x0. Then subtract the latterone from the former one to ob-tain 3x30 − 3y30 = 0. From this,we get x0 = y0 and put thisback to the equation 4.1, then3x20 − 3x0 = 3x0(x0 − 1) = 0.Thus, x0 = 0, 1.

SOLUTION Let f(x, y) = x3 + y3 − 3xy. Then we have fx = 3x2 − 3y, fy =3y2 − 3x. If f(x, y) takes the local maxima at (x0, y0), then

fx(x0, y0) = 3x20 − 3y0 = 0 (4.1)

fy(x0, y0) = 3y20 − 3x0 = 0 (4.2)

Solve this for x0, y0. Then substitute the equation y0 = x20, which is derivedfrom the equation 4.1, to the equation 4.2. Then

3x40 − 3x0 = 3x0(x30 − 1) = 0.

Thus x0 = 0, 1. Hence, (x0 = 0, y0 = 0), (x0 = 1, y0 = 1).Now we apply the 2nd derivative test. Since fxx(x, y) = 6x, fxy(x, y) =

−3, fyy(x, y) = 6y, at (0, 0) we have

∆ = fxx(0, 0)fyy(0, 0)− (fxy(0, 0))2 = −9 < 0.

This shows that f(0, 0) is not extrema. Now at (1, 1), we have

∆ = fxx(1, 1)fyy(1, 1)− (fxy(1, 1))2 = 6 · 6− (−3)2 = 27 > 0

A = fxx(1, 1) = 6 > 0.

Thus, f(1, 1) = 1 + 1− 3 = −1 is the local minimum Exercise A


1. Find the 2nd degree Taylor polynomial of f at (a, b).．

(a) f(x, y) = x2y, (a, b) = (1, 1) (b) f(x, y) = cosxy, (a, b) = (1,π

2)

(c) f(x, y) = log (1− x+ y), (a, b) = (1, 1) (d) f(x, y) = xe2x+y, (a, b) = (0, 0)

2. Find the local extrema of the following functions.

(a) f(x, y) = 2x− x2 − y2 (b) f(x, y) = x2 − 6y2 + y3

(c) f(x, y) = x3 − 3x+ y (d) f(x, y) = x2 + xy + y2 − 3x− 3y

Exercise B

1. Find the local extrema of the following functions．

(a) f(x, y) = 2x2 + y2 − xy − 7y (b) f(x, y) =x

y2+ xy

(c) f(x, y) = x3 + y3 − 3xy (d) f(x, y) = (x2 + y2)2 − 2(x2 − y2)

2. Find the 2nd degree Taylor polynomial of f at (a, b)．

(a) f(x, y) = ex cos y, (a, b) = (0, 0)

(b) f(x, y) = log (x+ y2), (a, b) = (2, 1)

4.8. CONDITIONAL EXTREMA 29

4.8 Conditional Extrema

4.8.1 Implicit Functions

Implicit Functions From the equation 3x+ 2y + 1 = 0, we can find y in terms of x. In other

words, we can find y = −1− 3x

2. In general, given a quadratic function

f(x, y), if y = g(x) always satisfies f(x, g(x)) = 0, then we say the equationy = g(x) is an implicit function determined by the equation f(x, y) = 0.

Understanding Finding the implicit func-tion y = g(x) determinedby the equation f(x, y) =0 is the same as solving theequation f(x, y) = 0 for y.But for some type of f andsome x, there may not bea y satisfying the equationf(x, y) = 0.

Implicit Function Theorem Theorem 4.8 Suppose that f(x, y) is the class C1 at x0 = (x0, y0) in theregion D. If

f(x0, y0) = 0, fy(x0, y0) = 0

, then there is an unique implicit function y = g(x) determined by f(x, y) =0 in the neighborhood of x = x0 which satisfies

1. f(x, g(x)) = 0

2. g(x0) = y0

3. g(x) is C1,dy

dx= −fx(x, y)

fy(x, y)

Checkfx(x, y) is a funciton of x, y andy is a function of x. Thus wehave the following figure.

Thus, dfx(x,y)dx = fxx(x, y) +

fxy(x, y)dydx

NOTE In the neighborhood of the point satisfying fy(x, y) = 0, the implicit func-tion y exists and the implcit function is differentiable. Thus the total diffetentialof f is

fxdx+ fydy = 0

Now using the condition fy = 0,

dy

dx= −fx

fy

This is 3. Furthermore, differentiate the above equation with respect to x, wehave

d2y

dx2= −

dfxdx fy − fx

dfydx

f2y

Now we usedfxdx

= fxx + fxydy

dx= fxx − fxy

fxfy

dfydx

= fyx + fyydy

dx= fxy − fyy

fxfy

to obtain

d2y

dx2= −

(fxx − fxyfxfy)fy − fx(fxy − fyy

fxfy)

f2y

= −fxxf

2y − 2fxyfxfy + fyyf

2x

f3y

Implicit Functions

d2ydx2 = − fxxf

2y−2fxyfxfy+fyyf

2x

f3y

is

called 2nd order derivativeof implicit functionExample 4.19 Find the derivative of a implicit function y such as

dy

dx,d2y

dx2deteremined by

x2 + y2 = 3xy


SOLUTION Set f(x, y) = x2 + y2 − 3xy and take total differential of f(x, y),

df = fxdx+ fydy = (2x− 3y)dx+ (2y − 3x)dy = 0

Thus,dy

dx= −2x− 3y

2y − 3x.

Since fxx = 2, fxy = −3, fyy = 2,

d2y

dx2= −2(2y − 3x)2 + 6(2x− 3y)(2y − 3x) + 2(2x− 3y)2

(2y − 3x)3

=10(x2 − 3xy + y2)

(2y − 3x)2= 0

Example4-19

A problem of finding a differen-tiation of an implicit function issolved by taking total differen-tiatial Checkx2 + y2 = 3xy

Exercise 4.19 Find the derivative of a implicit function y such asdy

dx,d2y

dx2deteremined by

2x2 + 5xy − 3y2 = 1 Exercise4-19Since 4(5x − 6y)2 − 10(4x +5y)(5x − 6y) − 6(4x + 5y)2 =−98(2x2+5xy−3y2) and 2x2+5xy − 3y2 = 1.

SOLUTION Set f(x, y) = 2x2 + 5xy − 3y2 − 1 = 0 and find total differential off(x, y). Then

df = fxdx+ fydy = (4x+ 5y)dx+ (5x− 6y)dy = 0.

Thus,dy

dx= −4x+ 5y

5x− 6y.

d2y

dx2= −

(4 + 5 dydx )(5x− 6y)− (4x+ 5y)(5− 6 dydx )

(5x− 6y)2

= −4(5x− 6y)2 − 10(4x+ 5y)(5x− 6y)− 6(4x+ 5y)2

(5x− 6y)3=

98

(5x− 6y)3

Example 4.20 Find zx, zy for the implicit function z = g(x, y) determinedby the equation xy + yz + zx = 1. Example4-20

By writing z = 1−xyy+x , we can

find zx = −y(y+x)−(1−xy)(y+x)2 =

− y2+1(y+x)2 = − y+z

y+x .

SOLUTION For f(x, y, z) = xy + yz + zx− 1 = 0, set x = t, y = s. Then,

∂f

∂t=∂f

∂x· ∂x∂t

+∂f

∂z

∂z

∂x

∂x

∂t= 0

Thus,

∂z

∂x= −

∂f∂x

∂x∂t

∂f∂z

∂x∂t

= − y + z

y + x

Similarly,

∂z

∂y= −

∂f∂y

∂x∂t

∂f∂z

∂y∂t

= −x+ z

y + x

Exercise 4.20 Find zx, zy for the implicit function z = g(x, y) determined bythe equation x2 + y2 + z2 = 4.

4.8. CONDITIONAL EXTREMA 31

SOLUTION For f(x, y, z) = x2 + y2 + z2 − 4, set x = t, y = s.

∂f

∂t=∂f

∂x

∂x

∂t+∂f

∂z

∂z

∂x

∂x

∂t= 0

Thus,

∂z

∂x= −

∂f∂x

∂x∂t

∂f∂z

∂x∂t

= −2x

2z= −x

z

Similarly,

∂z

∂y= −

∂f∂y

∂x∂t

∂f∂z

∂y∂t

= −2y

2z= −y

z

y, z is an implicit function of x determined by f(x, y, z) = 0, g(x, y, z) = 0. Theorem 4.9 If an implicit function y, z of x is determined by the equa-tions f(x, y, z) = 0, g(x, y, z) = 0, then,

dy

dx= −fxgz − fzgx

fygz − fzgy.

Theorem4-9

To find dydx , we first find df, dg

and then eliminate dz.

Example 4.21 Finddy

dx,dz

dxfor the implicit functions y, z of x determined by

the equations x2 + y2 + z2 = 1, 2x+ 3y − z = 1.

SOLUTION Let f(x, y, z) = x2+y2+z2−1 = 0, g(x, y, z) = 2x+3y−z−1 = 0.Then totally differentiate f and g to get

df = 2xdx+ 2ydy + 2zdz = 0, dg = 2dx+ 3dy − dz = 0

Now delete dz to get

(2x+ 4z)dx+ (2y + 6z)dy = 0

CheckFrom the equations 2xdx +2ydy + 2zdz = 0, 2dx + 3dy −dz = 0, we delete dz by adding4zdx+ 6zdy − 2zdz = 0 to thefirst equation,2xdx + 4zdx + 2ydy + 6zdy =(2x+ 4z)dx+ (2y + 6z)dy = 0.

Thus,dy

dx= −2x+ 4z

2y + 6z

Similarly, delete dy to get

(6x− 4y)dx+ (6z + 2y)dz = 0

Thus,dz

dx= −6x− 4y

6z + 2y

Exercise 4.21 Finddy

dx,dz

dxfor the implicit functions y, z of x determined by

the equations xyz = 1, xy + yz + zx = 1. Checkyzdx + xzdy + xydz = 0, (y +z)dx+ (x+ z)dy+ (y+ x)dz =0. Eliminate dz by multiplying(y + z) to the former equationand multiplyimg xy to the lat-ter equation.yz(y + x)dx + xz(y + x)dy −(xy(y+z)dx−xy(x+z)dy) = 0.Simplifying, (yz(y+x)−xy(y+z))dx + (xz(y + x) − xy(x +z))dy = 0

SOLUTION Let f(x, y, z) = xyz−1 = 0, g(x, y, z) = xy+yz+zx−1 = 0. Thenfind total derivatives.

df = fxdx+ fydy + fzdz = yzdx+ xzdy + xydz = 0 (4.3)

dg = gxdx+ gydy + gzdz = (y + z)dx+ (x+ z)dy + (y + x)dz = 0(4.4)

Now using the equation 4.3 and the equation 4.4 to eliminate dz.

((y + x)yz − xy(y + z))dx+ ((y + x)xz − xy(x+ z))dy = 0


Thendy

dx= −y

2z − xy2

x2z − x2y=y2(x− z)

x2(z − y).

This time, using the equation 4.3 and the equation 4.4 to eliminate dy.

((x+ z)yz − xz(y + z))dx+ ((x+ z)xy − xz(y + x))dz = 0

z2(y − x)dx+ x2(y − z)dz = 0

Thusdz

dx=z2(x− y)

x2(y − z)

Exercise A

1. Find thedy

dx,d2y

dx2for the implicit functions y of x determined by the fol-

lowing functions．

(a) x− y2 = 1 (b) x2 + xy + 2y2 = 1 (c) x− ey = 0

(d) x3 − 3xy + y3 = 1

2. Find thedy

dx,dz

dxfor the implicit functions y, z of x determined by the

following functions．

(a) x2 + y2 + z2 = 4, x+ y + z = 1 (b) xyz = 1, x+ y + z = 1

Exercise B

1. Finddy

dx,d2y

dx2for the implicit function y of x determined by the following

equations.

(a) 2x2 + 5xy − 3y2 = 1 (b) y = ex+y (c) x2 − y2 = xy

(d) log√x2 + y2 = tan−1 y

x

2. Finddy

dx,dz

dxfor the implicit function y, z of x determined by the following

equations.

(a) x2 + y2 + z2 = 4, x2 + y2 = 4x (b) xyz = 1, xy + yz + zx = 1

3. Find the equation of the tangent line and the normal line to the curve2x2 + 5y2 = 12 at the point (1,

√2).

4. Find the equation of the tangent plane to the surface z = tan−1 y

xat the

point (1, 1, π2 )．Find the equation of the normal line through (1, 1, π2 ).

5. Find the local extrema of y = g(x) implicitly defined by the following equa-tions．

(a) 8x2 + 4xy + 5y2 = 36 (b) x2y + x+ y = 0

(c) x3 + y3 − 6xy = 0

4.9. LAGRANGE MULTIPLIER 33

4.9 Lagrange Multiplier

Lagrange Multiplier Theorem 4.10 If f(x, y) takes extrema at x, y given the constraintϕ(x, y) = 0, Then let

F (x, y, λ) = f(x, y)− λϕ(x, y)

and solve the follwing equations for x, y

Fλ(x, y) = −ϕ(x, y) = 0, Fx(x, y) = 0, Fy(x, y) = 0

Understanding The constraint is ϕ(x, y) =0. Thus the set ofreal numbers (x, y) :ϕ(x, y) = 0 is a closedbounded set. Thus ifz = f(x, y) is continuous,then z = f(x, y) takes ei-ther the maximum or min-imum. Implicit Function

f(x, y) is the class C1・・,1. f(x0, y0) = 02. fy(x0, y0) = 0.

NOTE Under the constraint ϕ(x, y) = 0, we find the extrema of f(x, y). Asϕ(x, y) = 0, if ϕy(x, y) = 0, then we can find the implicit function y = g(x)satissfying ϕ(x, y) = 0. If a function z = f(x, g(x)) takes the extrema at (x0, y0),then y0 = g(x0) and zx = 0. Thus

fx(x0, y0) + fy(x0, y0)g′(x0) = 0.

Now differentiate the both sides of the equation ϕ(x, y) = 0 by x. Then

ϕx(x0, y0) + ϕy(x0, y0)g′(x0) = 0.

Thus

fx(x0, y0)− fy(x0, y0)ϕx(x0, y0)

ϕy(x0, y0)= 0.

Letfy(x0, y0)

ϕy(x0, y0)= λ

Then there exists λ satisying

fx(x0, y0)− λϕx(x0, y0) = 0

CheckSolve for g′(x), we have

g′(x) = −ϕx(x0,y0)ϕy(x0,y0)

. Now

substitute g′(x) intofx(x0, y0) + fy(x0, y0)g

′(x0) =0. Then fx(x0, y0) −fy(x0, y0)

ϕx(x0,y0)ϕy(x0,y0)

= 0.

Gradient ∇ϕ(x, y) = (ϕx, ϕy) iscalled gradient.

If we think of this theorem geometrically. Since ϕ(x, y) = 0, the vector∇ϕ(x, y) = (ϕx(x, y), ϕy(x, y)) is orthogonal to the curve ϕ(x, y) = 0. Let C bethe curve defined by the constraint ϕ(x, y) = 0 and (x(t), y(t)) is a point on C.If a function z = f(x, y) takes the extrema at (x(t0), y(t0)), then

df(x(t), y(t))

dt|t0 = fx(x(t), y(t)

dx

dt+ fy(x(t), y(t))

dy

dt|t0

= (fx(x(t0), y(t0)), fy(x(t0), y(t0))) · (dx

dt,dy

dt) |t0= 0.

Thua, the vector∇(f(x(t0), y(t0)) is orthogonal to the curve C. Then∇ϕ(xt0 , yt0)and ∇f(xt0 , yt0) are parallel. Thus, there exists λ so that

∇f(x, y) = λ∇ϕ(x, y)

Parallel Note that ∇ϕ(x0, y0)and ∇f(x0, y0)are parallel, then∇ϕ(x0, y0)×∇f(x0, y0) =

k

∣∣∣∣ ϕx ϕyfx fy

∣∣∣∣ = 0. Example 4.22 Find the maxima and minima of the following function withthe constraint ϕ(x, y) = 0.1. ϕ(x, y) = x2 + y2 − 4, f(x, y) = 3x2 + 4xy + 3y2

2. ϕ(x, y) = x2 − xy + y2 − 1, f(x, y) = xy


1. ϕ(x, y) = x2 + y2 − 4 = 0, f(x, y) = 3x2 + 4xy + 3y2. Let

F (x, y, λ) = 3x2 + 4xy + 3y2 − λ(x2 + y2 − 4)

Then

Fλ = 0 ⇒ x2 + y2 − 4 = 0 (4.5)

Fx = 0 ⇒ 6x+ 4y − 2xλ = 2(3− λ)x+ 4y = 0 (4.6)

Fy = 0 ⇒ 4x+ 6y − 2yλ = 4x+ (6− 2λ)y = 0 (4.7)

Example4-22

Using Fx = 0, Fy = 0, andϕ(x, y) = 0, first find a λ.

