part iv vector calculus - ime-uspgorodski/teaching/mat2219-2016/math217_p4… · part iv: vector...

Math 217 - Calculus 3 for Chemical EngineeringFall 2011

Instructor: Abdullah Hamadeh

Part IV

Vector calculus

1 Vector fields

Historically, vector calculus was developed in order to understand a number of physical phenomena occurring innature, e.g., gravitation, electromagnetism, fluid flow andheat transfer. Note that these phenomena require theconcept of afield which exists in space and which can be used to explain their behavior. You are already familiarwith the idea thatforcesare vector quantities: In order to specify a forceF, you need to specify both itsmagnitude,or strength, as well as itsdirectionof action.

In this final part of the course, we shall be concerned withvector-valued functions of several variablesthat willhave the form

F : Rm → Rn, m,n> 1

Usually, theinput to F will be a point(x,y,z) in physical space. This point will also be denoted asr , which is alsoused to represent the position vector of the point. Theoutputof F will most often be one of the following:

1. The velocityv = (v1, · · · ,vn) of a particle moving inRn, or

2. The forceF = (F1, · · · ,Fn) exerted on a particle at the pointr .

Therefore, for a vector fieldF(r), where the input to the vector fields isr = xi + yj + zk, there is a vectorFassociated with each point(x,y,z) in space. In other words,eachof the componentsFi of the vectorF is amultivariable functionFi = Fi(r) = Fi(x,y,z), of the form we saw in Part 2.

It is useful to be able to visualize these vector fields. For example, a velocity vector field could give an idea ofhow a fluid is travelling in space. The vector fieldF(x,y) = (F1(x,y),F2(x,y)) in R

2 is graphically represented bydrawing an arrow representing the vector

F1(x,y)i +F2(x,y)j

that starts at the point(x,y). Let us look at some examples.

1. The vector fieldF(x,y) = i. This is a quite trivial field – at each point(x,y) we simply draw the vectori, aunit vector pointing in the positivex-direction.

One possible physical interpretation of this vector field isthe velocity fieldv(x,y) of a thin layer of fluid thatis moving horizontally over the surface of a table with constant velocity.

2. The vector fieldF(x,y) = xi +yj . At each pointP with coordinates(x,y), we see thatF(x,y) is the positionvector ofP. If we draw such arrows at pointsP (see below), we obtain a set of arrows which point awayfrom the origin and which get larger and larger as we move awayfrom the origin.

1

Part IV: Vector calculus 2

x

y

0 1 2 3-1-2-3

The vector fieldF(x,y) = i.

x

y

The vector fieldF(x,y) = xi +yj .

Here is another planar vector field (i.e., a vector field in theplaneR2) that will be important in future applications:

F(x,y) =−yi +xj .

A sketch of the field is given below. You should compute a few values ofF(x,y) for convenient values of(x,y) inorder to convince yourself that the vector field behaves in this way.

Obviously, the vector field suggests some kind of rotationalmotion around the origin. But the magnitude of thefield vectors, as represented by the lengths of the arrows, appears to increase. In fact,

‖ F(x,y) ‖=√

x2+y2.

One might wonder where such a vector field would correspond toa realistic situation: How can you have rotation,in which the magnitude of the rotation, i.e., the velocity, increases as you move away from the origin. It is not toohard to come up with some situations, for example, the rotation of a turntable (or a merry-go-round platform) atconstant angular velocity. Each point on the turntable travels along a circular trajectory around the origin. And allpoints take the same timeT to execute one cycle of rotation. This means that points thatare farther away fromthe origin must travel faster. The vector field pictured above describes, up to a constant factor, the velocity field ofparticles on a turntable that rotates about the origin with constant angular speed. More on this later.

Let us conclude with a look at a vector field of major importance in physics: the gravitational force.

1.1 Gravitational force fields

Suppose that a massM is located at the originO in R3. And suppose that there is another massm situated at point

P with coordinatesr = (x,y,z). (You can refer to the relevant figure in the set of illustrations of vector fields.) Whatdo we know about gravitation?

Math 217 - Lecture 15 Fall 2011


y

x

The vector fieldF =−yi+xj .

1. First, we know that the force exerted byM onmpoints in the direction fromm to M. In this case, it points inthe direction of−r , wherer is the position vector ofm. (For convenience, we have placedM at the origin.)Therefore, the gravitational force exerted byM onm will have the form

FMm =− g r (1)

whereg(x,y,z) will be a positive quantity that isnot necessarily constant.

2. Secondly, we know that the magnitude ofFMm is inversely proportional to thesquareof the distance betweenM andm. In other words,

‖ FMm ‖=K

‖ r ‖2 =Kr2 , (2)

whereK is the constant of proportionality and, for convenience, weuse the notationr =‖ r ‖.

3. The magnitude ofFMm is also proportional to each ofM andm.

We shall first consider the effect of distance on the force, for fixed M andm. From Eq. (1), we have

‖ FMm ‖= gr (3)

Comparing (3) and (2), it follows that

g=Kr3 . (4)

Therefore, from (1), we have that

FMm =−Kr3 r . (5)

We now use Fact No. 3 to deduce thatK = GMm, whereG is a proportionality constant. The final result is wellknown to you:

FMm(r) =−GMm

r3 r , (6)

whereG is the so-calledgravitational constant. Note that this result can also be written as

FMm(r) =−GMm

r2 r , (7)

wherer =rr

denotes the unit vector pointing in the direction ofr .

Eq. (6) is a very compact notation for the vector force field. If we express it in Cartesian coordinatesx, y andz, theresult is rather complicated looking. First of all, acknowledging thatr = xi +yj +zk, we have

r =‖ r ‖=√

x2+y2+z2 (8)



so that

FMm(x,y,z) =−GMm

(x2+y2+z2)3/2[xi +yj +zk] (9)

This expression will be useful later.

What does this vector field look like? As we reasoned above, at each pointP with coordinatesr = (x,y,z), theforce exerted by massM on massm will point toward the origin. So all arrows point toward the origin. As wemove away from the origin, however, the lengths of the arrowsget smaller. Here is a rough sketch:

x

y

z

The gravitational force field vectorF =−GMm

r3 r .

It is often convenient to define thegravitational fieldf that exists due to the presence of the massM at the origin:It is the gravitational force per unit mass exerted byM, i.e.

f =1m

FMm =−GMr3 r . (10)

Then the force exerted on a massm at pointP with positionr is given by

F = mf. (11)

1.1.1 Electrostatic force fields

The electrostatic force on a “test point charge”q due to the presence of a point chargeQ at the origin is

F(r) =Qq

4πε0r3 r , (12)

whereε0 is known as thepermittivity of free space. Note thatF is (1) repulsive whenQ andq have the same signand (2) attractive when they have opposite signs, as expected.

The electric field due to the presence of chargeQ at the origin is the electrostatic force per unit charge, i.e.,

E(r) =Q

4πε0r3 r , (13)

so that the force exerted on a chargeq at positionr is given by

F = qE. (14)

In the caseQ> 0, the arrows representing the electric fieldE will point outward, as sketched below.



x

y

z

The electric field vectorE =Q

4πεr3 r , for Q> 0.

2 Integration along curves inRn

In this section, we shall be integrating scalar-valued functions f (x,y,z) and vector-valued functionsF(x,y,z) alongcurvesin R

3. The motivation for doing these things will become clear as the discussion progresses.

In what follows, it will be important to keep in mind that we shall be working on curvesC in R3 that will be

parameterized in terms of a single variable as follows

r(t) = (x(t),y(t),z(t)), a≤ t ≤ b. (15)

Let us briefly recall the working definition of the Riemann integral of the single-variable functionf (x) over theinterval[a,b]:

1. We partition the intervalI = [a,b] into n subintervals of length∆x = b−an using the partition pointsxk =

a+k∆x, k= 0,1,2, · · · ,n.

2. From each subintervalIk = [xk−1,xk], pick a sample pointx∗k and evaluatef (x∗k).

3. Form the Riemann sum

Sn =n

∑k=1

f (x∗k)∆x

Now letn→∞, implying that∆x→ 0. If f (x) is a sufficiently “smooth” function (i.e. piecewise continuous),then

limn→∞

Sn = S=

∫ b

af (x) dx

2.1 Line integrals of vector-valued functions of multiple variables

The integration of a vector fieldF over a curveC is a commonly encountered application of vector calculus. Inaddition to other applications that will be encountered later in the course, the line integral can be used to computethe work done by a nonconstant force acting on a particle thatmoves along a curve. It can also be used to measurethe circulation of a rotating fluid or a magnetic field.