SOLUTION

CheckFx = 6x+4y−λ(2x) = 2x(3−λ) + 4yFy = 4x + 6y − λ(2y) = 4x +2y(3− λ)

(x, y) = (0, 0) does not satisy the equation 4.5. Then the condition that theequations 4.6 and 4.7 have solutions (x, y) not equal to = (0, 0),∣∣∣∣ 3− λ 2

2 3− λ

∣∣∣∣ = (3− λ)(3− λ)− 22. = λ2 − 6λ+ 9− 4 = 0

Thus, λ2 − 6λ+ 5 = 0 and λ = 1, 5.For λ = 1, by the equation 4.6, 4x + 4y = 0 and y = −x. Substitute

y = −x into the equation 4.5, we have x2 + x2 − 4 = 0 and x = ±√2. Since

y = −x, we have (x, y) = (√2,−

√2), (−

√2,√2). With these x, y, the value of

3x2 + 4xy + 3y2 is 4.For λ = 5, by the equation 4.6, −4x + 4y = 0 and y = x. Substitute y = x

into the equation 4.5. Then x2+x2−4 = 0 which implies x = ±√2. Since y = x,

(x, y) = (√2,√2), (−

√2,−

√2). With these x, y, the value of 3x2 +4xy+3y2 is

20.

Check

3(√2)2 + 4

√2(√2) + 3(

√2)2 =

6 + 8 + 6 = 20

On the other hand, (x, y) : x2 + y2 = 4 is closed bounded region, and onthis region, 3x2 + 4xy + 3y2 is continuous, thus takes the extrema. these arealso local extrema. Thus by the result above, the extrema must be 4 or 20.Therefore, the maximum value is 20 and the minimum value is 4

2. Note that ϕ(x, y) = x2 − xy + y2 − 1 = 0, f(x, y) = xy. Then let

F (x, y, λ) = xy − λ(x2 − xy + y2 − 1)

Fλ = 0 ⇒ x2 − xy + y2 − 1 = 0 (4.8)

Fx = 0 ⇒ y − λ(2x− y) = −2λx+ (1 + λ)y = 0 (4.9)

Fy = 0 ⇒ x− λ(−x+ 2y) = (1 + λ)x− 2λy = 0 (4.10)

CheckFx = y − λ(2x − y) = −2λx +(1 + λ)y = 0Fy = x − λ(−x + 2y) = 0 =(1 + λ)x− 2λy = 0

System of Equation A sysytem of 1st order lin-ear equationax+ by = 0cx+ dy = 0

has no

solution if and only if∣∣∣∣ a bc d

∣∣∣∣ = 0

Now (x, y) = (0, 0) does not satisy the equation 4.8. Then the condition forthe equation 4.9), (4.10) has the solution (x, y) = (0, 0),∣∣∣∣ −2λ 1 + λ

1 + λ −2λ

∣∣∣∣ = −2λ(−2λ)− (1 + λ)2 = 4λ2 − λ2 − 2λ− 1

= 3λ2 − 2λ− 1 = (3λ+ 1)(λ− 1) = 0

Thus, λ = − 13 , 1.

For λ = − 13 , by the equation 4.9, 2

3x + 23y = 0 and y = −x. Substitute

y = −x into the equation 4.8. Then x2 + x2 + x2 − 1 = 0. From this, x = ± 13 .

Since y = −x, (x, y) = ( 13 ,−13 ), (−

13 ,

13 ). Now the value of xy for these values of

x, y is − 19 .

For λ = 1, by the equation 4.9, 2y−2x = 0 and y = x. Substitute y = x intothe equation4.8. Then x2 − x2 + x2 − 1 = 0. From this, x = ±1. Since y = x,(x, y) = (1, 1), (−1,−1). Now the valuesof xy for these values of x, y is 1.

4.9. LAGRANGE MULTIPLIER 35

On the other hand, (x, y) : x2 − xy + y2 = 1 is closed bounded region,and on this region, x2 − xy + y2 is continuous, thus takes the extrema. theseare also local extrema. Thus by the result above, the extrema must be − 1

9 or 1.Therefore, the maximum value is 1 and the minimum value is − 1

9

Exercise 4.22 Find the shortest distance from (1,−1, 2) to the surface x2 − 2y + 2z − 10 = 0.

SOLUTION Let f(x, y, z) be the distance from the point (1,−1, 2) to an arbi-trary point on the surface. Then minimize f2.

f2 = (x− 1)2 + (y + 1)2 + (z − 2)2

x2 − 2y + 2z − 10 = 0

Let

F (x, y, z, λ) = (x− 1)2 + (y + 1)2 + (z − 2)2 − λ(x2 − 2y + 2z − 10)

Then

Fλ = 0 ⇒ x2 − 2y + 2z − 10 = 0 (4.11)

Fx = 0 ⇒ 2(x− 1)− 2xλ = 0 (4.12)

Fy = 0 ⇒ 2(y + 1) + 2λ = 0 (4.13)

Fz = 0 ⇒ 2(z − 2)− 2λ = 0 (4.14)

By the equations 4.12, 4.13, 4.14, we have

x =1

1− λ, y = −1− λ, z = 2 + λ

Putting these values into 4.11,

(1

1− λ)2 − 2(−1− λ) + 2(2 + λ)− 10 = 0

Thus (1

1− λ

)2

= 4(1− λ)

Now solve this for λ to get

λ = 1−(1

2

)2/3

Thus,

x = 22/3, y = −2 +

(1

2

)2/3

, z = 3−(1

2

)2/3

Note that this point is on the surface. Since the value of f2 is not boundedabove. Thus this is the local minimum. Furthermore, it is only one. Thus, it isthe minimum value

Check( 11−λ )

2−2(−1−λ)+2(2+λ)−10 = ( 1

1−λ )2+2+2λ+4+2λ−

10 = ( 11−λ )

2 − 4 + 4λ Check(1

1−λ

)2= 4(1−λ), 4(1−λ)3 =

1. (1− λ)3 = 14 . 1− λ = ( 14 )

13 .

λ = 1− ( 12 )2/3.

Exercise A

1. Find the maxima and the minimua of f(x, y) with the constraint g(x, y) = 0．

(a) g(x, y) = x2 + y2 − 1, f(x, y) = x2 + 3y2

(b) g(x, y) = x2 + y2 − 4, f(x, y) = x2 + y2 − 2x

(c) g(x, y) = x2 + y2 − 1, f(x, y) = xy + x+ y

2. Find the shortest distance from the origin to the curve x2 + xy + y2 = 14．


Exercise B

1. Find the local extrema of the functions y = g(x) implicitly defined by thefollowing functions ．

(a) 8x2 + 4xy + 5y2 = 36 (b) x2y + x+ y = 0 (c) x3 + y3 − 6xy = 0

2. Find the maxima and minima of f(x, y) with the constraint g(x, y) = 0．

(a) g(x, y) = x2 + y2 − 1, f(x, y) = xy3

(b) g(x, y) = x3 + y3 − 6xy, f(x, y) = x2 + y2

(c) g(x, y) = x2 − xy + y2 − 1, f(x, y) = xy

3. Find the maxima of xy for if the point P(x, y) moves on the line 2x+3y = 12．

4. Find the maxima and minima of x2 + 2y2 + 3z2 for if the point P(x, y, z)moves on the surface x2 + y2 + z2 = 1.

Chapter 5

Multiple Integrals5.1 Double Integrals

Double Integral Over Rectangular Region Let z = f(x, y) be a bounded function on the rectangular region W overthe xy-plane. Divide the rectangular region W by the straight lines par-allel to x-axis and y-axis and denote the partitioned small rectangles byW1,1,W1,2, . . . ,Wm,n. We denote this partition by ∆. Now for each Wij ,take an arbitrary point (xi, yj) and consider the sum of small rectangularparallelpiped S(∆). Let

S(∆) =

m∑i=1

n∑j=1

f(xi, yj)|Wij |

where |Wij | is the area ofWij and |∆| is the longest diagonal ofWij . If S(∆)approaches the same value as |∆| approaches 0 independent of the partitionand the choice of (xi, yj), then f(x, y) is called Double Integrable onW .

lim∆→0

S(∆) = limm→∞n→∞

m∑i=1

n∑j−1

f(xi, yj)|Wij | =∫∫

W

f(x, y)dxdy

small parallepiped

small rectangle

SumS(∆) is written as∑nj=1

∑mi=1 f(xi, yj)|Wij |.

Thus add the small rectan-gular parallelpiped in thedirection of x-axis, then addin the direction of y-axis isthe same as add the smallrectangular parallelpiped inthe direction of y-axis, thenadd in the direction of x-axis.This is the basic concept of therepeated integrals.

Repeated Integrals Over Rectangle Theorem 5.1 (Fubini Theorem) Let W = (x, y) : a ≤ x ≤ b, c ≤ y ≤d. If f(x, y) is continuous on W , then∫∫

W

f(x, y)dxdy =

∫ b

a

∫ d

c

f(x, y)dydx =

∫ d

c

∫ b

a

f(x, y)dxdy NOTE Fix y and conside the integration of f(x, y) from a to b with respect tox. Then we have ∫ b

a

f(x, y)dx

Now integrate this from c to d with respect to y to obtain∫ d

c

(∫ b

a

f(x, y)dx

)dy or

∫ d

c

∫ b

a

f(x, y)dxdy

This is called repeated integral.

37

38 CHAPTER 5. MULTIPLE INTEGRALS

Example 5.1 Let W = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 2. Evaluate the followingrepeated integral

∫∫Wx2ydxdy.

SOLUTION∫∫W

f(x, y)dxdy =

∫ 1

0

( ∫ 2

0

x2ydy)dx =

∫ 1

0

x2( ∫ 2

0

ydy)dx

=

∫ 1

0

x2([y2

2

]20

)dx

=

∫ 1

0

2x2dx =[23x3]10=

2

3

Example5-1

We evaluate∫ 2

0x2ydy by keep-

ing x as a constant.∫ 2

0

x2ydy = x2∫ 2

0

ydy

= x2[y2

2]20

= 2x2

Exercise 5.1 Let W = (x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 1. Evaluate the followingrepeated integrals

∫∫Wx sinπydxdy.

Exercise5-1It is possible to evaluate∫ 1

0

( ∫ 2

0x sinπydx

)dy.

SOLUTION∫∫W

x sinπydxdy =

∫ 2

0

( ∫ 1

0

x sinπydy)dx

=

∫ 2

0

x( ∫ 1

0

sinπydy)dx

=

∫ 2

0

x([

− 1

πcosπy

]10

)dx

=

∫ 2

0

x(− 1

π(−1− 1)

)dx

=

∫ 2

0

2

πxdx =

[ 2π(x2

2)]20=

2

π(4

2) =

4

π

Repeated Integrals Over Closed Bounded Regions Let z = f(x, y) be a function defined on the closed bounded region Ω onxy-plane. Let W be a rectangular region containing Ω. Now divide therectangular region W by the straight lines parallel to x-axis and y-axis anddenote the partitioned small rectangles by W1,1,W1,2, . . . ,Wm,n.

Figure 5.1: PartitionLet f(x, y) be the function on W defined by

f(x, y) =

f(x, y) (x, y) ∈ Ω

0 (x, y) ∈W − Ω

If f(x, y) is integrable on W , then we say f(x, y) is integrable on Ω and theintegration of f(x, y) on Ω is expressed as follows:∫∫

W

f(x, y)dxdy =

∫∫Ω

f(x, y)dxdy

5.2. REPEATED INTEGRALS 39

Double Integral Formula Theorem 5.2 Suppose that f(x, y), g(x, y) are continuous on Ω. Then wehave the followings.1. Let a, b be constants. Then∫∫

Ω

af(x, y) + bg(x, y)dxdy = a

∫∫Ω

f(x, y)dxdy + b

∫∫Ω

g(x, y)dxdy

2. If Ω = Ω1 ∪ Ω2, then∫∫Ω

f(x, y)dxdy =

∫∫Ω1

f(x, y)dxdy +

∫∫Ω2

f(x, y)dxdy

3. |∫∫

Ω

f(x, y)dxdy| ≤∫∫

Ω

|f(x, y)|dxdy

Understanding If the region Ω is not arectangular region, thenconsider the rectangle con-taining Ω. For the rectan-gle inside of Ω, use f(x, y)as it is given. For therectangular region outsideof Ω, we set f(x, y) = 0.This way we can define re-peated integral over Ω.

Linearity Theorem5-2-1. is calledlinearity of double in-tegral. Theorem5-2-2.

5.2 Repeated Integrals

Figure 5.2: Vertically Simple Region Figure 5.3: Horizontal simple

The closed region Ω bounded by a ≤ x ≤ b, ϕ1(x) ≤ y ≤ ϕ2(x) is called Verti-cally simple region, The closed region Ω bounded by c ≤ y ≤ d, ψ1(y) ≤ x ≤ψ2(y) is called Horizontally simple region.

Vertically Simple If a vertical line does notcross the same curve morethan once.

Horizontal Simple If a horizontal line doesnot cross the same curvemore than once.

Iterated Integral Theorem 5.3 1. If f(x, y) is continuous on the closed region Ω(5.2)bounded by a ≤ x ≤ b, ϕ1(x) ≤ y ≤ ϕ2(x), then∫∫

Ω

f(x, y)dxdy =

∫ b

a

∫ ϕ2(x)

ϕ1(x)

f(x, y)dydx

2. If f(x, y) is continuous on the closed region Ω(5.2) bounded by c ≤y ≤ d, ψ1(y) ≤ x ≤ ψ2(y), then∫∫

Ω

f(x, y)dxdy =

∫ d

c

∫ ψ2(y)

ψ1(y)

f(x, y)dxdy

EvaluationTo evaluate a double integral,use either vertically simple re-gion or horizontally simple re-gion.

Example 5.2 Given the region Ω as in the figure 5.4, evaluate the followingdouble integral ∫∫

Ω

(x2 − y)dxdy Example5-2

When a line is drawed verti-cally, it will not intersect thesame curve more thatn once.Thus the region is verticallysimple region.


Figure 5.4: −x2 ≤ y ≤ x2

SOLUTION When a horizontal line is drawed to the region Ω, it intersects thecurve more than once. But when a vertical line is drawed to the region Ω, itdoes not intersect more than once. Thus the region is vertically simple region.Now fix x. Then the region is in between the curve y = −x2 and the curvey = x2. Thus we have −x2 ≤ y ≤ x2. Next free x to get −1 ≤ x ≤ 1. Thus Ω isexpressed as follows.

Example5-2

Ω = (x, y) : −1 ≤ x ≤ 1, −x2 ≤ y ≤ x2

Therefore, ∫∫Ω

(x2 − y)dxdy =

∫ 1

−1

(

∫ x2

−x2

(x2 − y)dy)dx

=

∫ 1

−1

[(x2y − 1

2y2)

]x2

−x2

dx

=

∫ 1

−1

[(x4 − 1

2x4)− (−x4 − 1

2x4)]dx

=

∫ 1

−1

2x4dx =

[2

5x5]1−1

=4

5

Exercise 5.2 Give the region Ω as in the figure 5.5, evaluate the followingdouble integral. ∫∫

Ω

(x1/2 − y2)dxdy

Figure 5.5: x2 ≤ y ≤ x1/4

Exercise5-2

In the above figure, summingthe small rectangles to the y-axis direction. Then x2 ≤ y ≤x1/4. Now to fill the region us-ing these vertically long rectan-gles, we need to sum 0 ≤ x ≤ 1. SOLUTION This region is both vertically simple region and horizontally simple

region. We first evaluate the integral by using vertically simple region.Ω can be expressed by the following.

Ω = (x, y) : 0 ≤ x ≤ 1, x2 ≤ y ≤ x1/4


Thus ∫∫Ω

(x1/2 − y2)dxdy =

∫ 1

0

∫ x1/4

x2

(x1/2 − y2)dydx

=

∫ 1

0

[x1/2y − 1

3y3]x1/4

x2

dx

=

∫ 1

0

(2

3x3/4 − x5/2 +

1

3x6)dx

=

[8

21x7/4 − 2

7x7/2 +

1

21x7]10

=8

21− 2

7+

1

21=

1

7

This time, we evaluate the integral by using horizontally simple region. Ωcan be expressed by the following.

Example5-2

By horizontally simple region,y4 ≤ x ≤ y1/2. Then to fill theregion Ω, we get 0 ≤ y ≤ 1.

Ω = (x, y) : 0 ≤ y ≤ 1, y4 ≤ x ≤ y1/2

Then ∫∫Ω

(x1/2 − y2)dxdy =

∫ 1

0

∫ y1/2

y4(x1/2 − y2)dxdy

=

∫ 1

0

[2

3x3/2 − y2x

]y1/2y4

dy

=

∫ 1

0

(2

3y3/4 − y5/2 +

1

3y6)dy

=

[8

21y7/4 − 2

7y7/2 +

1

21y7]10

=8

21− 2

7+

1

21=

1

7

We obtain the same result as expected Change the order of integration

Interchage from vertically simple region to horizontally simple region or viceversa. Then corresponding integral change the order of integration.

Example 5.3 Evaluate the following double integral.∫ 1

0

∫ 1

y

ey/xdxdy

SOLUTION We can not evaluate the integral

∫ 1

y

ey/xdx. Then by the change

the order of integration, the region Ω is given by

Ω = (x, y) : 0 ≤ y ≤ 1, y ≤ x ≤ 1

Thus it is given by the horizontally simple region. Now express Ω by the verti-cally simple region.

Ω = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x


Then

Vertically simple

∫ey/xdy = xey/x∫ 1

0

∫ 1

y

ey/xdxdy =

∫ 1

x=0

(

∫ x

y=0

ey/xdy)dx

=

∫ 1

0

[xey/x]x0 dx

=

∫ 1

0

(xe− x)dx =

[(e− 1)x2

2

]10

=e− 1

2

Exercise 5.3 Evaluate the following double integral.∫ 1

0

∫ 1

x

ey2

dydx

Exercise5-3∫ey

2

dy is known for non-integrable function.

Express∫ 1

0

∫ 1

xey

2

dydx by∫ 1

0

( ∫ 1

xey

2

dy)dx. Then the

range of the integration of xand the range of integration ofy become clear.

SOLUTION Note that∫ey

2

dy is known for non-integrable. Thus it is impos-

sible to integrate

∫ 1

0

∫ 1

x

ey2

dydx in this order. Thus, interchange the order of

integration. Since

Ω = (x, y) : 0 ≤ x ≤ 1, x ≤ y ≤ 1

is given by the vertically simple region. Then express Ω by the horizontallysimple region.

Ω = (x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ y

Thus, ∫ 1

0

∫ 1

x

ey2

dydx =

∫ 1

y=0

(

∫ y

x=0

ey2

dx)dy

=

∫ 1

0

[xey2

]y0 dy

=

∫ 1

0

yey2

dy =

[1

2ey

2

]10

=e− 1

2

Exercise A

1. Evaluate the following double integrals．

(a)

∫∫Ω

xdxdy, Ω : −1 ≤ x ≤ 1, 0 ≤ y ≤ 3

(b)

∫∫Ω

(2x+ 3y)dxdy, Ω : 0 ≤ x ≤ 1, 0 ≤ y ≤ x

(c)

∫∫Ω

(1 + x+ xy)dxdy, Ω : 0 ≤ y ≤ 1, y2 ≤ x ≤ y

(d)

∫∫Ω

sin(x+ y)dxdy, Ω : 0 ≤ x ≤ π

2, 0 ≤ y ≤ π

2

(e)

∫∫Ω

x3ydxdy, Ω : 0 ≤ x ≤ 1, 0 ≤ y ≤ x

2. Interchange the order of integration．

(a)

∫ 1

0

∫ x

x2

f(x, y)dydx (b)

∫ 1

0

∫ y2

0

f(x, y)dxdy (c)

∫ 2

0

∫ 3−x

x2

f(x, y)dydx


3. Answer the following questions．

(a) Find the volume of solid bounded by the following surface under thesurface z = x+ y and above the triqngle (0, 0), (0, 1), (1, 0)

(b) Find the volume of solid bounded by the following surface under thesurface z = 2x+ 3y and qbove the square (0, 0), (0, 1), (1, 0), (1, 1).

(c) Find the volume of the solid bounded above by the surface z = x2+y2

and below by the plane x2 + y2 ≤ 1

Exercise B


(a)

∫∫Ω

x2dxdy, Ω : −1 ≤ x ≤ 1, 0 ≤ y ≤ 3

(b)

∫∫Ω

ex+ydxdy, Ω : 0 ≤ x ≤ 1, 0 ≤ y ≤ x

(c)

∫∫Ω

√xydxdy, Ω : 0 ≤ y ≤ 1, y2 ≤ x ≤ y

(d)

∫∫Ω

(4− y2)dxdy, Ω is bounded by y2 = 2x and y2 = 8− 2x.

2. Interchange the order of integrals.

(a)

∫ 1

0

∫ x2

x4

f(x, y)dydx (b)

∫ 1

0

∫ y

−yf(x, y)dxdy (c)

∫ 4

1

∫ 2x

x

f(x, y)dydx


(a)

∫ 1

0

∫ 1

y

ey/xdxdy (b)

∫ 1

0

∫ 1

x

ey2

dydx (c)

∫ 1

0

dy

∫ √y

y

sinx

xdx


5.3 Change of Variables

Change of Variable Formula Theorem 5.4 By the one-to-one correspondence Φ(u, v) =(ϕ(u, v), ψ(u, v)), the region Γ is mapped into Ω. Furthermore,ϕ(u, v), ψ(u, v) are the class C1 with respect to u, v. Now supposethat Jacobian of x, y with respect to u, v

J(u, v) =∂(x, y)

∂(u, v)= det

(ϕu(u, v) ϕv(u, v)ψu(u, v) ψv(u, v)

)is never be 0 on Γ. Then the following is true for the continuous functionf(x, y) on Ω.∫∫

Ω

f(x, y)dxdy =

∫∫Γ

f(ϕ(u, v), ψ(u, v))|J(u, v)|dudv

Determinant

det

(ϕu(u, v) ϕv(u, v)ψu(u, v) ψv(u, v)

)=

ϕu(u, v)ψv(u, v) −ϕv(u, v)ψu(u, v)is the determinant of thematrix.Jacobian J can be negative.

NOTE Let Ω be the region on xy-plane and Γ be the region on uv-plane.Suppose that Φ is a map from Γ to Ω satisfying

x = ϕ(u, v) = 2u

y = ψ(u, v) = 2v

Then (x

y

)=

(2 00 2

)(u

v

)= Φ

(u

v

)and Φ is invertible matrix. Thus there exists the inverse of Φ such that(

u

v

)= Φ−1

(x

y

)=

1

4

(2 00 2

)(x

y

).

Invertible Matrix A matrix A is called in-vertible matrix if thereexists a matrix X suchthat AX = XA = E.Write X = A−1.

Inverse Matrix A =

(a bd d

), A−1 =

1ad−bc

(d −b−c a

)

Consider the rectangle with 4 vertices (x, y), (x + ∆x, y), (x, y + ∆y), (x +∆x, y +∆y).

Figure 5.6: Φ−1

Then the area of the rectangle is ∆x∆y. Now correspondence area of uv-

plane is given by

(u

v

)= Φ−1

(x

y

)

5.3. CHANGE OF VARIABLES 45(x, y) → (x2 ,

y2 ) = (u, v)

(x+∆x, y) → (x+∆x2 , y2 ) = (u+∆u, v)

(x, y +∆y) → (x2 ,y+∆y

2 ) = (u, v +∆v)

(x+∆x, y +∆y) → (x+∆x2 , y+∆y

2 ) = (u+∆u, v +∆v)

Thus the area ∆u∆v of uv-plane is

∆u∆v = (u+∆u− u)(v +∆v − v)

=

(x+∆x

2− x

2

)(y +∆y

2− y

2

)=

1

4∆x∆y

Thus∆x∆y = 4∆u∆v = |ϕuψv − ϕvψu|∆u∆v = |J(u, v)|∆u∆v.

This J(u, v) is the jacobian.

Jacobian If x = ϕ(u, v) = 2u, y =ψ(u, v) = 2v, then the ja-cobian J(u, v) is J(u, v) =

14 =

∣∣∣∣ 2 00 2

∣∣∣∣. This is the

same as

∣∣∣∣ ϕu ϕvψu ψv

∣∣∣∣. Thusthe absolute value of jaco-bian is the ratio of area ofthe dxdy and dudv. Polar Coordinate Transformations

Theorem 5.5 To transform the polar coordinate (r, θ) to the rectangularcoordinate (x, y), since x = r cos θ, y = r sin θ,

J(r, θ) =

∣∣∣∣ ∂x∂r

∂x∂θ

∂y∂r

∂y∂θ

∣∣∣∣ = ∣∣∣∣ cos θ −r sin θsin θ r cos θ

∣∣∣∣ = r(cos2 θ + sin2 θ) = r

Thus ∫∫Ω

f(x, y)dxdy =

∫∫Γ

f(r cos θ, r sin θ)rdrdθ Example 5.4 Evaluate the following double integrals.

1.

∫∫Ω

xydxdy, Ω = (x, y) : x2 + y2 ≤ 1, x, y ≥ 0

2.

∫∫Ω

1

x2 + y2dxdy, Ω = (x, y) : 1 ≤ x2 + y2 ≤ 4, y ≥ 0

3.

∫∫Ω

(x− y)2dxdy, Ω = (x, y) : |x+ 2y| ≤ 1, |x− y| ≤ 1

SOLUTION 1. Ω is a circular region. Thus using polar coordinate, x =r cos θ, y = r sin θ. Then 0 ≤ x2 + y2 = r2 ≤ 1 and 0 ≤ r ≤ 1. Also, x =

r cos θ ≥ 0, y = r sin θ ≥ 0 and 0 ≤ θ ≤ π

2.

Example5-4-1.

Thus Ω is transformed to

Γ = (r, θ) : 0 ≤ θ ≤ π

2, 0 ≤ r ≤ 1

Then by Theorem5.5,∫∫Ω

xydxdy =

∫∫Γ

r cos θ︸︷︷︸x

r sin θ︸︷︷︸y

rdrdθ =

∫∫Γ

r3 cos θ sin θdrdθ

=

∫ π2

0

sin θ cos θdθ

∫ 1

0

r3dr =

[sin2 θ

2

]π2

0

·[r4

4

]10

=1

8

2. Ω is a washer region. Then by letting x = r cos θ, y = r sin θ, we have Example5-4-2.

1 ≤ x2 + y2 ≤ 4 and 1 ≤ r2 ≤ 4. Here since r > 0, 1 ≤ r ≤ 2. Also sincey = r sin θ ≥ 0, 0 ≤ θ ≤ π. Thus Ω is transformed into

Γ = (r, θ) : 0 ≤ θ ≤ π, 1 ≤ r ≤ 2

46 CHAPTER 5. MULTIPLE INTEGRALS∫∫Ω

1

x2 + y2dxdy =

∫∫Γ

1

r2|J(r, θ)|drdθ

=

∫ π

0

∫ 2

1

1

r2rdrdθ =

∫ π

0

∫ 2

1

1

rdrdθ

=

∫ π

0

dθ[log r]21 = π log 2

3. This double integral can be evaluated directly. But using the transformation Example5-4-3.

-1 -0.5 0.5 1x

-2

-1

1

2

y

-1 -0.5 0.5 1u

-1

-0.5

0.5

1

v

of variables is easier.

Let

x+ 2y = ux− y = v

. Then solve for x, y to get

x = u+2v

3y = u−v

3

Substitute this into

the condition of Ω. Then the point of Γ = (u, v) : |u| ≤ 1, |v| ≤ 1 correspondsone-to-one into point in Ω. Now

J(u, v) =∂(x, y)

∂(u, v)= det

(1/3 2/31/3 −1/3

)= −1

3

Thus, ∫∫Ω

(x− y)2dxdy =

∫∫Γ

v2|J(u, v)|dudv

=

∫ 1

−1

du

∫ 1

−1

v2

3dv = [u]1−1

[v3

9

]1−1

= 2(2

9) =

4

9

Exercise 5.4 Evaluate the following double integrals.

1.

∫∫Ω

√1− x2 − y2dxdy, Ω = (x, y) : x2 + y2 ≤ x

2.

∫∫Ω

e(y−x)/(y+x)dxdy, Ω = (x, y) : x+ y ≤ 1, x ≥ 0, y ≥ 0

SOLUTION 1. Ω is a circlular region . Thus use the polar coordinate x =r cos θ, y = r sin θ. Since 0 ≤ x2 + y2 = r2 ≤ r cos θ, r2 ≤ r cos θ and r cos θ ≥ 0.Now r cos θ ≥ 0 implies −π

2 ≤ θ ≤ π2 . Since 0 ≤ r2 ≤ r cos θ, 0 ≤ r ≤ cos θ.

Thus, Ω is transformed into

Γ = (r, θ) : −π2≤ θ ≤ π

2, 0 ≤ r ≤ cos θ

Then ∫∫Ω

√1− x2 − y2dxdy =

∫ π2

−π2

∫ cos θ

0

√1− r2rdrdθ

Let t = 1− r2. Then dt = −2r dr,r 0 → cos θ

t 1 → 1− cos2 θ = sin2 θ. Thus,

∫ cos θ

0

√1− r2rdr = −1

2

∫ sin2 θ

1

t12 dt = −1

2· 23

[t3/2

]sin2 θ

1

= −1

3(sin3 θ − 1) =

1

3(1− sin3 θ)

Exercise5-4-1.

0.2 0.4 0.6 0.8 1x

-0.4

-0.2

0.2

0.4

y

Check∫ π2

0sin3 θ dθ = 2!!

3!! =23 Exercise5-4-2

|J(u, v)| = |∂(x,y)∂(u,v) |

= |∣∣∣∣ xu xvyu yv

∣∣∣∣ |= |∣∣∣∣ 1

212

12 − 1

2

∣∣∣∣ |= | − 1

2 | =12

Thus,∫∫Ω

√1− x2 − y2dxdy =

1

3

∫ π2

0

(1− sin3 θ)dθ =1

3(π

2− 2

3)

5.3. CHANGE OF VARIABLES 47

2. Let u = x + y, v = y − x. Then check to see where the region Ω = (x, y) :x+ y ≤ 1, x ≥ 0, y ≥ 0 map into.Since u = x, v = −x, the line y = 0 maps to u = −v.Since u = y, v = y, the line x = 0 maps to v = uA line x+ y = 1 maps to u = x+ y = 1

Ω is mapped to

Γ = (u, v) : 0 ≤ u ≤ 1,−u ≤ v ≤ u

Thus

I =

∫∫Ω

e(y−x)/(y+x)dxdy =

∫∫Γ

ev/u|J(u, v)|dudv

=1

2

∫ 1

0

∫ u

u

ev/udvdu =1

2

∫ 1

0

[uev/u]u−udu

=1

2

∫ 1

0

(ue− ue−1)du =1

2[u2e

2− u2

2e]10 =

1

4(e− 1

e)

Exercise A


(a)

∫∫Ω

(x2 + y2)dxdy, Ω = (x, y) : x2 + y2 ≤ 4

(b)

∫∫Ω

1

(x2 + y2)dxdy, Ω = (x, y) : 1 ≤ x2 + y2 ≤ 4, y ≥ 0

(c)

∫∫Ω

y2dxdy, Ω = (x, y) : x2 + y2 ≤ 1

(d)

∫∫Ω

(x+ y)dxdy,Ω = (x, y) : 0 ≤ x+ y ≤ 1, |x− y| ≤ 1

(e)

∫∫Ω

√4− x2 − y2dxdy, Ω = (x, y) : x2 + y2 ≤ 2x

2. Draw th graph of the region Gamma which is the image of Ω = (x, y) :0 ≤ x+ y ≤ 1, 0 ≤ x− y ≤ 1 by the transformation u = x− y, v = x+ y．Then using this transformation, evaluate

∫∫Ω(2x+ 3y)dxdy．

3. Evaluate∫∫

Ω2xdxdy, where Ω = (x, y) : 0 ≤ x− y ≤ 1, 0 ≤ x+ 2y ≤ 1.

Exercise B


(a)

∫∫Ω

x2dxdy, Ω = (x, y) : x2 + y2 ≤ 4

(b)

∫∫Ω

log (x2 + y2)dxdy, Ω = (x, y) : 1 ≤ x2 + y2 ≤ 4

(c)

∫∫Ω

e(y−x)/(y+x)dxdy, Ω = (x, y) : x+ y ≤ 1, x ≥ 0, y ≥ 0

(d)

∫∫Ω

ex2+y2dxdy, Ω = (x, y) : 1 < x2 + y2 < 4

(e)

∫∫Ω

√1− x2 − y2dxdy, Ω = (x, y) : x2 + y2 ≤ 1

(f)

∫∫Ω

(1− x− 2y)dxdy, Ω = (x, y) : x ≥ 0, y ≥ 0, x2 + y2 ≤ 1


2. Using the tranformation u = x+ y, v = x− y, evaluate the following doubleintegral.∫∫

Ω

(x2 + y2)e−x+ydxdy, Ω = −1 ≤ x+ y ≤ 1,−1 ≤ x− y ≤ 1

5.4. IMPROPER DOUBLE INTEGRALS 49

5.4 Improper Double Integrals

The double integrals treated so far are the case where a function is bounded onthe bounded region. Now consider the case where Ω is not bounded.