In what follows, we shall motivate our treatment with reference to the computation of work done by a force. Assuch, it will be useful to recall a fundamental formula from physics. Suppose that aconstantforceF = F1i +F2jis applied to a massm that is constrained to move only in thex-direction. The mass is moved from positionx= a



to x= b. In this case, it is easy to see that the total work done by the force isW = F1(b−a). It is the componentof the forceF in the direction of motion that “does the work.” This is an example of the more general formula thatcan be used inRn:

W = F ·d, (16)

whereF is theconstantforce acting on an object andd is the displacement vector−→PQ of the mass, whereP andQ

are, respectively, the initial and final positions of the mass. There is another assumption behind this formula:thatthe actual motion of the mass was along the straight linePQ.

Now let us consider a more general case: A forceF(r), which is not necessarily constant in space, is acting on amassm, as the mass moves along a curveC from pointP to pointQ as shown in the diagram below. The goal is to

m

xy

z

P

Q

F(x(t))x(t)

compute the total amount of workW done by the force. Clearly Eq. (16) does not apply here. But the fundamentalidea is to break up the motion into tiny pieces over which we can use (16) as an approximation. We then “sum up,”i.e., integrate, over all contributions to obtainW.

We assume that the curveC can be parametrized, i.e.,

r(t) = (x(t),y(t),z(t)), t ∈ [a,b], (17)

so thatr(a) is pointP andr(b) is pointQ. Now divide the parameter interval[a,b] into n subintervals of length∆t = (b−a)/n: We do this by defining the partition pointstk = a+k∆t, k= 0,1,2, · · · ,n. These points define a setof n+1 pointsPk = r(tk) = (x(tk),y(tk),z(tk)), k= 0,1, · · · ,n that lie on the curveC, with P0 = P andPn = Q.

We now come to the major point of this procedure. We shall maketwo approximations in the computation of thework ∆Wk done by the forceF in moving massm from pointPk−1 to pointPk:

1. We shall assume that the force acting on the mass during this time interval isconstant. From each parametersubinterval[tk−1, tk], pick a sample pointt∗k so that

(x∗k,y∗k,z

∗k) = r(t∗k ) = (x(t∗k ),y(tk∗),z(t

∗k ))

is a sample point on the curve extending fromPk−1 to Pk Now evaluate the force vectorF at this samplepoint, i.e.,F(r(tk)). This is the constant force that we shall assume acts on the mass fromPk−1 to Pk. (Fornvery large, we don’t expect the forceF to change appreciably over this subcurve.)

2. We shall also assume that the motion of the mass is along thestraight line segmentPk−1Pk.

These two approximations now allow us to use the “F ·d” formula in (16): the work∆Wk done by the force overthe subcurvePk−1Pk is approximated by

∆Wk ≈ F(r(t∗k )) ·−−−−→Pk−1Pk. (18)

We still need to express the displacement vector−−−−→Pk−1Pk in terms of the parametert, in order to be able to produce



an integration overt. Note that

−−−−→Pk−1Pk = ∆r(tk) (19)

= r(t∗k )− r(t∗k−1)

=r(t∗k )− r(t∗k−1)

∆t∆t

≈dr(t∗k )

dt∆t.

The final line is another approximation, but asn is increased, the length of the interval[tk−1, tk] decreases, and theapproximation gets better.

Thus the work∆Wk is now approximated by

∆Wk ≈ F(r(t∗k )) ·dr(t∗k )

dt∆t. (20)

The total work is now approximated as a sum over all subcurves:

W =n

∑k=1

∆Wk ≈n

∑k−1

F(r(t∗k )) ·dr(t∗k )

dt∆t. (21)

This is now a Riemann sum involving thescalar-valued function

f (r(t)) = F(r(t)) ·dr(t)

dt(22)

evaluated at the sample pointst∗k . In the limit n→ ∞, this Riemann sum converges to a definite integral:

W =∫ b

aF(r(t)) ·

dr(t)dt

dt = “∫

CF · dr . ” (23)

The term

dr =dr(t)

dtdt (24)

represents the infinitesimal displacement vector associated with the velocity vectordr(t)dt . The expression “

∫

CF ·dr ”

denotes the “line integral of the vector fieldF over the curveC”. But it will be the expression

W =

∫ b

aF(r(t)) ·

dr(t)dt

dt (25)

that will be useful for the practical computation of these line integrals.

Examples:

1. Evaluate the line integral∫

C F ·dr whereF = xyzi +y2j +zk along the curver(t) = (t, t, t) for 0≤ t ≤ 1.

This parametrization produces a straight line that starts at (0,0,0) and ends at (1,1,1).

Step 1: Evaluate the velocity vector:dr(t)dt = (1,1,1).

Step 2: EvaluateF at points on the curve, using the parametrization. Droppingthe unit vectorsi, j , andkfor convenience, we simply writeF = (xyz,y2,z) so that

F(r(t)) = (x(t)y(t)z(t),y(t)2,z(t)) = (t3, t2, t). (26)



Step 3: Now construct the dot product that will appear in the integrand:

F(r(t)) ·dr(t)

dt= (t3, t2, t) · (1,1,1) = t3+ t2+ t. (27)

We may now evaluate the line integral:

∫

CF ·dr =

∫ 1

0(t3+ t2+ t)dt =

14+

13+

12=

1312

. (28)

2. Evaluate the integral∫

C F ·dr whereF= xyzi+y2j +zk as in Example 1, but the curve is nowr(t) = (t, t2, t2),0≤ t ≤ 1

This parametrization produces a parabolic curve line that also starts at (0,0,0) and ends at (1,1,1).

Step 1: Evaluate the velocity vector:dr(t)dt = (1,2t,2t).

Step 2: EvaluateF at points on the curve, using the parametrization. Here,

F(r(t)) = (x(t)y(t)z(t),y(t)2,z(t)) = (t5, t4, t2). (29)


F(r(t)) ·dr(t)

dt= (t5, t4, t2) · (1,2t,2t) = t5+2t5+2t3 = 3t5+2t3. (30)

We may now evaluate the line integral:

∫

CF ·dr =

∫ 1

0(3t5+2t3)dt =

12+

12= 1. (31)

Note that this result differs from that of Example 1. The lineintegrals had the same endpoints but differentpaths. There is no guarantee that the results will be the same.

2.1.1 A comment on the notations used for line integrals of vector fields

There is another standard notation used for such line integrals. Firstly, the vector fieldF is written explicitly incomponent form as follows,

F(x,y,z) = P(x,y,z)i +Q(x,y,z)j +R(x,y,z)k. (32)

The infinitesimal vector element of displacementdr is also written in component form as

dr = dxi +dyj +dzk. (33)

Formally, the dot product of these two vectors becomes

F ·dr = (P,Q,R) · (dx,dy,dz) = Pdx+Qdy+Rdz, (34)

which is then inserted into the integral to give∫

CF ·dr =

∫

CPdx+Qdy+Rdz. (35)

This “P,Q,R” notation is often employed in calculus texts, possibly because of the use of line integrals in PhysicalChemistry, where changes to a system are often applied to each variable separately: For example, a container ofgas may first be subjected to a change in pressure, keeping temperature and volume constant – this could be the“dx” integration, then a change in temperature, keeping pressure and volume constant – the “dy” integration, etc..

We shall largely continue to use only the more general “F ·dr ” notation.



2.2 Line integrals of vector-valued functions over closed curves

Consider the situation sketched in Figure 1. The vector fieldF could represent the velocity field of fluid particlesmoving in a region of the plane. A few curves are also sketched, and we are interested the line integral ofF overthese curves.

AB

C

D

E

F

F(r)

Figure 1: The vector fieldF and curves in space.

Without even calculating these line integrals explicitly,we can get an idea of their values, keeping in mind thatthe line integral of a vector fieldF sums over the projection ofF in the direction of the tangent vectorv at eachpoint on the curve. For example, regarding curveCAB that starts atA and ends atB, its tangent vectors appear to bepointing in roughly the same direction asF. As such, we would expect the line integral overCAB to be a positivenumber.

On the other hand, the tangent vectors of curveCCD, that starts atC and ends atD, seem to be pointing in directionsopposite toF. As such, we would expect the line integral overCCD to be a negative number.

And for the curveCEF starting atE and ending atF , its tangent vectors seem to be roughly perpendicular toF, sowe would expect the line integral overCEF to be very small in value, i.e., close to zero.

Of greater importance will be the case when the curves areclosed, which will lead to the concepts ofcirculationandoutward flux.

2.2.1 Circulation of a vector field around a closed curveC in R2

We have seen that ifF is a vector field andC is a curve that travels from pointA to pointB, then we can define theline integral

∫

CF ·dr (36)

What would happen if curveC started atA, travelled around for a while and then ended back atA – in other words,A= B? In physics and other applications, one is often concerned with line integrals of vector fields oversimple,closed curves C. “Simple” means nonintersecting. “Closed” means that the curve has no endpoints – you pick anystarting point on the curve and you’ll eventually arrive back at it. Such line integrals are often denoted as follows

∮

CF ·dr . (37)

In what follows, we consider the line integral of a planar vector field F around a simple closed curveC in R2.