Improper Integral The sequence of bounded closed regions Ω1,Ω2, · · · ,Ωn, · · · in Ω satisfy

Ω1 ⊂ Ω2 ⊂ · · · ⊂ Ωn ⊂ · · ·

Furthermore, every subset of Ω is a subset of Ωn. Then if f(x, y) is inte-grable on the region Ωn,

limn→∞

∫∫Ωn

f(x, y)dxdy

exists. Then we say f(x, y) is integrable on Ω and define∫∫Ω

f(x, y)dxdy = limn→∞

∫∫Ωn

f(x, y)dxdy

Understanding If a function z = f(x, y) isnot bounded on a region,then we say the double in-tegral is improper inte-gral of the 1st kind. Ifthe region is not bounded,then we say the double in-tegral is improper inte-gral of the 2nd kind. Ifa function is not boundedon unbounded region, thenwe say the double integralis improper integral ofthe 3rd kind.

Example 5.5 Evaluate the following double integrals.

1. I =

∫∫Ω

1√y2 − x2

dxdy, Ω = (x, y) : 0 ≤ x ≤ y ≤ 1

2. I2 =

∫∫Ω

e−(x2+y2)dxdy, Ω = (x, y) : x, y ≥ 0

Example5-5-1.0 ≤ x ≤ 1, 0 ≤ y ≤ 1, x ≤ y.

SOLUTION 1. Using horizontally simple region, we have Ω = (x, y) : 0 ≤y ≤ 1, 0 ≤ x ≤ y. Then f(x, y) = 1√

y2−x2is discontinuous at y = x. Thus let

Ωn = (x, y) : 1n ≤ y ≤ 1, 0 ≤ x ≤ y − 1

n. Then

I = limn→∞

∫ 1

y= 1n

( ∫ y− 1n

x=0

1

(y2 − x2)1/2dx)dy

= limn→∞

∫ 1

y= 1n

( ∫ y− 1n

0

1

(y2 − x2)1/2dx)dy

= limn→∞

∫ 1

1n

([sin−1 x

y]y− 1

n0

)dy = lim

n→∞(sin−1 y − 1/n

y)(1− 1

n) =

π

2 Example5-5-2.

Create Ωn so that Ω ⊂ Ωn as ngoes to infinity.

2. The region Ω is not bounded. So, consider the sequence of closed boundedregions Ωn.

Ωn = (x, y) : x2 + y2 ≤ n2, x, y ≥ 0The figure of Ωn is given by the figure 5.7.

Figure 5.7: Sequences


For Ωn, use the polar coordinate x = r cos θ, y = r sin θ Checkx2 + y2 = r2 ≤ n2 implies thatr ≤ n. Since x, y ≥ 0, 0 ≤ θ ≤π2 .

∫∫Ωn

e−(x2+y2)dxdy =

∫ π2

0

dθ

∫ n

0

e−r2

rdr

=π

2

[−e

−r2

2

]n0

=π

4(1− e−n

2

)

Thus, by letting n→ ∞, we find I2 and I2 = π/4

∫∞0e−x

2

dx =√π2

I =

∫∞0e−x

2

dx, I =∫∞0e−y

2

dy. Then I2 =∫∞0e−x

2

dx∫∞0e−y

2

dy =∫∞0

∫∞0e−x

2−y2dydximplies

I =√π2 .

Consider a function f(x, y) is not bounded on Ω.

Exercise 5.5 Evaluate the following double integrals.

1. I =

∫∫Ω

dxdy

(x+ y)3/2, Ω = (x, y) : 0 < x ≤ 1, 0 < y ≤ 1

2. I =

∫∫Ω

tan−1 y

xdxdy, Ω = (x, y) : x2 + y2 ≤ 1, x ≥ 0, y ≥ 0 Exercise5-5-2.

Create Ωn so that Ω is a subsetof Ωn as n goes to infinity.

SOLUTION f(x, y) = 1(x+y)3/2

is discontinuous at (0, 0). Then create Ωn so

that (0, 0) is not included in Ωn.

Ωn = (x, y) : 1n≤ x ≤ 1,

1

n≤ y ≤ 1

Then we get figure 5.9.

Figure 5.8: Sequences of Ωn

Thus,∫∫Ωn

dxdy

(x+ y)3/2=

∫ 1

1/n

dx

∫ 1

1/n

dy

(x+ y)3/2=

∫ 1

1/n

[− 2

(x+ y)1/2

]11n

dx

= 2

∫ 1

1/n

[(x+1

n)−

12 − (x+ 1)−

12 ]dx

= 2

[2(x+

1

n)

12 − 2(x+ 1)

12

]11n

= 4[2(1 +1

n)

12 −

√2− (

2

n)

12 ]

Therefore, as n→ ∞ we can find I = 4(2−√2)

Check∫ 1

1/ndy

(x+y)3/2=∫ 1

1/n(x+ y)−3/2 dy.

This is the integral with re-spect to y. Thus we treat x

as a constant. Now∫ 1

1/n(x +

y)−3/2 dy =[−2(x+ y)−

12

]11n Exercise5-5-2.

tan−1 yx is not bounded at x =

0. Using polar coordinatetransformation x = r cos θ, y =r sin θ, the region Ω can be cov-ered by taking θ goes from 0to π

2 and the distance from thepole ranges from 0 to 1. Thus,Γ = (r, θ) : 0 ≤ θ ≤ π

2 , 0 ≤r ≤ 1.

2. f(x, y) = tan−1 yx is bounded except on x-axis.

5.4. IMPROPER DOUBLE INTEGRALS 51

Figure 5.9: Sequence of Regions

Using the polar coordinate, since 0 ≤ x2 + y2 = r2 ≤ 1, 0 ≤ r ≤ 1. Sincex ≥ 0, y ≥ 0, 0 ≤ θ ≤ π

2 . Thus Ω maps to

Γ = (r, θ) : 0 ≤ θ ≤ π

2, 0 ≤ r ≤ 1

Now let Γn = (x, y) : 0 ≤ θ ≤ π2 − 1

n ,1n ≤ r ≤ 1. Then,

tan−1 y

x= tan−1

(r sin θ

r cos θ

)= tan−1(tan θ) = θ

and is bounded on Γn. Thus,

I =

∫∫Ω

tan−1 y

xdxdy = lim

n→∞

∫∫Γn

θrdrdθ

= limn→∞

∫ π2 − 1

n

θ=0

θdθ( ∫ 1

1n

θrdr)= limn→∞

[θ22

]π2 − 1

n

0

( [r22

]11n

)=

π2

8· 12=π2

16

Exercise A

1. Evaluate the following improper integrals．

(a)

∫∫Ω

ey/xdxdy, Ω = (x, y) : 0 < x ≤ 1, 0 ≤ y ≤ x

(b)

∫∫Ω

x√1− x− y

dxdy, Ω = (x, y) : x+ y < 1, x ≥ 0, y ≥ 0

(c)

∫∫Ω

log(x2 + y2)dxdy, Ω = (x, y) : x, y ≥ 0, x2 + y2 ≤ 1

Exercise B

1. Evaluate the following improper integrals．

(a)

∫∫Ω

dxdy

(y2 − x2)1/2, Ω = (x, y) : 0 ≤ x < y ≤ 1

(b)

∫∫Ω

e−(x+y)dxdy, Ω = (x, y) : x ≥ 0, y ≥ 0

(c)

∫∫Ω

tan−1(y

x)dxdy, Ω = (x, y) : x, y ≥ 0, x2 + y2 ≤ 1

(d)

∫∫Ω

1

x2y2dxdy, Ω = (x, y) : x ≥ 1, y ≥ 1


(e)

∫∫Ω

dxdy√x− y2

, Ω = (x, y) : 0 ≤ x ≤ 1, y2 ≤ x

(f)

∫∫Ω

dxdy

1 + (x2 + y2)2, Ω = (x, y) : −∞ < x, y <∞

2. Using thee example 5.5, show that Γ

(1

2

)=

∫ ∞

0

x−12 e−xdx =

√π．

5.5. APPLICATION OF DOUBLE INTEGRALS 53

5.5 Application of Double Integrals

5.5.1 Area

Area For f(x, y) > 0, the double integral

∫∫Ωf(x, y)dxdy represents the volume

of a solid whose base area is Ω and the height is f(x, y). If f(x, y) = 1, thenthe double integral

∫∫Ωdxdy can be thought of the volume of solid whose

base area is Ω and the height is 1. Now ignore the unit, then we can thinkof the area of Ω. Thus

Ωbase area =

∫∫Ω

dxdy

Area by Double Integral If the region is given by thecartesian coordinate, thenthe area of the region canbe evaluated without us-ing double integrals. Forthe region bounded by thecurves given by the po-lar coordinates, it is mucheasier to use double inte-gral.

Example 5.6 Find the area of the region Ω bounded by the curve x2 = 4y andthe line 2y − x− 4 = 0

SOLUTION First find the intersection of two curves. Let x2 = 4y = 2x + 8.Then x2 − 2x− 8 = (x− 4)(x+ 2) = 0 which implies x = −2, 4.

Now using the vertical simple ・・Ω = (x, y) : −2 ≤ x ≤ 4, x2

4 ≤ y ≤ x+42 ・

・・・,∫∫Ω

dxdy =

∫ 4

−2

dx

∫ x+42

x2

4

dy =

∫ 4

−2

[x+ 4

2− x2

4]dx

=

[x2

4+ 2x− x3

12

]4−2

=16− 4

4+ 2(4 + 2)− 1

12(64 + 8)

= 3 + 12− 6 = 9

Exercise 5.6 Find the area of the region Ω that lies inside the cardioid r = 1 + cos θbut outside the circle r = 1.

Exercise5・—6

SOLUTION First find the intersection of r = 1 + cos θ and r = 1. Then sincecos θ = 0, θ = −π

2,π

2. Thus by change of variables, x = r cos θ, y = r sin θ, the

region Ω is transformed into the region Γ = (r, θ) : −π2 ≤ θ ≤ π

2 , 1 ≤ r ≤1 + cos θ. Thus∫∫

Ω

dxdy =

∫∫Γ

rdrdθ = 2

∫ π2

0

dθ

∫ 1+cos θ

1

rdrdθ

= 2

∫ π2

0

[r2

2

]1+cos θ

1

dθ = 2

∫ π2

0

2 cos θ + cos2 θ

2dθ

=

∫ π2

0

2 cos θdθ +

∫ π2

0

cos2 θdθ = 2 +π

4


Checkcos 2θ = 2 cos2 θ − 1 impliescos2 θ = 1+cos 2θ

2∫ π/20

cos2 θ dθ =∫ π/20

1+cos 2θ2 dθ =

12

[θ + 1

2 sin 2θ]π/20

= 12 (π2 ) =

π4

5.5.2 Surface Area

Surface Area Suppose that z = f(x, y) is a function of the class C1 on the closed boundedregion Ω. Then

S = (x, y, f(x, y) : (x, y) ∈ Ω

is called smooth surface. The area of S is given by

S =

∫∫Ω

√f2x + f2y + 1dxdy.

where Ω is the projection of S onto xy-plane. NOTE

Figure 5.10: Surface Area

The vector N orthogonal to the small rectangle on the surface is given by(fx, fy,−1). Let θ be the angle between the vector (fx, fy,−1) and the vectore1 which is orthogonal to xy-plane. Then ∆S cos θ is equal to the ∆Ω, a smallarea of xy-plane.Thus, ∆S = ∆Ωsec θ. Note that

N · e1 = (fx, fy,−1) · (0, 0, 1) =√f2x + f2y + 1 cos θ


Thus we can express the surface area S as the following double integral.

S =

∫∫Ω

√f2x + f2y + 1dxdy.

Dot Product The dot product of a vec-tor A = (a1, a2, a3) and avector B = (b1, b2, b3) isgiven by a1b1+a2b2+a3b3and is written as A · B.A ·B = |A||B| cos θ.

Example 5.7 Find the surface area of the following surface.

z2 = x2 + y2 with 0 ≤ x2 + y2 ≤ 1, z ≥ 0

Example5-7SOLUTION To find the surface area, we need to find the Ω which is a projectionof the surface z = f(x, y). Since z ≥ 0, the surface is z =

√x2 + y2. Now the

region Ω is given by Ω = (x, y) : x2 + y2 ≤ 1. Thus, the surface area is givenby the following double integral

S =

∫∫Ω

√z2x + z2y + 1dxdy.

Since

zx =2x

2√x2 + y2

=x√

x2 + y2, zy =

2y

2√x2 + y2

=y√

x2 + y2

,

z2x + z2y + 1 =x2 + y2

x2 + y2+ 1 = 2.

Thus,

S =

∫∫Ω

√2dxdy

Note that Ω is a circular region of the radius 1. Thus by using the polar coor-dinate,

Γ = (r, θ) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1

Thus the surface area is

S =√2

∫∫Γ

rdrdθ =√2

∫ 2π

0

∫ 1

0

rdrdθ

=√2

∫ 2π

0

[r2

2

]10

dθ =√2

∫ 2π

0

1

2dθ =

√2

2[θ]

2π0 =

√2π

Exercise 5.7 Find the surface area of sphere x2 + y2 + z2 = 1 cut by cylinderx2 + y2 = x.

Exercise5-7

SOLUTION Note that x2 + y2 + z2 = 1 implies z = ±√

1− x2 − y2. Then findthe following surface area and double it

z = f(x, y) =√1− x2 − y2,Ω = (x, y) : x2 + y2 ≤ x


S = 2

∫∫Ω

√f2x + f2y + 1dxdy

= 2

∫∫Ω

√(

x2

1− x2 − y2+

y2

1− x2 − y2+ 1)dxdy

= 2

∫∫Ω

1√1− x2 − y2

dxdy

Checkx2 + y2 + z2 = 1 implies z =±√1− x2 − y2.

fx =∂(√

1−x2−y2)∂x =

−2x

2√

1−x2−y2= − x√

1−x2−y2,

fy =∂(√

1−x2−y2)∂y =

−2y

2√

1−x2−y2= − y√

1−x2−y2

Let x2 + y2 = r2, x = r cos θ.Then x2 + y2 ≤ x is trans-formed to 0 ≤ r2 ≤ r cos θ.Thus, 0 ≤ r ≤ cos θ. There-fore, −π

2 ≤ θ ≤ π2

√1− cos2 θ =

√sin2 θThen for

−π2 ≤ θ ≤ 0, sin θ ≤ 0. Thus√sin2 θ = | sin θ|.

Now using polar coordinate, Ω is mapped into

Γ = (r, θ) : −π2≤ θ ≤ π

2, 0 ≤ r ≤ cos θ

Since |J(r, θ)| = r,

S = 2

∫∫Γ

1√1− r2

rdrdθ = 2

∫ π2

−π2

( ∫ cos θ

0

r√1− r2

dr)dθ

Let t = 1− r2 Then dt = −2rdr,r 0 → cos θ

t 1 → 1− cos2 θ = sin2 θ. Thus,

S = 2

∫ π2

−π2

( ∫ sin2 θ

1

1

−2√tdt)dθ

= 2

∫ π2

−π2

( ∫ sin2 θ

1

t−1/2

−2dt)dθ = 2

∫ π2

−π2

[− t

12

]sin2 θ

1dθ

= 2

∫ π2

−π2

(−√

1− cos2 θ + 1)dθ = 2

∫ π2

−π2

(1− | sin θ|)dθ

= 4

∫ π2

0

(1− sin θ)dθ = 4(π

2− 1) = 2π − 4

5.5.3 Volume of Solid

Understanding f(x, y)−g(x, y) representsthe height of the rect-angular solid and ∆x∆yrepresents the base area.Thus ∆V = (f(x, y) −g(x, y))∆x∆y.

Volume of Solid Given a closed bounded region Ω and continuous functions z = f(x, y), z =g(x, y) on Ω, and suppose that f(x, y) ≥ g(x, y) on Ω. Then the volume ofsolid bounded by the lines parallel to z-axis through the boundary ∂Ω andthe surfaces f, g is given by

V =

∫∫Ω

[f(x, y)− g(x, y)]dxdy Example 5.8 Find the volume of the solid common to the intersecting cylin-ders x2 + y2 = a2, x2 + z2 = a2 (a > 0) .

SOLUTION Projection of the solid bounded by two cylinders onto xy-plane isx2 + y2 = a2 and x2 = a2. Thus, Ω = (x, y) : x2 + y2 ≤ a2. Also x2 + z2 = a2

implies z = ±√a2 − x2. Thus the upper surface is

√a2 − x2 and the lower

surface is√a2 − x2 over Ω. From this, the volume of the solid is set up by∑∑

2√a2 − x2∆x∆y, where ∆x∆y represents the small rectangle of base area


and 2√a2 − x2 represents the height of a small solid cylinder. Now use vertically

simple region for Ω. Then

Ω = (x, y) : −a ≤ x ≤ a,−√a2 − x2 ≤ y ≤

√a2 − x2

Therefore the volume of the solid is

V = 2

∫∫Ω

√a2 − x2dxdy = 2

∫ a

−a

( ∫ √a2−x2

−√a2−x2

√a2 − x2dy

)dx

= 8

∫ a

0

( ∫ √a2−x2

y=0

√a2 − x2dy

)dx

= 8

∫ a

0

[√a2 − x2y

]√a2−x2

0dx

= 8

∫ a

0

(a2 − x2)dx = 8

[a2x− x3

3

]a0

=16a3

3

Example5-8

Take a point in Ω and evaluatethe value of z to find out whichsurfaces upper or lower surface.

Exercise 5.8 Find the volume of the solid bounded by x2 + y2 ≤ a2 and 0 ≤z ≤ x.

Exercise5-8

SOLUTION Note that the solid is bounded by the surface goes through theboundary of Ω = (x, y) : x2 + y2 ≤ a2, x ≥ 0 and parallel to the z-axis,and the plane z = 0, z = x. Using the polar coordinate to express Ω. Sincex = r cos θ > 0,

Γ = (r, θ) : −π2≤ θ ≤ π

2, 0 ≤ r ≤ a

Thus,

V =

∫∫Ω

xdxdy =

∫∫Γ

r cos θ(r)drdθ

= 2

∫ π2

0

∫ a

0

r2 cos θdrdθ = 2

∫ π2

0

cos θdθ

∫ a

0

r2dr

= 2 [sin θ]π20

[r3

3

]a0

=2a3

3

The limit of integration is con-stant, we can express the dou-ble integral as the product oftwo single integral.

Exercise A

1. Find the area of the region satisfying the following conditions．

(a) The region enclosed by r = 1− cos θ

(b) Inside of the curve r = 3 cos θ and outside of the curve r = 32 .

(c) Inside of the curve r = 3 cos θ and outside of the curve r = cos θ

2. Find the surface area of the following surface．

(a) z2 = x2 + y2 with 0 ≤ x2 + y2 ≤ 1, z ≥ 0.

(b) x+ y + z = 2 with x ≥ 0, y ≥ 0, z ≥ 0.

(c) z = xy with x2 + y2 = a2 (a > 0).

3. Find the volume of the following solid.

(a) z = x2 + y2 with z = 2y

(b) z = x2 + y2 with x2 + y2 ≤ 1

Exercise B


1. Find the area of the region enclosed by the following curves.

(a) x = cos3 t, y = sin3 t (0 ≤ t ≤ π

2)

(b) r = a cos 3θ (a > 0) (c) y =8

x2 + 4and y =

x2

4.

2. Find the surface area of the following surface．

(a) Sphere x2 + y2 + z2 = a2 with the radius a.

(b) z = xy with x2 + y2 ≤ a2

(c) x2 + z2 = a2 and x2 + y2 = a2.

(d) The surface generated by rotating y = mx (0 ≤ x ≤ k) about x-axisfor (m > 0).

3. Find the volume of the solid given by the following conditions.．

(a) x2 + y2 ≤ a2 with 0 ≤ z ≤ x

(b) 0 ≤ z ≤ 1− x2, x ≤ 1− y2, x ≥ 0, y ≥ 0.

(c) x2 + y2 + z2 ≤ a2 and x2 + y2 ≤ ax.

(d) z = 1−√x2 + y2 and z = x, x = 0.

5.6. TRIPLE INTEGRALA 59

5.6 Triple Integrala

Triple Integrals Let T be a closed bounded region in xyz space and the projection of Tonto xy-plane be Ωxy. Then T is expressed as

T = (x, y, z) : (x, y) ∈ Ωxy, ψ1(x, y) ≤ z ≤ ψ2(x, y)

Thus, ∫∫∫T

f(x, y, z)dxdydz =

∫∫Ωxy

( ∫ ψ2(x,y)

ψ1(x,y)

f(x, y, z)dz)dxdy

Furthermore, if

Ω = (x, y) : a ≤ x ≤ b, ϕ1(x) ≤ y ≤ ϕ2(x)

then∫∫∫T

f(x, y, z)dxdydz =

∫ b

a

(∫ ϕ2(x)

ϕ1(x)

(∫ ψ2(x,y)

ψ1(x,y)

f(x, y, z)dz

)dy

)dx

Understanding In the triple integral,it is important to knowwhich direction the smallcuboid ∆x∆y∆z shouldbe stacked. If the closedbounded region T isbounded by the twosurfaces z = ϕ1(x, y)and z = ϕ2(x, y), thenit is better to stack thecuboid into the directionof z and in this case,∫ ϕ2(x,y)

ϕ1(x,y)f(x, y, z)dz is

innermost integral. Notethat z = ϕ1(x, y) is thesurface under the surfacez = ϕ2(x, y).

NOTE Consider a small rectangular solid ∆x∆y∆z. Then as in figure5.11, fillthe solid by piling up these small rectangular solid to the direction of z-axis.

Figure 5.11: Triple Integral

Then the height of the long rectangular solid can be expressed by∫ ψ2

ψ1dz =

(ψ2 − ψ1), the volume of the small long rectangular solid is (ψ2 − ψ1)∆x∆y.Thus ∫∫∫

T

f(x, y, z)dxdydz =

∫∫Ωxy

[

∫ ψ2(x,y)

ψ1(x,y)

f(x, y, z)dz]dxdy.

This way, a triple integral can be reduced a double integral.

Example 5.9 For T = (x, y, z) : 0 ≤ x ≤ y ≤ z ≤ 1. evaluate the followingtriple integrals. ∫∫∫

T

dxdydz

Example5-9

Pile up small rectangular solidint the direction of z, the lowersurface is z = y and the uppersurface is z = 1.

SOLUTION The projection of T = (x, y, z) : 0 ≤ x ≤ y ≤ z ≤ 1 onto thexy-plane is the region Ωxy = (x, y) : 0 ≤ x ≤ y ≤ 1. Now using vertical simple


region, Ωxy = (x, y) : 0 ≤ x ≤ 1, x ≤ y ≤ 1. Its volume is given by

V =

∫∫Ω

∫ 1

z=y

dzdxdy =

∫ 1

x=0

∫ 1

x

[z]1ydydx

=

∫ 1

0

∫ 1

x

(1− y)dydx =

∫ 1

0

[y − y2

2

]1x

dx

=

∫ 1

0

(1− 1

2− (x− x2

2))dx =

[x

2− x2

2+x3

6

]10

=1

2− 1

2+

1

6=

1

6

Exercise 5.9 For T = (x, y, z) : 0 ≤ x ≤ y ≤ z ≤ 1, evaluate the followingtriple integral.

∫∫∫T

ex+y+zdxdydz

Exercise5-9

Check

To evaluate∫ 1

z=yex+y+zdz,

the variables other thatz are treated as con-stant.

∫ 1

z=yex+y+zdz =

[ex+y+z]1z=y = ex+y+1 − ex+2y

SOLUTION The projection of T = (x, y, z) : 0 ≤ x ≤ y ≤ z ≤ 1 onto xy-planeis the region Ωxy = (x, y) : 0 ≤ x ≤ y ≤ 1. Using vertical simple region,Ωxy = (x, y) : 0 ≤ x ≤ 1, x ≤ y ≤ 1. Thus

V =

∫∫Ω

∫ 1

z=y

ex+y+zdzdxdy =

∫ 1

x=0

∫ 1

x

[ex+y+z]1ydydx

=

∫ 1

0

∫ 1

x

(ex+y+1 − ex+2y)dydx

=

∫ 1

0

[ex+y+1 − 1

2ex+2y

]1x

dx

=

∫ 1

0

(ex+2 − 1

2ex+2 − e2x+1 +

1

2e3x)dx

=

[1

2ex+2 − 1

2e2x+1 +

1

6e3x]10

=1

2e3 − 1

2e3 +

1

6e3 − (

1

2e2 − 1

2e+

1

6)

=1

6e3 − 1

2e2 +

1

2e− 1

6=

1

6(e− 1)3


transformation Theorem 5.6 The transformation Φ :

x = ϕ(u, v, w)y = ψ(u, v, w)z = ξ(u, v, w)

is one-to-one

mapping of points in the close bounded region T in xyz-space into pointsin the closed bounded region T ′ in uvw-space. Suppose that ϕ, ψ, ξ are theclass C1 and

J = J(u, v, w) =

∣∣∣∣∣∣xu xv xwyu yv ywzu zv zw

∣∣∣∣∣∣ = 0.

If f(x, y, z) is continuous on T , then∫∫∫T

f(x, y, z)dxdydz

=

∫∫∫T ′f(ϕ(u, v, w), ψ(u, v, w), ξ(u, v, w))|J |dudvdw.

Determinant Cofactor expansion of 1strow.∣∣∣∣∣∣xu xv xwyu yv ywzu zv zw

∣∣∣∣∣∣ =

xu

∣∣∣∣ yv ywzv zw

∣∣∣∣ −

xv

∣∣∣∣ yu ywzu zw

∣∣∣∣ +

xw

∣∣∣∣ yu yvzu zv

∣∣∣∣ Cylindrical Coordinates

Corollary 5.1 Let (x, y, z) be a point of the cylinder in the rectangular co-

ordinates. Now express this by the cylindrical coordinates,

x = r cos θy = r sin θz = z

J(r, θ, z) =

∣∣∣∣∣∣cos θ −r sin θ 0sin θ r cos θ 00 0 1

∣∣∣∣∣∣ = r

Thus ∫∫∫T

f(x, y, z)dxdydz =

∫∫∫T ′f(r cos θ, r sin θ, z)dz|r|drdθ

CheckJ(r, θ, z) =∣∣∣∣∣∣cos θ −r sin θ 0sin θ r cos θ 00 0 1

∣∣∣∣∣∣ =

r cos2 θ + r sin2 θ = r

CheckThe projection of (r, θ, z) ontoxy-plane is x = r cos θ, y =r sin θ.

Figure 5.12: Cylindrical Coordinates

Example 5.10 For T = (x, y, z) : x2 + y2 ≤ 1,−√1− x2 − y2 ≤ z ≤

√1− x2 − y2,

evaluate the triple integral

I =

∫∫∫T

1√1− x2 − y2

dxdydz

Example5-10

SOLUTION In the cylindrical coordinates, a point (r, θ, z) is expressed by thedistance r from the origin , the angle θ from the polar axis , and z. Thus, by


letting x = r cos θ, y = r sin θ, x2 + y2 ≤ 1 is written as r2 ≤ 1. a circle withthe radius 1. Thus, (r, θ) : 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 1 and −

√1− x2 − y2 =

−√1− r2. From this, the region T is mapped into the region

T ′ = (r, θ, z) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π,−√

1− r2 ≤ z ≤√1− r2

I =

∫∫∫T ′

1√1− r2

rdzdrdθ =

∫ 2π

0

( ∫ 1

0

( ∫ √1−r2

−√1−r2

r√1− r2

dz)dr)dθ

=

∫ 2π

0

( ∫ 1

0

[ rz√1− r2

]√1−r2√1−r2dr

)dθ =

∫ 2π

0

∫ 1

0

2rdr = 2π ·[r2]10= 2π

Exercise 5.10 For T = (x, y, z) : x2 + y2 ≤ a2, 0 ≤ z ≤ x, evaluate the fol-lowing triple integral. ∫∫∫

T

dxdydz Exercise5-10SOLUTION The region T is enclosed by the cylinder x2+y2 ≤ a2 and the surfacez = 0, z = x. Use the cylindrical coordinate, x = r cos θ, y = r sin θ, z = z. Then

T = (r, θ, z) : −π2≤ θ ≤ π

2, 0 ≤ r ≤ a, 0 ≤ z ≤ r cos θ

Thus

I =

∫ π2

−π2

∫ a

0

∫ r cos θ

0

dzrdrdθ =

∫ π2

−π2

∫ a

0

[z]r cos θ0 rdtdθ

=

∫ π2

−π2

∫ a

0

r2 cos θdrdθ = (

∫ π2

−π2

cos θdθ)(

∫ a

0

r2dr)

= 2 [sin θ]π20

[r3

3

]a0

=2a3

3

Spherical Coordinates Corollary 5.2 The first coordinate, ρ, is the distance from the origin tothe point (x, y, z). The second coordinate, ϕ, is the angle measured fromthe positive z-axis. The third coordinate, θ, is the angle measured from the

x-axis. Then

x = sinϕ cos θy = ρ sinϕ sin θz = ρ cosϕ

Jacobian is

J(ρ, ϕ, θ) =

∣∣∣∣∣∣sinϕ cos θ ρ cosϕ sin θ −ρ sinϕ sin θsinϕ sin θ ρ cosϕ cos θ ρ sinϕ sin θcosϕ −ρ sinϕ 0

∣∣∣∣∣∣ = ρ2 sinϕ

∫∫∫T

f(x, y, z)dxdydz

=

∫∫∫T ′f(ρ sinϕ cos θ, ρ sinϕ sin θ, ρ cosϕ)ρ2 sinϕdρdϕdθ

NOTE


Figure 5.13: Spherical coordinates

Example 5.11 For T = (x, y, z) : x2 + y2 + z2 ≤ 1, evaluate the followingtriple integral.

I =

∫∫∫T

dxdydz√1− x2 − y2 − z2

SOLUTION Note that the region T is the sphere of the radius 1. Now usespherical coordinates to express (ρ, ϕ, θ). The angle θ is measured from the x・・-axis, so to cover the sphere, 0 ≤ θ ≤ 2π. The angle ϕ is measured from thez-axis, so to cover the sphere, 0 ≤ ϕ ≤ π. The distance ρ is measured from theorigin, so to cover the sphere, 0 ≤ ρ ≤ 1. Thus, the region T is mapped into theregion T ′

T ′ = (ρ, ϕ, θ) : 0 ≤ ρ ≤ 1, 0 ≤ ϕ ≤ π, 0 ≤ θ ≤ 2π

Thus

I =

∫∫∫T ′

ρ2 sinϕ√1− ρ2

dρdϕdθ =

∫ 2π

0

dθ

∫ π

0

sinϕdϕ

∫ 1

0

ρ2√1− ρ2

dρ

Now note that the integral with respect to ρ is improper integral. So, we let

ρ = sin t. Then dρ = cos tdt andρ 0 → 1t 0 → π

2

. Thus

∫ 1

0

ρ2√1− ρ2

dρ =

∫ π/2

0

sin2 t cos t√1− sin2 t

dt =

∫ π/2

0

sin2 t cos t

cos tdt

=

∫ π/2

0

1− cos 2t

2dt =

1

2

[t− sin 2t

2

]π/20

=π

4

Therefore,

I =π

4

∫ 2π

0

dθ

∫ π

0

sinϕdϕ =π

4· 2π · 2 = π2

Check∫ π/20

sin2 t cos tcos t dt =∫ π/2

0sin2 t dt.

Exercise 5.11 For T = (x, y, z) : x2

22+y2

32+z2

42≤ 1, evaluate the following

triple integral.

I =

∫∫∫T

(x2 + y2 + z2) dxdydz Exercise5-11SOLUTION

T = (x, y, z) : x2

22+y2

32+z2

42≤ 1


is ellipsoid. To use spherical coordinates, we use the following change of vari-ables. Let

x = 2u, y = 3v, z = 4w

Then the region T is mapped into the region T ′ = (u, v, w) : u2+v2+w2 ≤ 1.Then the Jacobian is

J(u, v, w) = | ∂(x, y, z)∂(u, v, w)

| =

∣∣∣∣∣∣xu xv xwyu yv ywzu zv zw

∣∣∣∣∣∣ =∣∣∣∣∣∣2 0 00 3 00 0 4

∣∣∣∣∣∣ = 24

Thus

I =

∫∫∫T

(x2 + y2 + z2)dxdydz =

∫∫∫T ′(4u2 + 9v2 + 16w2)(24)dudvdw

Now we use spherical coordinates. By the transformation

u = ρ sinϕ cos θ, v = ρ sinϕ sin θ, w = ρ cos θ

T ′ is mapped into T ′′.

T ′′ = (ρ, ϕ, θ) : 0 ≤ ρ ≤ 1, 0 ≤ ϕ ≤ π, 0 ≤ θ ≤ 2π.