The convention is that the integration alongC is performed in the counterclockwise direction so that the regionDenclosed byC lies always to the left ofC as we move along the curve. Let’s recall that this line integral sums upthe projection of the vector field onto the unit tangent vector T to the curve. Assuming that we can parametrize



this curve asr(t), a≤ t ≤ b,

∮

CF ·dr =

∫ b

aF(r)(t)) ·

dr(t)dt

dt (38)

=∫ b

aF(r)(t)) ·

dr(t)dt∣

∣∣

dr(t)dt

∣∣∣

∣∣∣∣

dr(t)dt

∣∣∣∣dt

=∫ b

aF(r)(t)) · T(t) ds

=∮

f ds

where f (r(t)) = F(r(t)) · T(t) is the projection ofF in the direction of the unit tangent vector to the curveC at r(t):

T(t1)

T(t2)

T(t3)

F(r(t3))

T(t4)

F(r(t4))

F(r(t1))

F(r(t2))

C

Starting at any pointP on the curveC, the orientation of the tangent vectorT will change as we travel alongC. Inone traversal ofC, the net rotation of the tangent vector is 2π. This is quite clear whenC is a circle. As such, wesay that the line integral in (38) is thecirculation of the vector fieldF around the closed curve C.

If the vector fieldF is roughly parallel over the regionD enclosed by curveC, then we expect the line integral tobe small in value – in some regions of the curve,F points in the same direction asT and in others, it points inthe opposite direction. In other words, the vector field exhibits very little circulation. Let’s consider a very simplecase:

F = Ki (39)

andC=CR is the circle of radiusRcentered at the origin:

y

x

CR

R

F = K i

1. Parametrization of curveCR: r(t) = (Rcost,Rsint), 0≤ t ≤ 2π.

2. Velocity: dr(t)dt = (−Rsint,Rcost).

3. EvaluateF on curve:F(r(t)) = (K,0).

4. Integrand:F(r(t)) · dr(t)dt = (K,0) · (−Rsint,Rcost) =−KRsint.



Thus∮

F ·dr =−KR∫ 2π

0sint dt = 0. (40)

As expected, the line integral is zero.

Here is another situation that we expect will produce a zero result:C=CR as before and

F = Kr = Kxi +Kyj (41)

In this case, the vectorF on the curve is perpendicular to the tangent vector:

x

y

R

CR

F = Kr



3. EvaluateF on curve:F(r(t)) = K(x,y) = K(Rcost,Rsint).

4. Integrand:F(r(t)) · dr(t)dt = K(Rcost,Rsint) · (−Rsint,Rcost) =−KRcost sint +KRcost sint = 0.

Thus∮

F ·dr =∫ 2π

00 dt = 0. (42)

As expected, the line integral is zero.

Finally, let’s consider the vector fieldF = K(−yi +xj) (43)

as sketched below.

You will recall that this is the velocity vector field of a diskthat is revolving about the origin with angular frequencyω = K In this case,F points in the direction of the unit tangent vectorT at every point on the curveCR. As such,we expect a nonzero result:



3. EvaluateF on curve:F(r(t)) = K(−y,x) = K(−Rsint,Rcost).

4. Integrand:F(r(t)) · dr(t)dt = K(−Rsint,Rcost) · (−Rsint,Rcost) = KR2sin2 t +KR2cos2 t = KR2.

Thus∮

F ·dr =∫ 2π

0KR2 dt = 2πKR2. (44)

As expected, the line integral is nonzero.



x

y

CR

R

2.2.2 An important application to physics

We now consider an important application of line integrals of vector fields in physics.

Proposition: A massm moves inR3 under the influence of a forceF according to Newton’s LawF = ma. Themass moves fromA to pointB along a trajectoryr(t) which we shall denote as curveCAB. Then the workW doneby the forceF alongCAB is equal to the change in kinetic energy of the mass, i.e.

W = EK(B)−EK(A) = ∆EK , (45)

whereEK denotes the kinetic energy of the particle.

Proof: The total work done by the force is

W =∫

CAB

F ·dr (46)

=∫ b

aF(r(t)) ·

dr(t)dt

dt,

wherea andb denote the times that the particle is atA andB, respectively. Butdr(t)dt = v(t), the velocity of the

mass. At all points on the trajectory, Newton’s Law is obeyed, implying that

F(r(t)) = ma(t) = mdr(t)

dt. (47)

We substitute this result into the work integral:

W =∫ b

am

dv(t)dt

·v(t) dt (48)

The product rule applied to the dot product yields

ddt

(v(t) ·v(t)) =dv(t)

dt·v(t)+v(t) ·

dv(t)dt

= 2dv(t)

dt·v(t) (49)

ordv(t)

dt·v(t) =

12

ddt

(v(t) ·v(t)) =12

ddt

|v(t)|2 . (50)



Therefore

W =12

m∫ b

a

ddt

|v(t)|2 dt (51)

=12

m|v(b)|2−12

m|v(a)|2

= EK(b)−EK(a)

= ∆EK .

3 Integrals of flux across curves and surfaces

3.1 Outward flux of a vector field across a closed curveC in R2

Let us now return to the line integral in (38), replacing the unit tangent vectorT with theunit outward normalnto the curveC. This is the unit vector at a pointr(t) on the curve which is perpendicular to the unit tangent vectorT(t) and which points outward from the regionD enclosed byC, as shown below:

The result is the line integral that we denote as

∮

CF · n ds=

∫ b

aF(r(t)) · n(t)

∣∣∣∣

dr(t)dt

∣∣∣∣

dt. (52)

Clearly, this line integral adds up the projection of the vector field F onto the outward normal along the curveC.The result is thetotal outward fluxof F across the closed curveC. If F represents the velocity field of a fluid, thenthe total outward flux measures the net rate of fluid escape from the regionD through the curveC per unit time.

The practical calculation of the outward flux integral is notdifficult – the only complication is that you have todetermine the outward unit normal vectorn(t) to the curveC. This is easily done from a knowledge of the velocity

vector dr(t)dt . You first construct the unit tangent vector as follows:

T(t) =1

∣∣∣

dr(t)dt

∣∣∣

dr(t)dt

= (T1(t),T2(t)). (53)

Note thatT21 +T2

2 = 1. There are two unit vectors that are perpendicular toT(t) – we’ll define them as

n1 = (T2(t),−T1(t)), n2 =−(T2(t),−T1(t)). (54)

You choose the vector that pointsoutward.

n(t4)

F(x(t3)

C

n(t3)

F(x(t4))

F(x(t1))

F(x(t2))

n(t1)

n(t2)



Example 1. Find the flux integrals through the circle CR of radius R, centered on the origin for the vector fields

1. F = Kr = Kxi +Kyj

2. F =−Kyi +Kxj

where K is a positive constant.

1. In this case, the vectorF on the curve is normal to the curve, as shown in Figure 2.

x

y

R

CR

F = Kr

Figure 2: The vector fieldF = Kxi +Kyj

(a) Parametrization of curveCR: r(t) = (Rcost,Rsint), 0≤ t ≤ 2π.

(b) Unit tangent to curve:dr(t)

dt∣∣∣

dr(t)dt

∣∣∣

= (−sint,cost).

(c) Normal to curve: We have two choices for the normal,n = (cost,sint) andn = (−cost,−sint). Byconvention, we choose the first of these vectors since it points ‘outwards’ with respect to the closedcurveCR.

(d) EvaluateF on curve:F(r(t)) = K(x,y) = K(Rcost,Rsint).

(e) Integrand:F(r(t)) · n = K(Rcost,Rsint) · (cost,sint) = KRcos2 t +KRsin2 t = KR.

(f) Elemental segmentsds: Sincedsdt =

∣∣dr

dt

∣∣, thends=

∣∣dr

dt

∣∣dt = |(−Rsint,Rcost)|dt = Rdt

Thus∮

F · nds=∫ 2π

0KR2 dt = 2πKR2 (55)

2. The vector fieldF = K(−yi +xj) (56)

which is sketched in Figure 3

is tangential to the curveCR. In other words, it does not crossCR at any point, and we may therefore expectits flux integral acrossCR to be zero. To see this, we proceed as follows

(a) Parametrization of curveCR: r(t) = (Rcost,Rsint), 0≤ t ≤ 2π.

(b) Unit tangent to curve:dr(t)

dt∣∣∣

dr(t)dt

∣∣∣

= (−sint,cost).

(c) Normal to curve: We have two choices for the normal,n = (cost,sint) andn = (−cost,−sint). Byconvention, we choose the first of these vectors since it points ‘outwards’ with respect to the closedcurveCR.



x

y

CR

R

Figure 3: The vector fieldF =−Kyi +Kxj

(d) EvaluateF on curve:F(r(t)) = K(x,y) = K(−Rsint,Rcost).

(e) Integrand:F · n = K(−Rsint,Rcost) · (cost,sint) = 0.

(f) Elemental segmentsds: Sincedsdt =

∣∣dr

dt

∣∣, thends=

∣∣dr

dt

∣∣dt = |(−Rsint,Rcost)|dt = Rdt

Thus∮

F · nds=∫ 2π

00 dt = 0 (57)

3.2 Surface integrals of vector functions

In the previous section, we examined the outward flux of a vector field through a closed curveC that enclosed aregionD in the plane. In this section, we extend this idea toR

3, namely, the flux of a vector fieldF through asurfaceS.

We’ll develop this idea by means of some simple steps. First,consider a surfaceD that lies parallel to thexy-planeas sketched in Figure 4, left. Now suppose that theconstantvector fieldF = vk, with v > 0 a constant, is alsodefined at all points inR3. We shall suppose that this vector field represents the constant velocity of fluid particlesthat are moving in the positivez-direction. A subset of these particles move through regionD – those that liedirectly below regionD.

v∆t

∆S

∆V=v∆S∆t

D

z

xy

∆A

D

z

xy

F=vk

∆x

∆y

Figure 4: A surfaceD, parallel to thexy-plane. Left: flow of vector fieldF throughD, representing the velocity ofa fluid. Right: the small volume∆V of fluid that passes vertically throughD in time ∆t.

We first ask the question: How much fluid flows through surfaceD during a time interval∆t? Consider a tinyrectangular element of the surface area ofD, labelled∆S in Figure 4, right. SinceD is parallel to thexy-plane,



∆S= ∆A = ∆x∆y centered at a point(x,y) in D. After a time∆t, the fluid particles situated in this element willhave moved a distancev∆t upward. The volume of fluid that has passed through the element ∆SonD is the volumeof the box of base area∆Sand heightv∆t:

v∆t∆A. (58)

This box is sketched in Figure 4, right.

The total volume∆V of fluid that has passed through regionD over the time interval∆t is obtained by summingup over all area elements∆S in D:

∆V = v∆tx

D

dS= v∆tS(D), (59)

whereS(D) denotes the area ofD. Of course, this is a rather trivial result: the volume of fluid passing throughDis simply the volume of the solid of base areaS(D) and heightv∆t. Dividing both sides by∆t, we have

∆V∆t

= vS(D). (60)

In the limit ∆t → 0, we have the instantaneous rate of change of the volume of fluid passing through regionD, orsimply therate of fluid flowthrough regionD:

dV(t)dt

= vS(D). (61)

This quantity is thefluxof the vector fieldF through regionD.

Suppose now that the fluid is moving at a constant speedv through regionD but not necessarily at right angles toit, i.e., not necessarily parallel to its normal vectork. We shall suppose that

F = v1i +v2j +v3k, |F|= v (62)

and letγ denote the angle betweenF and the normal vectork.

In this case, the fluid particles that pass through the tiny element∆Safter a time interval∆t form a parallelepipedof base area∆Sand heightv∆t cosγ, as sketched in Figure 5.

v∆t cos γ

∆S D

z

xy

F

γ

∆V=v∆t∆S cos γ

Figure 5: Flow of vector fieldF through surfaceD, representing the velocity of a fluid. The small volume∆V offluid passes throughD at an angleγ in time ∆t.

The volume of this box isvcosγ∆t∆S (63)



The total volume∆V of fluid that has passed through regionD over the time interval∆t is obtained by summingup over all area elements∆S in D:

∆V = vcosγ∆tx

D

dS= vcosγ∆tS(D), (64)

Dividing both sides by∆t, we have∆V∆t

= vcosγS(D). (65)

In the limit ∆t → 0, we obtain the flux of the vector fieldF through regionD:

V ′(t) = vcosγS(D). (66)

We shall rewrite this flux as follows,V ′(t) = F · n S(D), (67)

wheren = k is the normal vector to surfaceD. Note that this general case includes the first case,γ = 0. And in thecase thatγ = π/2, there is no flow through the regionD, so the flux is zero.

Of course, the above results have been rather trivially obtained since (i) the vector fields are constant and (ii) theregionD is flat. Let us now generalize the first case, i.e., the vector field F is assumed to benonconstantoverregionD, i.e.,

F(x,y) = v1(x,y)i +v2(x,y)j +v3(x,y)k. (68)

In this case, the total volume∆V of fluid that has passed through regionD over the time interval∆t is obtained bysumming up over all area elements∆S in D:

∆V = ∆tx

D

F(x,y) · n dS= ∆t∫

Dv3(x,y) dS (69)

Once again dividing by∆t and taking the limit∆t → 0, we obtain the total flux ofF through regionD:

V ′(t) =x

D

F(x,y) · n dS (70)

We now arrive at the final generalization which will cover allsurfaces inR3: we do not require the surface to beflat, as was regionD in the plane, but rather a general surfaceS in R

3 – for example, a portion of a sphere, orperhaps the entire sphere. Very briefly, we first divideS into tiny infinitesimal piecesdS. We then construct anormal vectorn to each surface elementdSat a point indS, as sketched in Figure 6.

We then form the dot product of the vector fieldF at that point with the normal vectorn. This will represent thelocal flux of F through the surface elementdS. To obtain the total flux through the surfaceS, we add up the fluxesof all elementsdS– an integration overS that is denoted as

x

S

F · n dS (71)

As a matter of convention

• often, the vector quantityndS is combined into one term,dS, so the the flux integral of a vector fieldFthrough a surfaceS, which has norma vectorn is written as either one of

x

S

F · ndSorx

S

F ·dS

• when the flux integral is through aclosedsurfaceS, the notation

S

F · ndS

is often used



dS

F

n

FFF

F

F

ˆ

n

n n

n

n

n n

n

n n

n

n

Figure 6: The vector fieldF, crossing a surface with elemental surface areadSand normal vectorsn.

• for surface integrals across closed surfaces, we always measure the fluxout of the closed surface. For thisreason, the normal to the surface is taken to point outwards with respect to the closed surface.

Example 2. Evaluate the flux integral of the vector fieldF = r =[

x y z]T

through the sphere S, which is ofradius R and centered on the origin.

The sphere of interest is illustrated in Figure 7, left. A vector that is normal to this sphere is the position vectorr . A unit vector normal to the sphere is thereforen = r

|r | . Therefore the integrand of the surface integral isF · n = r · r

|r | = |r |. On the surfaceS, this integrand is simplyR.

To find the elemental surface areadS, we use spherical polar coordinates to divide the surface ofthe sphere.As illustrated in Figure 7, right, a small ‘rectangular’ elementδS of the surface area of the sphere has ‘height’Rδφ and ‘width’ Rsinφδθ . Therefore the area of this element isδS= R2sinφδφδθ (compare this with theelemental volumeδV in spherical polar coordinates), and the flux ofF through just one of these elements isF · nδS= R3sinφδφδθ . To obtain the flux through the entire volume, we sum togetherthe flux through each ofthe elementsδS that make up the surface. In the limit asδS→ 0, this summation becomes the surface integral

S

F · ndS=∫ 2π

0

∫ π

0R3sinφdφdθ

=∫ 2π

02R3dθ

= 4πR3



δθ

δφ

z

y

x

Rsin φ δS

φR

Rsin φ δθ

δS

R

z

y

x

Rδφ

Figure 7: Finding the elemental surface areaδS in terms of polar coordinates.

4 The divergence and the curl of a vector field

4.1 Divergence of a vector field

It is convenient to define the following operator, called the“del” operator,

∇ =∂∂x

i +∂∂y

i +∂∂z

k (72)

This operator could also be written as an ordered triple of operators, i.e.

∇ =

(∂∂x

,∂∂y

,∂∂z

)

. (73)

Now, an operator acts on suitable mathematical objects. We have already seen an example of the action of the deloperator – it can act on a scalar-valued functionf (x,y,z) to produce the associatedgradient vector∇ f :

∇ f =∂ f∂x

i +∂ f∂y

j +∂ f∂z

k (74)

But operators can also act on vectors. For example, we could form the scalar product of thedel operator with avector fieldF:

∇ ·F =

(∂∂x

i +∂∂y

i +∂∂z

k)

· (F1i +F2j +F3k) (75)

=∂F1

∂x+

∂F2

∂y+

∂F3

∂z.

This is known as thedivergenceof the vector fieldF. In some books, it is also written as “divF”. It is a scalarquantity, as should be the case when the dot product of two vectors is taken. Note that we can also define thedivergence of vector fields inR2 by omitting thek terms.

Examples:



1. ∇ ·C = 0 for any constant vectorC(x,y,z) =C1i +C2j +C3k.