Thus,

I = 24

∫∫∫T ′′

(4u2 + 9v2 + 16w2)dudvdw

= 96

∫∫∫T ′′u2dudvdw + 216

∫∫∫T ′′v2dudvdw + 384

∫∫∫T ′′w2dudvdw

Since |J | = ρ2 sinϕ,

96

∫∫∫T ′′u2dudvdw = 96

∫ 2π

0

∫ π

0

∫ 1

0

ρ2 sin2 ϕ cos2 θ|J |dρdϕdθ

= 96

∫ 2π

0

cos2 θdθ

∫ π

0

sin3 ϕdϕ

∫ 1

0

ρ4 dρ

= 96 · π · 43· 15=

4 · 9615

π =128

5π

Check∫ π0sin3 ϕ dϕ =∫ π

0sin2 ϕ sinϕ dϕ =∫ π

0(1 − cos2 ϕ) sinϕ dϕ =

2∫ 1

0(1−t2)dt = 2

[t− t3

3

]10= 4

3

216

∫∫∫T ′′v2dudvdw = 216

∫ 2π

0

∫ π

0

∫ 1

0

ρ2 sin2 ϕ sin2 θ|J |dρdϕdθ

= 216

∫ 2π

0

sin2 θdθ

∫ π

0

sin3 ϕdϕ

∫ 1

0

ρ4 dρ

= 216 · π · 43· 15=

4 · 21615

π =288

5π

∫ 2π

0sin2 θ dθ =∫ 2π

01−cos 2θ

2 dθ =12

[θ − sin 2θ

2

]2π0

= π

∫ 2π

0cos2 θ dθ =∫ 2π

01+cos 2θ

2 dθ =12

[θ + sin 2θ

2

]2π0

= π 384

∫∫∫T ′′w2dudvdw = 384

∫ 2π

0

∫ π

0

∫ 1

0

ρ2 cos2 θ|J |dρdϕdθ

= 384

∫ 2π

0

sin θ cos2 θdθ

∫ π

0

sinϕdϕ

∫ 1

0

ρ4 dρ

= 384 · π · 2 · 15=

768

5π.

Therefore,

I = (128

5+

288

5+

768

5)π =

1184

5π


Mass Suppose the density σ = f(x, y, z). Then the mass m of the solid T is givenby

m =

∫∫∫T

f(x, y, z)dxdydz

Especially when the density σ = f(x, y, z) = 1,∫∫∫T

dxdydz

We can think of this as the volume of T . Example 5.12 Find the volume of the tetorahedoron in the figure.

SOLUTION To evaluate the triple integral, first find the projection of T ontoxy-plane. The projection Ωxy is the triangle region bounded by x = 0, y = 0, y ≤1− x. Now expressing by vertically simple region.

Ωxy = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1− x

ThusT = (x, y, z) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1− x, 0 ≤ z ≤ 1− x− y

Example5-12

Ωxy

If we know a surface is givenby z = 1 − x − y, then by let-ting z = 0, we can find Ωxy =(x, y) : x+ y = 1.

V =

∫∫Ωxy

[

∫ 1−x−y

0

dz]dxdy =

∫ 1

0

∫ 1−x

0

(1− x− y)dydx

=

∫ 1

0

[(1− x)y − y2

2

]1−x0

dx =

∫ 1

0

(1− x)2

2dx

= −[(1− x)3

6

]10

=1

6

In this question, the volume is simply given by the base area × the height ÷

3. Thus1

2× 1

3=

1

6.

Exercise 5.12 In Example5.11, when the density is given by σ(x, y, z) = xy,find the mass.

SOLUTION

m =

∫∫∫T

xydxdydz =

∫ 1

0

∫ 1−x

0

∫ 1−x−y

0

xydzdydx


Now ∫ 1−x−y

0

xydz = xy(1− x− y) = x(1− x)y − xy2

Then∫ 1−x

0

∫ 1−x−y

0

xydzdy =

∫ 1−x

0

(x(1− x)y − xy2)dy

=

[1

2x(1− x)y2 − 1

3xy3]1−x0

=1

6x(1− x)3

Thus

m =

∫ 1

0

1

6x(1− x)3dx =

1

6

∫ 1

0

(x− 3x2 + 3x3 − x4)dx

=1

6

[1

2x2 − x3 +

3

4x4 − 1

5x5]10

=1

120

Check∫ 1−x−y0

xydz =

xy∫ 1−x−y0

dz = xy [z]1−x−y0 =

xy(1− x− y)

Centroid of Plane consider the system of n particles, P1,P2, . . . ,Pn. Now the cartesian coor-dinates of Pi is Pi(xi, yi) and the mass of the particle Pi is mi. Now draw aline perpendicular to the x-axis so that the rotation moment is equal. Thus

n∑i=1

mi(xi − x) = 0

or

x =

∑ni=1mixi∑ni=1mi

Similarly,

y =

∑ni=1miyi∑ni=1mi

The point (x, y) is called centroid of a system of particles.


Figure 5.14: Centroid of a system of particles

If every point P(x, y) in the closed bounded region Ω is given a densityσ(x, y), we partition Ω into ∆ : Ω1,Ω2, . . . ,Ωn and select an arbitrary pointPi(xi, yi) in Ωi. Consider the centroid (x∆, y∆) of particles P1,P2, . . . ,Pn.

Let the area of Ωi be ∆Si. Then

x∆ =

∑ni=1 σ(Pi)xi∆Si∑ni=1 σ(Pi)∆Si

, y∆ =

∑ni=1 σ(Pi)yi∆Si∑ni=1 σ(Pi)∆Si

Now by letting |∆| → 0, x∆, y∆ converge to x, y. Thus,

x =1

m

∫∫Ω

σ(x, y)xdxdy, y =1

M

∫∫Ω

σ(x, y)ydxdy

where m =∫∫

Ωσ(x, y)dxdy represents mass.

Centroid of Solid If the density σ = ρ(x, y, z) is given to each point in the closed region T ,we can find (x, y, z) as follows.

x =1

m

∫∫∫T

ρ(x, y, z)xdxdydz, y =1

m

∫∫∫T

ρ(x, y, z)ydxdydz

z =1

m

∫∫∫T

ρ(x, y, z)zdxdydz, m =

∫∫∫T

ρ(x, y, z)dxdydz


Example 5.13 Find the centroid of the figure below if the density at eachpoint is σ(x, y, z) = xy.

SOLUTION

x =1

m

∫∫∫T

xyxdxdydz

=1

m

∫ 1

0

∫ 1−x

0

∫ 1−x−y

0

x2ydzdydx

=1

m

∫ 1

0

∫ 1−x

0

x2y(1− x− y)dydx

=1

m

∫ 1

0

[(x2(1− x)y2

2− x2y3

3)

]1−x0

dx

=1

m

∫ 1

0

x2(1− x)3

6dx =

1

6m

∫ 1

0

(x2 − 3x3 + 3x4 − x5)dx

=1

6m

[x3

3− 3x4

4+

3x5

5− x6

6

]10

=1

360m

Since by Exercise5.12, m =1

120. Thus, x =

1

3.

We next find y. Interchange x and y, we get the same figure. Thus y = 13 .

Similarly, z = 16

Check∫ 1−x−y0

x2ydz =[x2yz

]1−x−y0

= x2y(1− x− y) Check[x2(1−x)y2

2 − x2y3

3

]1−x0

=

3x2(1−x)3−2x2(1−x)36 =

x2(1−x)36 = x2(1−3x+3x2−x3)

6 =x2−3x3+3x4−x5

6

Exercise 5.13 Find the centroid of the right cone with the radius a and theheight h if the density at each point is constant.

Exercise5-13

SOLUTION By symmetry, x = y = 0. So we only need to find z.

z =1

V

∫∫∫T

zdxdydz


V = 13πa

2h. Note that the triangle with the base a and the height h and thetriangle with the base r and the height k is similar. Thus,

h

a=k

r

and k = hra . Therefore,

T ′ = (r.θ, z) : 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π,hr

a≤ z ≤ h

Check

Thus ∫∫∫T ′z|J |dzdrdθ =

∫∫∫T ′zrdzdrdθ

=

∫ 2π

0

∫ a

0

∫ h

hra

zdzrdrdθ

=

∫ 2π

0

∫ a

0

[z2

2

]hhra

rdrdθ

=

∫ 2π

0

∫ a

0

1

2(h2 − h2r2

a2)rdrdθ

=1

2

∫ 2π

0

dθ

∫ a

0

(h2r − h2r3

a2)dr

= π

[h2r2

2− h2r4

4a2

]a0

= π(h2a2

2− h2a4

4a2)

=πh2a2

4.

Therefore,

z =πh2a2

4· 3

πa2h=

3h

4

Moment of Inertia Let σ = σ(x, y, z) be the density at each point (x, y, z) of region. Thenthe moment of inertia of R about x-axis, y-axis, z-axis is give by thefollowing.

Ix =

∫∫∫R

σ(y2 + z2)dxdydz Iy =

∫∫∫R

σ(x2 + z2)dxdydz

Iz =

∫∫∫R

σ(x2 + y2)dxdydz Exercise A

1. Evaluate the following triple integrals．

(a)

∫ a

0

∫ b

0

∫ c

0

dxdydz (b)

∫ 1

0

∫ x

0

∫ y

0

ydzdydx

2. Express the followings using triple integrals.

(a) The mass of the ball x2 + y2 + z2 ≤ r2 provided the density is pro-portional to the distance from the origin．

(b) The mass of the cone z = 1 and z =√x2 + y2 provided that the

density is proportional to the distance from the origin．

(c) The volume common to z = 4− x2 − y2 and z = 2 + y2．


3. Find the center of mass of the following closed region bounded by the fol-lowing curves．

(a) y = x and y = x2 provided the density is constant. (b) x2 = 4y andx− 2y + 4 = 0.

(c) y = x2 − 2x and y = 6x− x2.

Exercise B

1. Evaluate the following triple integrals for T = (x, y, z) : 0 ≤ x ≤ y ≤ z ≤ 1．

(a)

∫∫∫T

dxdydz (b)

∫∫∫T

ex+y+zdxdydz

2. Evaluate the following triple integrals.．

(a)

∫∫∫T

dxdydz, T = (x, y, z) :√x2 + y2 ≤ z ≤ 3

(b)

∫∫∫T

(x2 + y2 + z2)dxdydz, T = (x, y, z) : x2

a2+y2

b2+z2

c2≤ 1

3. Find the center of mass of the following region．

(a) y = x and y = 6x− x2

(b) Semishpher x2 + y2 + z2 ≤ a2, z ≥ 0 provided the density is propor-tional to the distanace from the origin.

(c) The right circular cone with the bottom radius a and the hight h．

(d) ax ≤ x2 + y2 ≤ a2

(e) Find the center y, z of the trapezoid given in Example 5.13

(f) ax ≤ x2 + y2 ≤ a2 provided the density is proportional to the distancefrom the origin.

Appendix A

ANSWERS

0.2 INEQUALITIES

1.

(a) (−∞, 1] (b) (−∞,−1

5) (c) (−∞, 1) ∪ (2,∞)．

(d) (−1, 0) ∪ (1,∞) (e) (−3, 1) ∪ (3,∞) (f) (1, 2) ∪ (6,∞)．

2.

√x+ 1

x+ 2is larger．

3.

(a) (−2, 2) (b) (−9

4,−7

4) (c) (−5, 3)∪(3, 11) (d) (−∞,−4)∪(−1,∞)

4. |a− b|2 = (a− b)2 = a2 − 2ab+ b2 ≤ |a|2 + 2|a||b|+ |b|2 = (|a|+ |b|)2

1.1 Functions

1.

(a) f(1) = 1 (b) f(1) = 13 (c) f(1) = 2

2.

(a) Domain(f) = (−∞,∞) Range(f) = [−1,∞)．

(b) Domain(f) = [1,∞) Range(f) = [0,∞)．

(c) Domain(f) = (−∞,∞) Range(f) = [0, 1]．

3.

(a) The graph of y = f(x) + c is a parallel shift of the graph of y = f(x) byc into the positive y-axis direction．

(b) The graph of y = f(x− a) is a parallel shift of the graph of y = f(x) bya into the positive x-axis direction．

71

72 APPENDIX A. ANSWERS

4.

(a) (f g)(x) = 2x2 + 5 g(f(x)) = (2x+ 5)2 (b) (f g)(x) = x g(f(x)) = x

(c) (f g)(x) = x4

1 + x2g(f(x)) = x2(x+ 1)2

5.

(a) f−1(x) =x+ 4

7(b) y = 3

√x− 2− 1 (c) is not one-to-one．

6.

(a) odd function (b) odd function

1.2 Trigonometric Functions

1.

(a)π

6(b)

2π

9(c)

2π

52.

(a)π

3(b)

π

63.

(a) [0,π

6) ∪ (

11π

6, 2π] (b) (

π

6,π

2) ∪ (

7π

6,3π

2) (c) (

π

6,π

3) ∪ (

2π

3,5π

6)

4.

(a) 0 (b)π

6(c)

√3

2

1.3 Limits of Functions

1.

(a) −1 (b)1

2(c) 0 (d)

1

2(e) 6 (f) 5 (g) 0 (h) −1

9(i)

−1

2.

(a)2

3(b) −1 (c) 0 (d) −1

73

1.4 Continuous functions

1.

(a) −1 (b) 1 (c) 0 (d) 0 (e)1

2(f) does not exist (g) does

not exist

2.

(a) continous (b) removable discontinuity (c) continous (d) essen-tial discontinuity

3.

(a) f(1) = 2 (b) f(1) = 3

4.

x =13

16

1.5 Sequences

1.

(a) ∞ (b) 0 (c) ∞ (d) 1 (e) 0

2.

(a) bounded，monotonically decreasing (b) bounded，monotonically in-creasing

3.

(a)1

n!(b) 2n− 1 (c) n2

1.6 Transcendental functions

1.

(a) 0 (b) 発散 (c) 0

2.

(a) 3.00 (b) 2.76 (c) 4.40 (d) −2.30 (e) 3.225 (f) −0.92

3.

(a) e2 (b)1

e(c) e2, 1 (d) 1

2.1 Derivatives

1.

(a) 0 (b)1

2√x− 1

(c) − 2

x3

2.

(a) 1 (b) −6

3.

(a) y = −x− 1 (b) y = −12x+ 21 (c) y =1

4x+ 1

4.

(a) y′ = 55x4 − 18x2 (b) y′ =2

x3(c) y′ = 3x2 − 6x− 1 (d) y′ = − 1

(x− 2)2

(e) y′ =2(x2 + 3x+ 1)

(2x+ 3)2(f) y′ = −3x2 − x+ 1

x2(x− 2)2(g) y′ =

2(x4 − 1)

x3


2.2 Differential Formulas

1.

(a)dy

dx=

1

nx

1n−1 (b)

dy

dx=

1

2√x

2.

(a) y′ = 4008x(x2 + 1)2003 (b) y′ = 6(x2 +1

x2)2(x− 1

x3)

(c) y′ = 6(5x+ 3)(5x2 + 6x+ 2)2

3.

(a)dy

dx= 2(x− 1) (b)

dy

dx= −x

y4.

(a) 2x log x+ x (b) 3x2 sin 2x+ 2x3 cos 2x (c)2√

1− 4x2(d)

ex

2√ex + 1

(e) 3(sin (x+ 1))2 cos (x+ 1) (f) sin−1(2x) +2x√

1− 4x2

2.3 Higher order dirivatives

1.(a) y(0) = 4, v(0) = 3, a(0) = −2 (b) y(0) = 0, v(0) = −6, a(0) = 0(c) y(0) = 9, v(0) = − 9

2 , a(0) =92

2.

(a)1

(x2 + 1)3/2(b)

1

x(c) 2ex cosx

2.4 Mean value theorem

1.

(a) ξ =3

2(b) ξ =

√13

3(c) ξ =

1√2

2.(a) concave up on (−∞, 0)，concave down on (0,∞), local minimum 4 at

x = −1，local minimum 0 at x = 1, inflection point (0, 2)(b) concave up on (−∞, 0)，concave down on (0,∞), locla maximum −2 at

x = −1，local minimum 2 at x = 1

(c)上に凸 (−∞,−1)，下に凸 (−1,∞), x = −1−√3

3で極大値

2√3

9，x = −1 +

√3

3

で極小値 −2√3

9, 変曲点 (−1, 0)

(d) concave up on (−∞,−√3), (0,

√3)，concave down on (−

√3, 0), (

√3,∞),

local minimum−1

2at x = −1，local maximum

1

2at x = 1, (−

√3,−

√3

4), (0, 0), (

√3,

√3

4)

are points of inflection(e) concave up on (−2, 1)，concave down on (−∞,−2), (1,∞), local minimum

0 at x = −2 and x = 1, local maximum 94 at x = − 1

23.

(a) 400 (b)32√3

9(c)

64√2

3(d)

1

2

2.5 Curve sketching

1.

(a) x =1

3, y =

1

3(b) x = 2, y = x+ 2 (c) x = ±3, y = 0

75

2.(a) x = 0 vertical tangent． (b) x = 0 vertical cusp． (c) x = 2 vertical

cusp．

2.6 Limit of indeterminate form

1.

(a) 0 (b) −1 (c)1

4(d) log 2 (e) 0 (f) 1 (g) 0 (h) 2

3.1 Antiderivatives

1.

(a) x2 − 3x+ C (b) x5 + C (c)3

5x

53 + C (d) 2

√x+ C (e)− log | cosx|+ C

(f) tanx+ C (g)x

2− sin 2x

4+ C (h)

1

4log |x− 2

x+ 2|+ C (i)

1

2tan−1 x

2+ C

3.2 Integration by substitution

1.