2. If F = x2yzi +xz5sinyj +x2ysin4zk, then

∇ ·F =∂∂x

(x2yz)+∂∂y

(xz5siny)+∂∂z

(x2ysin4z) = 2xyz+xz5cosy+4x2ycos4z. (76)

3. If F = r = xi +yj +zk, thenF1 = x, F2 = y andF3 = zso that

∇ · r =∂x∂x

+∂y∂y

+∂z∂z

= 3. (77)

Recall the importance of the class of vector fieldsKr3 r in physics. WhenK = −GMm, we have the gravitational

force exerted on a massm at r by a point massM at the origin of a coordinate system. WhenK = Qq/(4πε0), wehave the electrostatic force exerted on a chargeq at r due to a point massQ at the origin. For convenience, we shallomit the multiplicative factorK. Let us first express this field in terms of Cartesian coordinates:

1r3 r =

1

(x2+y2+z2)3/2[xi +yj +zk] (78)

We have the final result (Exercise),

∇ ·1r3 r = div

1r3 r = 0, (x,y,z) 6= (0,0,0) (79)

This is an extremely important result! As we’ll see later, it reflects how the electrostatic and gravitational fieldsin R

3 naturally behave in the presence and absence of charges/masses, respectively.

Definition: A vector fieldF for which ∇ ·F = 0 for all points(x,y,z) ∈ D ⊆ R is said to beincompressibleoverD. (Some books also use the termsolenoidal.)

The vector fieldF =Kr3 r examined above is incompressible over the setR−(0,0,0).

4.1.1 Physical interpretation of the divergence

Consider a velocity fieldu(x,y,z) = uxi +uyj +uzk and a small rectangular volumeδV = δxδyδz, illustrated inFigure 8. Let us calculate the net flux ofu out of δV, given by

∮

u ·dA

Φ1 Φ2

Surface S1, area δA1

δxδy

δz

x0

Surface S2, area δA2

Figure 8: A small volumeδV.



Consider first the fluxesΦ1 andΦ2 through the surfacesS1 andS2. These are

Φ1 =∫

S1

u ·dA =−ux(x0)δA1 =−uxδyδz

Φ2 =∫

S2

u ·dA = ux(x0+δx)δA2 = ux(x0+δx)δyδz

This gives the total flux out ofδV from via surfacesS1 andS2 as

Φ1+Φ2 = [ux(x0+δx)−ux(x0)]δyδz=∂ux

∂xδV

Similarly, the pair of surfaces with normals aligned with the y-axis contribute a net flux of∂uy∂y δV, while those

aligned with thez-axis give a net flux of∂uz∂z δV. In total, the flux throughδV is

∮

u ·dA = (∂ux

∂x+

∂uy

∂y+

∂uz

∂z)δV

which is equivalent to∮

u ·dA = (∇ ·u)δV

This result holds for any small volumeδV, and not just a rectangular volume, and provides a physical interpretationof the divergence: the divergence∇ ·u at a point(x,y,z) is a measure of the net flux ofu out of a small volumeelement located at(x,y,z), as illustrated in Figure 9.

δV δV δV

∇• u= 0u

u

u

u

u

u

u

∇• u< 0∇• u> 0

Figure 9: The divergence measures the net outward flow of a vector fieldu through a small volumeδV. Left: thereis a net outward flow ofu through the volumeδV, yielding a positive divergence:∇ ·u > 0. Center: there is anet inward flow (i.e. a negative net outward flow) ofu through the volumeδV, yielding a negative divergence:∇ ·u < 0. Right: there is no net flow out of the volumeδV, what flows intoδV also flows out ofδV.

The velocity field associated with a fluid flow where the fluid has constant densityρ has zero divergence since theconservation of mass demands that what flows into a volume must also flow out of it. That is

∮

ρu ·dA = 0

and soρu, and henceu, has zero divergence:∇ ·u = 0. This is why, as mentioned previously, a field with zerodivergence is said to beincompressible.

In your course on electricity and magnetism, you will eventually encounter the equation

∇ ·E(r) =ρ(r)ε0

, (80)

whereE(r) is the electric field at a pointr andρ(r) is the charge density atr . In other words, if there is no charge ata point, then the divergence of the field will be zero there. Charge is responsible for net inflow/outflow (dependingupon the sign convention) of electric field.



Recall that the electric field at a pointr due to the presence of a point chargeQ at the origin isE =Q

4πε0r3 r . We

showed above that∇ ·E = 0 at all pointsr 6= (0,0,0). At (0,0,0), however, the divergence “blows up,” i.e., isinfinite. This is because the charge densityρ at (0,0,0) is infinite – we are assuming that a charge Q exists at apoint with no volume. Of course, a point charge is a mathematical abstraction that does not correspond to physicalreality. We’ll come back to this point later.



4.2 The curl of a vector field

Let us now take the vector product of the del operator with a vector fieldF:

∇×F =

(∂∂x

i +∂∂y

i +∂∂z

k)

× (F1i +F2j +F3k) (81)

=

∣∣∣∣∣∣

i j k∂∂x

∂∂y

∂∂z

F1 F2 F3

∣∣∣∣∣∣

=

(∂F3

∂y−

∂F2

∂z

)

i +(

∂F1

∂z−

∂F3

∂x

)

j +(

∂F2

∂x−

∂F1

∂y

)

k.

This is known as thecurl of the vector fieldF. In some books, it is also written as “curlF”. It is a vector quantity,as should be the case when the vector product of two vectors istaken. Note that we can also define the curl ofvector fields inR2 by settingF3 = 0.

Examples:

1. ∇×C = 0 for any constant vectorC(x,y,z) =C1i +C2j +C3k.

2. If F = r = xi +yj +zk, thenF1 = x, F2 = y andF3 = zso that

∇× r =

∣∣∣∣∣∣

i j k∂∂x

∂∂y

∂∂z

x y z

∣∣∣∣∣∣

=

(∂z∂y

−∂y∂z

)

i +(

∂x∂z

−∂z∂x

)

j +(

∂y∂x

−∂x∂y

)

k =~0.

3. Now considerF =−yi +xj +0k. Then

∇×F =

∣∣∣∣∣∣

i j k∂∂x

∂∂y

∂∂z

−y x 0

∣∣∣∣∣∣

=

(∂0∂y

−∂x∂z

)

i +(

∂ (−y)∂z

−∂0∂x

)

j +(

∂x∂x

−∂ (−y)

∂y

)

k = 2k.

We examined this vector field in the previous lecture in the plane,z= 0. Since there is nok component, wecan translate this field in the positive and negativez directions to obtain the vector field inR3. A view fromthe positivez-axis looking down onto thexy-plane gives the same picture as shown earlier: In thexy-plane,this vector field can represent the velocity field of a rotating disk (see next section). In this three-dimensionalcase, it could represent the velocity field of a rotating cylinder.

4.2.1 Physical interpretation of the curl

Consider a two-dimensional circular disk of radiusR, lying parallel to thexy-plane, centered on(x,y) = (0,0),rotating about thez-axis with angular velocityΩk, as illustrated in Figure 10. Each point(x,y) on the disk hasthe velocityu(x,y) = Ω[−yi+xj ] (by considering the time derivative of the parameterized curve r(t) = cos(Ωt)i+sin(Ωt)j , it can be verified that this vector field yields the angular velocity of the disk asΩk). We shall next

1. calculate∮

Cu ·dr for the curveC : |r |= R,



y

x

A top view of the vector fieldF =−yi+xj +0k

2. calculate∇×u

3. show that(∇×u)z(πR2) =∮

Cu ·dr = 2Ω, where(∇×u)z is thek component of∇×u.

Ωk

R

x

y

Figure 10: A disk of radiusR, rotating with angular velocityΩk.

1. The curveC, being the locus of points a distanceR away from the origin, can be parameterizedr(t) =(x(t),y(t)) = (Rcos(t),Rsin(t)), 0≤ t ≤ 2π. This yieldsdr = (−Rsin(t)dt,Rcos(t)dt) and

u ·dr = (−ΩRsin(t),ΩRcos(t)) · (−Rsin(t)dt,Rcos(t)dt) = ΩR2

This yields∮

Cu ·dr = 2πΩR2.

2. ∇×u =

(∂ (Ωx)

∂x−

∂ (−Ωy)∂y

)

k = 2Ωk

3. From the above two calculations, we see that(∇×u)z(πR2) =∮

Cu ·dr = 2Ω

We can generalize the above calculations for a two-dimensional vector fieldu. Consider a small areaδA of thedomain of the vector field, with unit normaln, encircled by a curveC. If δA is small enough that∇×u is constantover the area, then it can be shown that

(∇×u) · nδA=∮

Cu ·dr



n

C

Area δA

Figure 11: A small areaδA with normaln encircled by a curveC.

The above calculations also showed that(∇×u)z, the component of the curl normal to the disk, is equal to twicethe angular velocity of the disk. This result can also can be extended to show that, in general,(∇×u) · n is twicethe angular velocityΩ of the disk.

Suppose the disk which we have just analyzed is now a circularlump of fluid in a field of flowing fluid, and thatthe flow is parallel to thexy-plane. Suppose this flow field is such that at any point(x,y) in the flow, the velocityof the flow is

u(x,y) = Ω[−yi +xj ]

Into the flow, we drop a small wooden cross, as illustrated in Figure 12, left. The wooden cross would, like thedisk, exhibit translation, flowing around along the curveC, but it would alsorotatewith angular velocityΩk. Ingeneral,the curl of a vector field can be used to measure the amount and direction of rotation that arises from thevector field.

Irrotational field u: ∇ × u=0Rotational field u: ∇ × u ≠ 0

u

u u

u u

u u

u

Figure 12: Left: wooden crosses dropped into the fluid velocity field u = Ω[−yi + xj ] exhibit rotation along with

translation along the flow. Right: wooden matchsticks dropped into the irrotational fieldu = Ω[

− yx2+y2 i + x

x2+y2 j]

do not rotate along flow.

In cases such as the vector fieldu =−

yx2+y2 i +

xx2+y2 j

it can be verified that∇×u = 0. If this were a fluid flow, it would therefore give rise to pure translation, and norotation. The wooden cross dropped into the flow would therefore have an angular velocity of zero, al illustratedin Figure 12, right.

As another example, compare the two vector fields

u1 =−yi and u2 =−i

which are illustrated in Figure 13. The fieldu1 is such that∇×u1 = 1k, and we would therefore expect the woodencross dropped in to a fluid flow that behaves asu1 to rotate. This is indeed the case, and can be explained by thefact that the speed of the fluid flow increases withy, causing the top of the wooden cross to move faster than the



u1=−yi: Translational and rotational motion, ∇ × u1 =1k u2=−i: Pure translational motion: ∇ × u2 = 0

y=2 y=2

Slow flow

Fast flowy=3

y=1

y=3

y=1

x

y

x

y

Figure 13: Left: wooden crosses dropped into the fluid velocity field u1 = −yi exhibit rotation along with trans-lation along the flow. Right: wooden matchsticks dropped into the irrotational fieldu2 = −i do not rotate alongflow.

bottom. On the other hand, the fieldu2 is such that∇×u2 = 0. This field is thus irrotational, and the wooden crossflowing along it will exhibit no rotation, but pure translational motion.

In general therefore, we can characterize the motion of eachlump of fluid in a two-dimensional flow as a meanvelocity u, giving the lump translational motion, plus a mean angular velocity Ωk giving the lump rotationalmotion as illustrated in Figure 14. In a frame of reference moving with u, the above analysis has shown that in athat(∇×u)z = 2Ω.

Ω

uδA

Ωu

Figure 14: Flow of a lump of fluid can be viewed as a translational motion of velocityu along with rotationalmotion of angular velocityΩk.

4.3 Some properties of div and curl

Here are a few properties that follow naturally from the definitions of div and curl which involve derivatives. Inthe following, f : R3 → R is a scalar field andF : R3 → R

3 is a vector field.



∇( f +g) = ∇ f +∇g

∇ · (F+G) = ∇ ·F+∇ ·G

∇× (F+G) = ∇×F+∇×G

∇ · ( f F) = f ∇ ·F+F ·∇ f

∇× ( f F) = f ∇×F+∇ f ×F

These results can be derived from straightforward application of the definitions.

Here is another important result that is left as an exercise.

div ( curl F) = ∇ · (∇×F) = 0 (82)



5 The Divergence Theorem (Gauss’ theorem)

There are well-established methods to perform integrationof scalar- and vector-valued functions over surfaces.(See Sections 14.7 and 14.8 of the text.) Of course, time (or lack thereof) does not permit discussion of thesemethods. In special cases, involving closed surfacesS, it may be possible to compute the net outward flux bymeans of a volume integration, thanks to Gauss’ Divergence Theorem, one of the most important theorems ofvector calculus.

Gauss’ Divergence Theorem is very important in the study oftransport phenomena, e.g., fluid flow, heat transfer,as we have alluded to in the study of the two-dimensional divergence theorem. Its importance lies not so much inthe ability to compute surface integrals, but rather in the consequences of its applications to tranport, as we shallsee below.

In this section, we are concerned with the outward flux of a vector field (through a smooth/piece-wise smooth)surfaceS that encloses a regionV ⊂ R

3. Typically, in physics such surfaces are spheres, boxes, parallelpipeds orcylinders.

Main Result:

S is a “nice” surface enclosing regionV, piece-wise smooth, and always has an outward unit normal vector n. F isC1 overV (i.e., differentiable).

The Divergence Theorem states that:

S

F · ndS

︸︷︷︸

sur f ace integral

=y

V

∇ ·FdV

︸︷︷︸

volume integral

(83)

assuming that∇ ·F exists at all points inV. The left hand side surface integral is the total flux ofF through theclosed surfaceS.

Gauss’s theorem, states that ‘the volume integral of∇ ·F over the volumeV is equal to the net flux ofF throughthe surfaceSenclosingV ’.

Gauss’s theorem can be interpreted in this way: Imagine a volumeV composed of a large number of small elemen-tal volumesδV, as illustrated in Figure 15, left. The surfaces of these elements are mostly composed of commoninterfaces between adjoining elements, but each element atthe edge of the volume will include small part of theexternal surfaceS as part of its own surface. For each elementδV, the net flux ofF is given by∇ ·FδV. In theinterior of the volume, the flux out of one element is equal to the flux into an adjacent element across the sharedsurface. All these contributions to the total volume integral cancel, as illustrated in Figure 15, right. It is only atthe surfaceS that the contributions survive and sum together to give the net flux from the whole volume.

Example: Let Sbe a surface with outward normaln and enclosing a regionV. Then ifF is a constant vector field,i.e.,

F =C1i +C2j +C3k, (84)

it follows that divF = 0. From the Divergence Theorem,

S

F · n dS=y

V

divF dV = 0. (85)

In other words, there is as much vector fieldF entering the regionV through surfaceSas is leaving it throughS.



δV

δA

S

Vector field F

n

Net outward flow:∇• F > 0

Net outward flow from δV1cancels net inward flow into δV2

δV1

δV2

Net inward flow:∇• F < 0ˆ

Figure 15: The Divergence Theorem (Gauss’ theorem).

Example: Let Sbe the sphere of radiusRcentered at the origin andF = zk. In this case divF = 1 so that

S

F · n dS=y

V

div F dV =43

πR3. (86)

You will note that the same result will be obtained for (i)F = xi and (ii) F = yj . In fact, there are many vectorfieldsF for which divF = 1. Can you come up with others?

Example: Let Sbe the surface that encloses the cylindrical regionx2+y2 ≤ 1, 0≤ z≤ 2. And letF = y2i +x2j +z2k. Then

div F =∂∂x

(y2)+∂∂y

(x2)+∂∂z

(z2) = 2z. (87)

Thus, by the Divergence Theorem,

S

F · ndS=y

V

divFdV =y

V

2z dV. (88)

We could compute this volume integral using straightforward integration – cylindrical coordinates would be themost convenient choice:

y

V

2zdV=∫ z=2

z=02z

[∫ r=1

r=0

∫ θ=2π

θ=0rdθdr

]

dz

=∫ z=2

z=02z[πr2]r=1

r=0dz

=[πz2]z=2

z=0 = 4π

5.1 The Divergence Theorem in the plane

For a vector fieldF, a special case of the divergence theorem can be stated if we imagine the volume over whichwe integrate∇ ·F to be flattened to give a subset of a plane,D. The surface through which we calculate the flux



of F then becomes a curveC that bounds the flattened volumeD. This would give a special case of the divergencetheorem that relates the integral of∇ ·F over the areaD, to the line integral ofF along the curveC that bounds theareaD.

More specifically, letF(x,y)=F1(x,y)i+F2(x,y)j be a vector field in the plane. LetC be a simple closed (piecewiseC1) curve that encloses a regionD. Let n denote the unit outward normal toC assumed to exist at all points on thecurve (except perhaps at a finite set of “corners”). Furthermore, assume that the divergence ofF is defined for allpoints inD, i.e.,

div F(x,y) = ∇ ·F(x,y) =∂F1

∂x(x,y)+

∂F2

∂y(x,y) (89)

is defined for all(x,y) ∈ D.

The Divergence Theorem in the plane then states that∮

CF · n ds=

x

D

div F dA. (90)

The left integral is aline integralaround the curve C – it measures thenet outward fluxof the vector fieldF throughthe closed curveC. The right integral is a double integrationover the region Denclosed byC.

Examples:

1. The vector fieldF = K(xi+yj) with C=CR, the circle of radiusRcentered at the origin(0,0). This problemwas examined in a previous section.