(a)−1

2cos (2x) + C (b)

1

2log(x2 + 1) + C (c)

1

2e2x + C (d) log | log x|+ C

(e)1

2ex

2

+ C (f)1

3sin3 x+ C (g) 2

((1 + x)5/2

5− (1 + x)3/2

3

)+ C

(h) log | sinx+ cosx|+ C (i) log(1 + ex) + C

3.3 Integration by parts

1.

(a) xex − ex + C (b) −x cosx+ sinx+ C (c)1

4e2x(2x− 1) + C

(d) (x2 − 2x+ 2)ex + C

(e) (2− x2) cosx+ 2x sinx+ C (f)1

3ex

3

(x3 − 1) + C

(g) x sinx+ cosx+ C

3.4 Integration of rational functions

1.

(a)1

x− 2+

−1

x+ 5(b)

−1

x+

1

x+ 1+

1

x− 1(c) 1 +

−4

x− 1+

7

x− 2

(d)3/4

x− 1+

1/2

(x− 1)2+

1/4

x+ 1(e) x3 + 4x2 + 12x+ 32 +

80

x− 2+

32

(x− 2)2

(f)1

x+

−x+ 1

x2 + 1

3.5 Integration of trigonometric functions

1.

(a)sin4 x

4+ C (b)

1

9sin3 3x+ C (c)

x

2+

sin 2x

4+ C (d) sinx− sin3 x

3+ C

(e)cos7 x

7− cos5 x

5+ C (f)

cosx

2− cos 5x

10+ C (g)

sinx

2− sin 3x

6+ C


(h)sinx

2+

sin 3x

6+ C (i)

tan2 x

2+ C (j)

1

2(secx tanx+ log | secx+ tanx|) + C

3.6 Integration of irrational functions

1.(a) 2

√x− 2 log |1 +

√x|+ C (b) x− 2

√x+ 2 log |1 +

√x|+ C

(c)1

2log |

√1− ex − 1√1− ex + 1

|+ C

(d)2

3(x− 1)

32 + 2

√x− 1 + C (e)

√x2 + 4 + C

(f)(x2 + 4)3/2

3− 4√x2 + 4 + C (g) log |x− 1 +

√x2 − 2x− 3|+ C

(h) 2(√x2 − 1− tan−1(

√x2 − 1)) + C

3.7 Definite integrals

1.

(a) limn→∞

1

n

n∑i=1

(i

n

)2

=1

3

2.

(a)10

3(b)

3

2− log 2 (c)

2

5(d) 0 (e) 1

3.(a) 5 (b) −2 (c) −1 (d) 0 (e) −4

4.(a) sinx (b) − cosx (c) 2

√sin 2x

5.

(a)1

2(b) log

3

2(c)

2

3

3.8 Evaluation of definite integrals

1.

(a) 0 (b) 0 (c)2

3(d) 0 (e)

25/2

3

3.9 Improper integrals

1.(a) 1 (b) ∞ (c) 1 (d) does not exist (e) 2 (f) ∞

3.10 Application of definite integrals

1.

(a)9

2(b)

1

12(c)

22

3(d)

4

3(e) 36 (f) 2

2.

(a)π

3(b)

1944π

5(c)

5π

14(d)

72π

53.

(a) 2√5 (b) 296 (c) 4

√3 + 2 log |2 +

√3| (d)

√2(e4π − 1)

77

4.1 Function of several variables

1.

(a) D(f) = (x, u) : xy ≥ 0, R(f) = [0,∞)

(b) D(f) = (x, y) : x+ y = 0, R(f) = (−∞, 0) ∪ (0,∞)

(c)D(f) = R2 − (0, 0), R(f) = (0,∞) (d)D(f) = R2 − (0, 0), R(f) = [0, 1]

(e) D(f) = (x, y) : xy < 1, R(f) = (−∞,∞)

(f) D(f) = (x, y, z) : x2 − y2 = 0, R(f) = (−∞,∞)

2.

(a) quadratic cone (b) elliptic paraboloid (c) parabolic cylinder (d)elliptic paraboloid

(e) elliptic cylinder (f) elliptic paraboloid (g) hyperbolic cylinder (h)hyperboloid of one sheet (i) hyperboloid of two sheets

4.2 Partial derivatives

1.

(a) fx = 6x− y, fy = −x+ 1 (b) fx = 2xe−y, fy = −x2e−y

(c) zx =x√

x2 + y2, zy =

y√x2 + y2

(d) zx = sin y, zy = x cos y

(e) zx =2y

(x+ y)2, zy =

−2x

x2 + y2

2.

(a) fx = 2ax+ 2by, fy = 2bx+ 2cy, fxx = 2a, fxy = 2b, fyy = 2c

(b) fx = 2(x+ y2 + z3), fy = 4y(x+ y2 + z3), fz = 6z2(x+ y2 + z3), fxx = 2fxy = 4y, fyy = 4(x+ y2 + z3), fyz = 12yz2, fzz = 12z(x+ y2) + 30z4

(c) fx = 3 cos (3x− 2y), fy = −2 cos(3x− 2y), fxx = −9 sin(3x− 2y)

fxy = 6 sin(3x− 2y), fyy = 4 sin(3x− 2y)

(d) fx = e2y, fy = 2xe2y, fxx = 0, fxy = 2e2y, fyy = 4e2y

4.3 Limit of functions of two variables

1.

(a) 1 (b) does not exist (c) 0

2.

(a) continuous (b) discontinuous　 (c) continuous

4.4 Total differential and tangent plane

1.

(a) ∇f = (3x2, 2y), df = 3x2dx+ 2ydy

(b) ∇f = (6x− y,−x+ 1), df = (6x− y)dx− (x− 1)dy

(c) ∇z = (2xy−2,−2x2y−3), dz = 2xy−2dx− 2x2y−3dy

(d) ∇z = (2xy, x2), dz = 2xydx+ x2dy

(e) ∇z = (ex cos y,−ex sin y), dz = ex cos ydx− ex sin ydy

2.

(a) 3x+ 2y − z = 4, (x, y, z) = (1, 1, 1) + (3, 2,−1)t

(b) x+ 2y − z = 3, (x, y, z) = (2, 1, 1) + (1, 2,−1)t　(c) 3x+ 5y − z = 4, (x, y, z) = (1, 1, 4) + (3, 5,−1)t


4.5 Directional derivatives

1.(a) −

√2 (b) − e√

22.

(a)√3 (b) 1　

3.

(a)3√10, (0, 1) (b) − 6√

10, (0,−1)

4.6 Partial differentiation of composite functions

1.(a) 8t+ 6t2 (b) 0 (c) cos 2t+ sin 2t (d) 12t11

2.(a) zu = 10u, zv = 10v(b) zu = 2u+ 2v + 2uv + v2 + 4uv2, zv = 2u+ 2v + 2uv + u2 + 4u2v　(c) zu = 2uv6, zv = 6u2v5

4.7 Extreme values of functions of two variables

1.(a) 1 + 2(x− 1) + y − 1 + (x− 1)2 + 2(x− 1)(y − 1)

(b)−π2(x− 1)− (y − π

2) +

1

2

(−π

2

4(x− 1)2 − 2(x− 1)(y − π

2)− (y − π

2)2)

(c) −(x− 1) + y − 1 +1

2

(−(x− 1)2 + 2(x− 1)(y − 1)− (y − 1)2

)(d) x+

1

2(4x2 + 2xy)

2.(a) f(1, 0) = 1 maximum (b) f(0, 4) = −32 minimum, f(0, 0) = 0 saddle

point　(c) f(1, 0) = −2 minimum, f(−1, 0) = 4 saddle point (d) f(1, 1) = −3

minimum

4.8 Implicit functions

1.

(a)dy

dx=

1

2y,d2y

dx2= − 1

8y3(b)

dy

dx= −2x+ y

x+ 4y,d2y

dx2=

−14(x2 + xy + 2y2)

(x+ 4y)3

(c)dy

dx= e−y,

d2y

dx2= e−2y (d)

dy

dx=x2 − y

x− y2,d2y

dx2=

2xy(x3 − 3xy + y3 + 1)

(x− y2)3

2.

(a)dy

dx=z − x

y − z,dz

dx=x− y

y − z(b)

dy

dx=y(z − x)

x(y − z),dz

dx=z(x− y)

x(y − z)　

4.9 Conditional extrema

1.(a) f(1, 0) = f(−1, 0) = 1 minimum, f(0, 1) = f(0,−1) = 3 maximum(b) f(2, 0) = 0 minimum, f(−2, 0) = 8 maximum

79

(c) f(0,−1) = −1 minimum, f(1√2,1√2) =

1

2+√2 maximum

2.1

4

5.2 Repeated integrals

1.

(a) 0 (b)7

6(c)

11

40(d) 2 (e)

1

122.

(a)

∫ 1

0

∫ √y

y

f(x, y)dxdy (b)

∫ 1

0

∫ 1

√x

f(x, y)dydx

(c)

∫ 1

0

∫ 2y

0

f(x, y)dxdy +

∫ 3

1

∫ 3−y

0

f(x, y)dxdy

3.

(a)1

3(b)

5

2(c)

π

2

5.3 Change of variables

1.

(a) 8π (b)3π

4(c)

π

8(d)

1

2(e)

8π

32.

(a)1

2(b)

1

3

5.4 Inproper integrals

1.

(a)e− 1

2(b)

8

15(c) −π

4

5.5 Application of double integrals

1.

(a)3π

4(b) 9(

π

12−

√3

8) (c) 2π

2.

(a) 2π (b)3√3

2(c)

2π

3

((a2 + 1)3/2 − 1

)3.

(a) π (b)π

2

5.6 Triple integrals

1.

(a) abc (b)1

122.

(a)

∫ 2π

θ=0

∫ π

ϕ=0

∫ a

ρ=0

kρ3 sinϕdρdϕdθ = kπa4

(b)

∫ 2π

θ=0

∫ π/4

ϕ=0

∫ secϕ

ρ=0

kρ3 sinϕdρdϕdθ =kπ

6(2

32 − 1)


(c)

∫∫Ω

∫ 4−x2−y2

z=2+y2dzdxdy =

∫ 2π

0

∫ 1

0

(2− 2r2)√2rdrdθ =

√2π

3.

(a) (x, y) = (1

2,2

5) (b) (x, y) = (1,

8

5) (c) (x, y) = (2, 4)

Appendix B

ANSWERS B

1.1 FUNCTIONS

1.(a) double-valued function (b) single-valued function

2.

(a) D(f) = [−2, 2] (b) D(h) = (−∞,−1

2) ∪ (0,

4

3]

3.(a) (f g)(x) = 2x2 + 1, (g f)(x) = 4x2 − 4x+ 2

(b) (f g)(x) =

x2 x < 01 + x 0 ≤ x < 1

(1 + x)2 x ≥ 1(g f)(x) =

2− x x ≤ 0−x2 0 < x < 11 + x2 x ≥ 1

4.

(a) y =1

x− 2 (b) y = −2 +

√x+ 6 (x ≥ −6) y = −2−

√x+ 6 (x ≥ −6)

5.(a) even function (b) odd function

6.(a) a product of even function and even function is even，a product of even

and odd functions is odd(b) even function is symmetric about y-axis．odd function is symmetric

about the origin．

1.2 Trigonometric Functions

1.(a)

(b)

(c)

1 = cos2α

2+ sin2

α

2

cosα = cosα

2+α

2= cos2

α

2− sin2

α

2

81

82 APPENDIX B. ANSWERS B

Thus，cos2

α

2= 1 + cosα

(d)

sin (α+ β) = sinα cosβ + cosα sinβ

sin (α− β) = sinα cosβ − cosα sinβ

Thus，

sinα cosβ =1

2[sin (α+ β) + sin (α− β)]

(e)

sin (A+B) = sinA cosB + cosA sinB

sin (A−B) = sinA cosB − cosA sinB

Let A+B = α, A−B = β. Then

A =α+ β

2, B =

α− β

2

，

sinα+ sinβ = 2[sinα+ β

2cos

α− β

2]

2.

(a) −√2

2(b)

√2 +

√6

4(c)

√2 +

√2

23.

(a) −π6

(b) π (c)π

4(d)

π

34.

Let u = sin−1 x, v = cos−1 x. Then x = sinu (−π2

≤ u ≤ π

2), x = cos v (0 ≤ v ≤ π)．

From this, x = sinu = cos v = sin(π2 − v)．Thus, u+ v =π

2.

5.

(a) y = sin−1 (−x) ⇔ −x = sin y (−π2

≤ y ≤ π

2) implies x = − sin y ⇔ x = sin (−y) ⇔

−y = sin−1 x⇔ y = − sin−1 x(b) y = cos−1 (−x) ⇔ −x = cos y (0 ≤ y ≤ π) implies x = − cos y ⇔ x = cos (π − y)

⇔ cos−1 x = π − y ⇔ y = π − cos−1 x

1.3 Limits

1.

(a) 12 (b)1

4(c) −2 (d) −8 (e) 1 (f)

2

3(g) 4

2.(a) 3 (b) 2 (c) 1 (d) 0

3.||f(x)| − 0| = |f(x)− 0|．For all ε > 0, there exists δ such that |x− a| < δ

implies ||f(x)| − 0| = |f(x)− 0| < ε

83

1.4 Continuous functions

1.

(a) does not exist (b) ∞ (c)√a (d) −

√2 (e) 1 (f) 0 (g)

does not exist

2.

continuous

3.

For a ∈ (0,∞), show limx→a√x =

√a．For all positive number ε，let

δ =ε√a，then

|√x−

√a| = 1√

x+√a|x− a| ≤ |x− a|√

a≤ ε

4.

(a) max = 11,min = −1 (b) min = 1

(c) max =

4− 2a a ≤ 04− 2a 0 < a < 4

0 a ≥ 4min =

0 a ≤ 0

−a2

4 0 < a < 4

−a2

4 a ≥ 45.

Put f(x) = 2 sinx− x. Then f(x) is continous at [π

2, π] and f(

π

2) = 2− π

2> 0.

Since f(π) = −π < 0, by the intermediate value theorem，there exists ξ in

[(π

2, π) satisfying f(ξ) = 0．

1.5 Sequences

1.

(a) ∞ (b) 0 (c) −1 (d) 1 (e) 0

2.

For a > 1, n√a > 1. Thus let n

√a = 1 + h, h > 0. Then

a = (1 + h)n = 1 + nh+ · · ·+ hn ≥ 1 + nh

Therefore

limn→∞

h = limn→∞

a− 1

n= 0

For a = 1, n√a = 1 for all n. Thus

limn→∞

n√a = 1

For 0 < a < 1, let b =1

a. Then b > 1 and

1 = limn→∞

n√b = lim

n→∞

1n√a

3.

(a) a (b) 3


1.6 Transcendental functions

1.(a) nonbounded (b) bounded

2.(a) 4 (b) 2

23

3.(a) 1 (b) e2 (c) 0 (d) 0

4.

(a) 1 (b) 1 (c)1

a(d) α (e) 1

2.1 Derivatives

1.(a)

cos(3(x+ h))− cos(3x)

h=

cos 3x(cos 3h− 1)

h− sin 3x sin 3h

h→ −3 sin 3x

(b)

(x+ 2 + h)n − (x+ 2)n

h=n(x+ 2)n−1h

h+ h(· · · ) → n(x+ 2)n−1

2.(a) df = 4x3dx (b) df = exdx

3.(a) f

′

+(0) = 1, f′

−(0) = −1 (b) f′

+(0) = 0, f′

−(0) = 0

(c) f′

+(0) = 1, f′

−(0) = −14.

(a)−3x2 + 2x+ 3

(x2 + 1)2(b) secx tanx (c) −cosecx cotx (d) −cosec2x

(e) x(x+ 2)ex (f) ex(sinx+ cosx) (g)ex(sinx− cosx)

sin2 x

2.2 Differentiation formulas

1.

(a)−1√1− x2

(b)1

1 + x2

2.

(a) x2√x3 + 2x+ 1

x2 − 3x+ 1

(2 +

1

2(

3x2 + 2

x3 + 2x+ 1− 2x− 3

x2 − 3x+ 1)

)(b) xx(log x+ 1)

(c) sinxx(log(sinx) +

x cosx

sinx

)(d) x1/x

(1

x2− log x

x2

)3.

(a) −cos t

sin t(b)

2t2 − t1/2

t3/2 + 24.

(a) 2x+5

2x3/2 (b) 2x3 sec2(2x) + 3x2 tan(2x) (c)

x√1− x2

+ sin−1(x)

(d)1− x2

(1 + x2)2 (e) x cos(x) + sin(x) (f) sin−1(x)

(g)2x

2 + 2x2 + x4(h) − sin(

√1 + 2x)√

1 + 2x

85

(i)x2

(cos(x) + x sin(x))2 (j) e2 x (2 cos(x)− sin(x)) (k)

1√A+ x2

(l) 2x cos(x2 + 1) (m) − 1

2√x+ 1

sin√x+ 1 (n) esin x cosx

2.3 Higher order derivatives

1.