The divergence of this vector field is simply the value 2K:

∂F1

∂x+

∂F2

∂y= K+K = 2K. (91)

Therefore, by the Divergence Theorem,∮

CF · n ds=

x

D

2K dA= 2KA(D) = 2KπR2, (92)

in agreement with our previous result.

Note that because of the constancy of the divergence in this case, the circleCR could be placed anywhere inthe plane and the net outward flux would be the same. In fact, for a general simple closed curveC,

∮

CF · n ds= 2KA(D), (93)

whereA(D) denotes the area of the regionD enclosed byC.

2. The vector fieldF = x2i + yj , with C the perimeter of the unit square in the first quadrant with vertices at(0,0),(1,0),(1,1) and(0,1).

div F = 2x+1, (94)

which is defined at all points in the plane. From the Divergence Theorem,∮

CF · n ds =

x

D

(1+2x) dA (95)

=x

D

dA+2x

D

x dA

= 1+2·12

= 2,



5.2 Applications of the Divergence Theorem in the Plane

5.2.1 The Continuity Equation (optional reading)

We now use the results of the previous section to derive an important equation that has widespread applicationsin fluid mechanics, theoretical physics and applied mathematics in general. Since we are using the DivergenceTheorem in the plane, the discussion is limited to idealizedtwo-dimensional materials. This in itself is oftensufficient for applications. However, the method describedbelow is easily extended to three dimensions – thesubject of the final lecture.

We consider the case of two-dimensional fluid flow. Letv(x,y, t) = v(r , t) represent the velocity field of a fluidmoving inR2. (We acknowledge that the field – in particular, the components of v – can change in time.) And letρ(r , t) be a scalar field representing the mass density at a pointr and timet. Then the vector field

F(r , t) = ρ(r)v(r , t), (96)

which is themomentumfield of the fluid, describes the rate of mass transfer, or “transport”, at a pointr at timet.The rate of mass transfer is‖ F(r , t) ‖ and the direction of flow isv(r , t).

Now consider afixedcurveC that encloses a bounded regionD in R2. At a given timet, the total amount of mass

in regionD is given by

M(t) =x

D

ρ(r , t) dA, (97)

where the integration is performed over the variablesr = (x,y) ∈ D. The instantaneous rate of change of mass isgiven by

M′(t) =dMdt

=ddt

x

D

ρ(r , t) dA=x

C

∂ρ∂ t

(r , t) dA. (98)

We were able to take the partial derivative into the integralsince the regionD is presumed to be fixed, henceindependent of time.

Since we assume that matter is neither created nor destroyedin region D, the rate of change of mass inD isdetermined only by the rate of entry/escape through the boundary curveC. By definition, the total outward flux ofF, given by

∮

CF · n ds, (99)

wheren is the unit outward normal toC, measures the outward flux ofF through curveC, hence the rate of escapeof mass throughC. By conservation of mass, it follows that

dMdt

=−∫

CF · n ds, (100)

or x

D

∂ρ∂ t

(r , t) dA=−∮

CF · n ds. (101)

We now assume that the Divergence Theorem can apply to the right-hand side, (i.e., divF exists at all points inD)so that the above equation becomes

x

D

∂ρ∂ t

(r , t) dA=−x

D

div F(r , t) dA. (102)

We now rewrite this equation as follows, also substitutingF = ρv,

x

D

[∂ρ∂ t

(r , t)+div [ρ(r , t)v(r , t)]]

dA= 0. (103)



We now assume that the integrand is continuous in the variablesx, y andt. Since Eq. (103) is assumed to apply toarbitrary surfacesC with enclosed regionsD, we can conclude that (see note at the end of this section)

∂ρ∂ t

(r , t)+div [ρ(r , t)v(r , t)] = 0. (104)

This may be written more simply as∂ρ∂ t

+∇ · (ρv) = 0. (105)

This important equation is known as thecontinuity equation. It is a conservation relation that represents the firststep in the analysis of fluid mechanics, continuum mechanicsand field theory.

If ρ is constant, then the continuity equation implies that

∇ ·v = 0, (106)

i.e.,v is incompressible.

Let us now return to the conclusion made in going from Eq. (103) to Eq. (104). The justification of this conclusionis called the “duBois-Reymond lemma”. A simple one-dimensional version might help:

Suppose we are given thatf (x) is continuous on[a,b] and that

∫ b

af (x) dx= 0. (107)

Of course, we can not conclude thatf (x) = 0 on[a,b]. But if we are given that

∫ d

cf (x) dx= 0 for all c,d such thata≤ c< d ≤ b, (108)

then we may prove thatf (x) = 0 identically on[a,b]. This is a one-dimensional version of the duBois-Reymond lemma.

5.2.2 The Heat Equation (optional reading)

We now consider the problem of heat transfer – once again in two dimensions. Suppose that a simple closed curveC encloses a regionD of a solid object. LetT(r , t) denote the temperature of the material at a pointr = (x,y) in D.It is a fact from physics that the total heat contained in regionD is given by

Q=x

D

σρT dA, (109)

whereρ(r) is the (areal) mass density andσ(r) is the specific heat. The rate of change of heat content inD withrespect to time will be given by

Q′(t) =x

D

σρ∂T∂ t

dA, (110)

where we have assumed thatρ andσ do not change in time.

The only way that heat can be transferred from regionD is through the boundary curveC. The rate of heat transferthrough curveC is given by the total outward flux

∮

Cq · n ds, (111)

whereq is theheat flux vectorthat describes heat flow. We know that heat travels from a region of higher tem-perature to one of lower temperature. Earlier we discussed the vector field termed the heat flux density, whichquantifies such transfer of heat:

q =−λ∇T, (112)



whereλ denotes the thermal conductivity of the material. (Note that the heat flows in the direction of steepestdescent ofT.) The total heat transfer throughC then becomes

∮

Cq · n ds=−

∮

Cλ∇T ds. (113)

Using the same conservation argument as for fluid flow in the previous section, the rate of change of heat in regionD must be equal to the negative outward flux of heat through the boundaryC:

Q′(t) =x

D

σρ∂T∂ t

dA=∮

Cλ∇T ds. (114)

For simplicity, we shall assume that the thermal conductivity λ is constant throughout the solid. We then employthe Divergence Theorem on the outward flux integral to obtain

∮

C∇T ds=

x

D

∇ · (∇T) dA=x

D

∇2T dA. (115)

where we have used the two-dimensionalLaplacianoperator

∇2() = ∇ ·∇() =∂ 2

∂ 2x()+

∂ 2

∂ 2y()

From Eq. (114), we have the result x

D

[

ρσ∂T∂ t

−λ∇2T

]

dA= 0. (116)

Since this result is assumed to hold for all closed curvesC and enclosed regionsD, we have, from the du Bois-Reymond lemma,

ρσ∂T∂ t

= λ∇2T. (117)

This is known as theheat equation. The solution of the heat equation leads to an equationT = T(x,y, t), whichdescribes the temperature at a timet of a point in the medium of interest with coordinates(x,y) - a spatial andtemporal description of how temperature varies in a medium as a result of the spread of heat throughout themedium.



6 Stokes’ theorem

Stokes’ theorem is the second major integral theorem of vector calculus and is a relationship between certain typesof line and surface integrals. We have previously seen the relationship

(∇×u) · nδA=∮

Cu ·dr (118)

which, for a vector fieldu says that the line integral around a small closed loop is equal to the component of thecurl normal to the elemental area enclosed by the loop. We canthink of this later quantity as the ‘flux’ of the vectorfield formed by the curl∇×u through the elemental areaδA.

Stokes’ theorem extends this relationship to integral form, relating the line integral ofu around a finite closedloopC to the flux of∇×u through any surface spanning the loop. The concept of a surface spanning a loopC isillustrated in Figure 16, where the surfacesS1 andS2 span the loopC. Note that a curveC in the three-dimensionalspace can be spanned by an infinite number of surfaces.

CS1

S2

u

u

Figure 16: Two surfaces,S1 andS2, spanning the curveC.

The statement of Stokes’ theorem is as follows: ifS is a surface that spans a closed curveC then∮

Cu ·dr =

∫

S(∇×u) ·dA

where the direction in whichC is traversed and the direction of the normal to the surfaceS are related by theright-hand rule, as illustrated in Figure 17 (for example, if the normal toSpoints upwards, the curveC is traversedcounterclockwise.