(a) Use mathematical induction．For n = 1，(sinx)′ = cosx = sin(x+ π2 ).

Now assume the claim is true for n = k and show that the claim is truefor n = k + 1．

(sinx)(k+1) =((sinx)(k)

)′=

(sin(x+

kπ

2)

)′

= cos(x+kπ

2) = sin(x+

nπ

2+π

2)

(b) For n = 1，(cosx)′ = − sinx = cos(x + π2 ). Assume that it is true for

n = k and show that true for n = k + 1．(cosx)(k+1) =

((cosx)(k)

)′=(cos(x+ nπ

2 ))′

= − sin(x + nπ2 ) = cos(x +

nπ2 + π

2 )

(c) For n = 1，[(1 + x)α]′ = α(1 + x)α−1. Next assume that claim is truefor n = k. Then show it is true for n = k + 1．

[(1 + x)α](k+1) =([(1 + x)α](k)

)′=(α(α− 1) · · · (α− k + 1)(1 + x)α−k

)′= α(α− 1) · · · (α− k + 1)(α− k)(1 + x)α−k−1

2.

(a) f (n)(x) =

−2x− 1 + (1− x)−2, n = 1−2 + 2(1− x)−3, n = 2n!(1− x)−n−1, n ≥ 3

(b) x2 sin(x+

nπ

2

)+ 2xn sin

(x+

(n− 1)π

2

)+ n(n− 1) sin

(x+

(n− 2)π

2

)(c) (

√2)nex sin (x+ (nπ/4))

2.4 Properties of functions

1.

(a) −1

3(b)

√1−

(2

π

)2

(c) e− 1

2.

f ′(x) = 1− sec2 x ≤ 0 and the equality holds only for x = 0．3.

(a) Let f(x) = log (1 + x)− x

1 + x. Then we have f(0) = 0. Thus for x > 0,

we only need to show f ′(x) > 0．

f ′(x) =1

1 + x− 1

(1 + x)2=

x

(1 + x)2> 0

(b) Let f(x) = x− tan−1 x. Then f(0) = 0. Thus for x > 0, we only needto show f ′(x) > 0．

f ′(x) = 1− 1

1 + x2=

x2

1 + x2> 0


Next, let g(x) = tan−1 x− x

1 + x2. Then，f(0) = 0. Thus for x > 0, we need

to show f ′(x) > 0．

f ′(x) =1

1 + x2− 1− x2

(1 + x2)2=

2x2

(1 + x2)2> 0

(c) Take logarithm to both sides，show π > e log π．Let f(x) = x− e log x.Then f(e) = 0．For x > e

f ′(x) = 1− e

x=x− e

x> 0

Thus f(π) = π − e log π > 04.

(a) local maximum 7 at x = 1，local minimum 3 at x = 3

(b) local minimum 0 at x = 0， local maximum4

e2at x = 2

-2 2 4 6

-2.5

2.5

5

7.5

10

12.5

-2 2 4 6 8 10

1

2

3

4

2.5 Curve sketching

1.

-10 -5 5 10

0.2

0.4

0.6

0.8

1

-3 -2 -1 1

-75

-50

-25

25

50

75

2.

2.6 Limits of indeterminate form

1.1.

(a)2

3(b) 0 (c) 1 (d) 0 (e)

−1

3(f) 1 (g)

√e (h) e−1

87

2.7 Taylor’s theorem

1.(a) f (n)(x) = cos (x+

nπ

2)より

|Rn| = |f(n)(θx)

n!xn| ≤ |

cos (θx+ nπ2 )

n!||x|n ≤ 1

n!|x|n → 0 (n→ ∞)

(b) f (n)(x) = (−1)n−1(1 + x)−n より

|Rn| = |f(n)(θx)

n!xn| ≤ | (−1)n−1(1 + θx)−n

n!||x|n → 0 (n→ ∞)

(c) f (n)(x) = α · (α− 1) · · · (α− n+ 1)(1 + x)α−n より

|Rn| = |f(n)(θx)

n!xn| ≤ |α · (α− 1) · · · (α− n+ 1)(1 + θx)α−n

n!||x|n → 0 (n→ ∞)

2.

(a) 1 (b)1

6(c)

1

2(d) 0

3. omitted

3.1 Antiderivatives

1.

(a) −3

2x−2 +

2

3x6 (b)

t4

4+ tan−1(t) (c) −2(tanx− x) (d)

2

7x

72 + 2x

12

(e) x+1

2tan−1 x

2(f) sin−1(

t

2) (g)

x

2+

sin 2x

4(h) log |t+

√t2 + 4|

(i)1

4log

∣∣∣∣2 + x

2− x

∣∣∣∣3.2 Integration by substitution

1.(a) −e2−x (b) − tan(1− x) (c) −

√1− x2 (d) sec(x)

(e) −e 1x (f)

2√1 + 3 tan(θ)

3(g) −2x− 2 cot(x) (h)

1

2(log(x))2

(i) tan−1(ex) (j)− cos(x2)

2

3.3 Integration by parts

1.

(a)x2 log x

2− x2

4(b) −ex(x2 + 2x+ 2) (c) x(log x)2 − 2(x log x− x)

(d)1

16(x+ 5)16 − 1

3(x+ 5)15 (e) cos(x) + x sin(x)

(f) −ex

2(cosx− sinx) (g) x log (1 + x2)− 2(x− tan−1 x)

(h)x2 tan−1 x

2− 1

2(x− tan−1 x) (i)

xn+1 log x

n+ 1− xn+1

n+ 1(j) −x3 cosx+ 3x2 sinx+ 6x cosx− 6 sinx (k) x coshx− sinhx

89

(k)√x (6 + x)

(−9

2+x

2+x2

3

)+

27

2log(3 + x+

√x (6 + x))

3.7 Definite integrals

1.

(a) −f(x) (b) f(x+ 1)− f(x) (c) 2x

∫ 2x

0

f(t)dt+ 2x2f(2x)

2.

(a)32

3(b)

1

4(c) 1 (d)

1

2− 1

2 e(e) log (

3

2)

3. omitted4.

(a)π

4=

∫ 1

0

dx

1 + x2≤∫ 1

0

dx

1 + xn< 1

(b)

1

2(n+ 1)<

∫ 1

0

xn

1 + xdx <

1

n+ 1<

1

n

5.

(a) log 2 (b) log (1 +√2) (c)

1

3

3.8 Evaluation of integrals

1.

(a)1

5(b) −2 + 2

√2 (c)

3π

16(d) 4 cos 1− 2 sin 1 (e) 0

(f)π

3−

√3

2(g) 1

2.

In =

∫ π2

0

sinn xdx =

∫ π2

0

sinn−1 x sinxdx =n− 1

nIn−2

Jn =

∫ π2

0

cosn xdx =

∫ π2

0

cosn−1 x cosxdx =n− 1

nJn−2


1.(a) 2 (b) divergence (c) −1

2.(a) 1 (b) convergence for α > 1, divergence for α ≤ 1 (c) convergence

for α > 1，divergence for α ≤ 13.

(a) convergence

√πΓ(1/4)

2Γ(3/4)(b) convergence −4

3.10 Application of integrals

1.

(a) 4 (b)3π

162.

(a) 4π2 (b) 5π2

3.


(a) 6 (b)1

6(c)

3π

24.1 Functions of several variables

1.(a) D(f) = R2 (b) D(f) = R2 \ (0, 0) (c) D(f) = (x, y) : xy < 1

4.2 Partial derivatives

1.(a) zx = 3x2 + y2, zy = 2xy + 3y2

(b) zx = ex sin y, zy = ex cos y

(c) zx =2x

x2 + y2, zy = x3 + 2xy

2.(a) zx = 3x2y + y2, zy = x3 + 2xy, zxx = 6xy, zxy = 3x2 + 2y, zyx = 3x2 + 2y, zyy = 2x

(b) (y2 + xy)exy , zy = (2xy − x2)e

xy , zxx = e

xy (x+ 2y)

zxy = exy (2y − x2

y), zyy = e

xy (2x− 2x2

y+x3

y2)

(c) zx =2x

1 + (x2 + y2)2, zy =

2y

1 + (x2 + y2)2, zxx =

2 + 2(x2 + y2)2 − 8x2(x2 + y2)

(1 + (x2 + y2)2)2

zyy =2 + 2(x2 + y2)2 − 8y2(x2 + y2)

(1 + (x2 + y2)2), zxy =

−8x(x2 + y2)

(1 + (x2 + y2)2)2

3.(a) Check to see fx(0, 0), fy(0, 0) exist．fx(x, 0) = 0

x2 = 0 implies fx(x, 0) =

0．Thus，fx(0, 0) = 0．fy(0, y) = y3

y2 = y．fy(0, y) = 1．Therefore，fy(0, 0) = 1．(b) f(x, 0) = log 1 = 0 implies fx(x, 0) = 0. Then，fx(0, 0) = 0. f(0, y) =

log(1 + y2) implies fy(0, y) =2y

1+y2．Thus，fy(0, 0) = 0.

4.3 Limit of functions of two variables

2(a) does not exist (b) does not exist (c) 0

3.(a) continuous (b) discontinuous (c) discontinuous

4.4 Total differentials

1.(a) df = 3x2y4dx+ 4x3y3dx, ∇f = (3x2y4, 4x3y3)

z = 3x+ 4y − 6,x− 1

3=y − 1

4=z − 1

−1(b) df = (3x2y + 2xy4)dx+ (x3 + 4x2y3)dy, ∇f = (3x2y + 2xy4, x3 + 4x2y3)

z = 5x+ 5y − 8,x− 1

5=y − 1

5=z − 2

−1(c) df = (2xye2x + 2x2ye2x)dx+ xe2xdy, ∇f = (2xye2x + 2x2ye2x, xe2x)

z = 4e2x+ e2y − 4e2,x− 1

4e2=y − 1

e2=z − e2

−1(d) df = y sin(xy)dx− x sin(xy)dy, ∇f = (−y sin(xy),−x sin(xy))−(x− 1) sin 1− (y − 1) sin 1− (z − cos 1) = 0,

x− 1

− sin 1=

y − 1

− sin 1=z − cos 1

−12.

91

(a) f(125, 17) ≈ 22.7 (b) f( 6π7 ,π3 ) ≈ 0.22

4.5 Directional derivatives

1.

(a)1 +

√3

2(b)

√3

22.

(a)1

8(b) −1 +

√3

23.

(a) in the direction of (1,1)√2 (b) in the direction of (−1,−1)0　

4.6 Partial differentiation of composite functions

1.

(a)2x

x2 + y2(1− 1

t2) +

2y

x2 + y2(2t− 1) (b) zx(2t) + zye

t

(c) 2zx + 8tzy (d) −6 sin t cos t2.

(a) zr =−3s

r2 + s2, zs =

3r

r2 + s2

(b) zr =−2(r − 1)

(r − 1)2 + s2+

2(r + 1)

(r + 1)2 + s2, zs =

−2s

(r − 1)2 + s2+

2s

(r + 1)2 + s2

(c) zr = 1, zs = −2r sin s cos s—bf 3.

∂z

∂r=

∂z

∂x

∂x

∂r+∂z

∂y

∂y

∂r

= zx cos θ + zy sin θ

∂z

∂θ=

∂z

∂x

∂x

∂θ+∂z

∂y

∂y

∂θ

= zx(−r sin θ) + zy(r cos θ)

= r(−zx sin θ + zy cos θ)

4.7 Extreme value of functions of two variables

1.(a) local minimum −14 at (1, 4) (b) no local extrema (c) local

minimum −1 at (1, 1) (d) local minimum −1 at (±1, 0)2.

(a) f(x, y) = 1 + x+1

2(x2 − y2)

(b) log 3 +1

3(x− 2) +

2

3(y − 1)− 1

18(x− 2)2 − 2

9(x− 2)(y − 1) +

2

9(y − 1)2 + · · ·

4.8 Implicit functions

1.

(a)dy

dx= −4x+ 5y

5x− 6yd2y

dx2= −4(5x− 6y)2 − 2(5)(4x+ 5y)(5x− 6y) + (−6)(4x+ 5y)2

(5x− 6y)3


(b)dy

dx=

ex+y

1− ex+y

d2y

dx2= −−ex+y(1− ex+y)2 − 2(−ex+y)(−ex+y)(1− ex+y) + (−ex+y)(−ex+y)2

(1− ex+y)3

(c)dy

dx=

2x− y

x+ 2yd2y

dx2= −2(−2y − x)2 − 2(−1)(2x− y)(−2y − x) + (−2)(2x− y)2

(−2y − x)3

(d)dy

dx=x+ y

x− y,d2y

dx2=

2(x2 + y2)

(x− y)3

2.

(a)dy

dx= −x− 2

y,dz

dx= −2

z(b)

dy

dx= −y

2(z − x)

x2(z − y),dz

dx= −z

2(y − x)

x2(y − z)3.

tangent line : t =x− 1

5=y − 5

−√2

normal line : t =x− 1√

2=y − 5

54.

tangent plane : z = −x+ y +π

2normal line : t =

x− 1

−1=y − 1

1=z − π

2

−15.

(a) local minimum −2√2 at x =

√1

2, local maximum 2

√2 at x = −

√1

2

(b) local maximum −1

2at x = 1, local minimum

1

2at x = −1

(c) local maximum 2 3√4 at x = 2 3

√2

4.9 Lagrange’s multiplier

1.(a) at x =

√22 , y = −2

√2 is local minimum，at x = −

√22 , y = 2

√2 is local

maximum．(b) at x = 1, y = − 1

2 is local maximum，at x = −1, y = 12 is local minimum．

(c) at x = 2 3√2, y = 2 · 2 2

3 is local minimum．2.

(a) maximum −3√3

16, maximum

3√3

16(b) at (0, 0), minimum value 0,

at (3, 3), maximum value 18

(c) at (±1,±1), maximum value 1, at ± 1√3,∓ 1√

3, minimum value −1

33.

64.

at (0, 0, 1), maximum value 3， at (1, 0, 0) minimum value 1

5.2 Repeated interals

1.

(a) 2 (b)e2

2− e+

1

2(c)

2

27(d)

512

152.

(a)

∫ 1

0

∫ y1/4

√y

f(x, y)dxdy

(b)

∫ 0

−1

∫ 1

−xf(x, y)dydx+

∫ 1

0

∫ 1

x

f(x, y)dydx

93

(c)

∫ 2

1

∫ y

1

f(x, y)dxdy +

∫ 4

2

∫ y

y2

f(x, y)dxdy +

∫ 8

4

∫ 4

y2

f(x, y)dxdy

3.

(a) e− 1 (b)e− 1

2(c) 1− sin 1

5.3 Change of variables

1.

(a) 4π (b) 4π[2 log 2− 34 ] (c)

1

4(e− e−1) (d) π(e4 − e) (e)

2π

3

(f) − 7

607.2

2e

3− 8

3e


1.

(a)π

2(b) 1 (c)

π2

16(d) 1 (e) π (f)

π2

22.

Let t = x12 . Then 2tdt = dx．Thus，

Γ(1

2) =

∫ ∞

0

x−12 e−xdx = 2

∫ ∞

0

e−t2

dt

Let I =

∫ ∞

0

e−x2

dx, I =

∫ ∞

0

e−y2

dt. Then

I2 =

∫∫ ∞

0

e−(x2+y2)dxdy

Note that by 5.5，I2 =π

4より Γ(

1

2) =

√π

5.5 Application of double integrals

1.

(a)3π

32(b)

a2π

2(c) 4π − 3

22.

(a) 4πa2 (b)2π

3[(1 + a2)

32 − 1] (c) 8a2 (d) πk2m

√m2 + 1

3.

(a)2a3

3(b)

54

105(c) −4a3

3(2

3− π

2) (d)

2

9

5.6 Triple integrals

1.

(a)1

6(b)

1

6(e− 1)3

2.

(a)27π(2

√2− 1)

2(b)

4

15πabc

(a2 + b2 + c2

)3.


(a) (x, y) = (1

2,2

5) (b) (x, y, z) = (0, 0,

2a

5) (c) in the direction of the

heighth

4

(d) (x, y) = (−a6, 0) (e) (x, y, z) = (

1

3,1

3,1

3) (f) (x, y) = (− 6a

5(3π − 2), 0)

Index

change the order of integration, 41contour, 4

dependent variable, 3double integrable, 37double integral formula, 39

entroid of a system of particles, 66extrema, 24

function of two variables, 3

graph, 3

harmonic function, 23horizontally simple region, 39

implicit funciton, 29independent variable, 3

Laplacian, 23linearity of double integral, 39local maximum, 24local minimum, 24

mass, 67moment of inertia, 69

normal vector, 16

partial derivative, 11partial differentiation, 11partially differentiable, 10

repeated integral, 37

smooth surface, 54

tangent plane, 16The second partial derivatives, 21total differential, 13totally differentiable, 13

vertically simple region, 39

95

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