Stokes’ theorem can be interpreted as follows. Imagine the surfaceS spanning the closed curveC to be dividedinto a large number of surface elementsδA = δAn, as illustrated in Figure 18, left. Suppose that each elementalarea spans a small closed loopCδ . As with (118), for each of these elements, the relation

(∇×u) · nδA=∮

Cδu ·dr (119)

holds. If we calculate the total flux of the curl∇×u through the entire surfaceS, then, from (119), this is equivalentto summing the circulation along the boundary of each elemental area making up the surfaceS. If we sum all thecirculations ofu along the boundary of each element of area, then all line integrals along the common boundarylines of adjoining surface elements that do not lie on the curveC will cancel (as in Figure 18, right), leaving onlythe line integrals along the curveC. This is because, during the line integration procedure on all the elementalareas, each boundary is traversed just once in each direction, and it is only along the boundary lineC, at the edge



n

S

C

ˆ

Figure 17: Orientation of the curveC with respect to the normal to the surfaceS.

of the surfaceS, that the contributions remain to give the total circulation along the curveC. In other words,calculating the total flux of∇×u through the entire surfaceS is equal to taking the circulation alongC, the curvethat is spanned byS, which is the statement of Stokes’ theorem.

It is important to note that Stokes’ theorem applies to any surfaceSspanning the loopC. If we imagine multiplesurfacesS1,S2, · · · spanning the same loopC (as in Figure 16), then the flux of∇×u through each of these surfacesis the same.

C

u

u

δA

δA1 δA2

δA3 δA4

Line integrals along commonboundaries of elemental areas cancel out

S

Figure 18: Stokes’ theorem.



Example 3. SupposeF = −yi +xj +2zk andB = ∇×F. Find the outward flux ofB through the northern hemi-sphere|r |= R (illustrated in Figure 19) using

1. Stokes’ theorem

2. Gauss’ Theorem

3. Direct integration of the flux ofB through the hemisphere.

R

S2 (the hemispherical 'shell')

S1 (the base of the hemisphere)

n1ˆ

n2ˆ

Curve C

Figure 19: The northern hemisphereS2 and the base of the hemisphereS1.

1. LetC be the curve that encloses the base of the hemisphereS2. Note thatC is a circle of radiusR centeredon the origin. We wish to find x

S2

B · n2 dS=x

S2

(∇×F) · n2 dS

(note that, since we want the flux coming out of the hemisphere, we taken2 pointing out of the hemisphericalshellS2). By Stokes’ theorem, the right hand side of this equation isequal to

x

S2

(∇×F) · n2 dS=∮

CF ·dr

With respect to the curveC, the normaln2 point ‘upwards’, and therefore when using Stokes’ theorem wemust calculate the line integral aroundC in the direction that obeys the right hand rule with respect to n2.This direction is the counterclockwise direction looking at C from ‘above’, and a parametrization of thecurveC that follows this direction is

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

= Rcos(t)i +Rsin(t)j +0k(120)

Note that if we wanted to integrate in the opposite direction, the parametrization would have been

r(t) = Rcos(t)i −Rsin(t)j +0k

Now with the parametrization (120)

dr(t) = [−Rsin(t)i +Rcos(t)j +0k]dt

On the curveC,

F =−y(t)i +x(t)j +2z(t)k

=−Rsin(t)i +Rcos(t)j +0k

This yieldsF ·dr = R2dt and therefore

x

S2

B · n2 dS=x

S2

(∇×F) · n2 dS=∮

CF ·dr =

∫ 2π

t=0R2dt = 2πR2



Another way of applying Stokes’ theorem would have been to use the fact that for any surface spanningCsuch asS1 or S2 x

S2

(∇×F) · n2 dS=x

S1

(∇×F) · n1 dS

Note thatn1 = 1k. The vectorn1 points in the positivez direction, because with respect the curveC it mustbe orientated in the same direction as the vectorn2.

Now ∇×F = 2k, and thereforex

S1

(∇×F) · n1 dS=x

S1

2 dS= 2πR2

.

2. We wish to finds

S2B · n2dS. Let Sbe the surface composed of bothS1 andS2 in Figure 19, enclosing the

hemispherical volumeV.

Sincev

SB ·dSgives the fluxout of the hemisphere,

S

B ·dS=x

S1

B · (−n1)dS+x

S2

B · n2dS (121)

Note that we have taken the normal to the surfaceS1 to be−n1 since that is the normal toS1 that points outof the volumeV. At the same time, we have usedn2 as the normal toS2 since that is the normal toS2 thatpoints out of the volumeV.

By Gauss’ theoremy

V

∇ ·B dV =

S

B ·dS

=x

S1

B · (−n1)dS+x

S2

B · n2dS(122)

Note that for any vector fieldF, we have the identity

∇ · (∇×F) = 0

Therefore∇ ·B and the left hand side of (122) both equal zero. From (122), wethen havex

S2

B · n2dS=x

S1

B · n1dS

The right hand side of this equation is easy to compute:B = 2k and thereforeB · n1 = 2 andx

S1

B · n1dS=x

S1

2dS= 2πR2

3. We wish to finds

S2B · n2dS. The position vectorr always points normal to a spherical surface centered on

the origin. Therefore on the surfaceS2 n2 =r|r | =

rR. SinceB = 2k, we haveB · n2 = 2z. Hence,

x

S2

B · n2 dS=x

S2

2R

z dS

In spherical polar coordinatesz= Rcosφ , whilst on the surfaceS2, dS= R2sinφ dθ dφ . This gives

x

S2

B · n2 dS=x

S2

2R

z dS=∫ π

2

0

∫ θ=2π

θ=0

2R(Rcosφ)(R2sinφ dθ dφ)

=∫ π

2

0

∫ θ=2π

θ=0R2sin(2φ) dθ dφ = 2πR2

∫ π2

0sin(2φ) dφ = 2πR2



6.1 Green’s theorem: Stokes’ theorem in the plane

As with the divergence theorem in the plane, a special case ofStokes’ theorem can be stated when the surface ofinterest is flattened to give a subsetD of a plane. When the flattened surfaceD is bounded by a closed curveC,then Stokes’ theorem relates circulation of a vector fieldF aroundC to the total flux of∇×F throughD. This isknown as Green’s theorem.

More specifically, letF(x,y) = F1(x,y)i +F2(x,y)j be a vector field inR2. Let C be a simple closed curve inR2

that encloses a simply connected regionD ⊂ R. Also assume that the partial derivatives∂F2

∂xand

∂F1

∂yexists at all

points(x,y) ∈ D.∮

CF ·dr =

x

D

[∂F2

∂x−

∂F1

∂y

]

dA, (123)

where the line integration is performed overC in a counterclockwise direction, withD lying to the left of the path.Note that:

1. The integral on the left is aline integral– the circulation of the vector fieldF over the closed curveC.

2. The integral on the right is adouble integralover the regionD enclosed byC.

Special case:If∂F2

∂x=

∂F1

∂y, (124)

for all points(x,y) ∈ D, then the circulation integral∮

C F ·dr is zero. But recall that the above equality conditionfor the partial derivatives is the condition forF to be conservative. In fact, the integrand in Green’s Theorem, Eq.(123) is thek component of the curl ofF. To see this, let us compute it:

curl F = ∇×F =

∣∣∣∣∣∣

i j k∂/∂x ∂/∂y ∂/∂z

F1(x,y) F2(x,y) 0

∣∣∣∣∣∣

= 0i +0j +[

∂F2

∂x−

∂F1

∂y

]

k.

It is important to keep in mind that that the curl of a vector field in R2 is a vector that points in thez-direction. This

is related to the convention of assigning avelocity vectorthat points along an axis of rotation using the right-handrule. With the above result, we can rewrite Green’s Theorem as

∮

CF ·dr =

x

D

[curl F]k dA, (125)

where[v]k denotes thek-component of vectorv. In this case, thek-component of curlF is its onlycomponent.

Examples:

1. The vector fieldF = −Kyi +Kxj . (This is the velocity vector field of a thin plate on thexy-plane that is

rotating about thez-axis with angular speeddθdt

= K.) Now letCR denote the circle of radiusR> 0 centered

at the origin. In the previous lecture, we computed the circulation integral ofF overCR to be∮

F ·dr = 2πKR2. (126)



This was done by a direct calculation of the line integral using the parametrization ofCR asr(t)= (Rcost,Rsint).Let us now compute this result using Green’s Theorem. HereF1 =−Ky andF2 = Kx so that

∂F2

∂x−

∂F1

∂y= K− (−K) = 2K. (127)

Then the double integral in Green’s Theorem is

x

D

∂F2

∂x−

∂F1

∂ydA= 2K

x

D

dA= 2KA(D) = 2KπR2, (128)

whereA(D) denotes the area of regionD (area of a circular region, radiusR).

It is sometimes misleading to present this example because people may associate the curl ofF, which is 2K,with the origin, about which the rotation is taking place. But the curl of this vector field is 2K everywhere.This means that for any simple closed curve in the plane,

∮

CF ·dr = 2KA(D), (129)

whereA(D) denotes the area of the regionD enclosed byC.

2. Compute the circulation of the vector fieldF =−y2i+xj around the circle of radius 1 centered at the origin.Here,F1 =−y2 andF2 = x. Then

x

D

∂F2

∂x−

∂F1

∂ydA =

x

D

(1+2y) dA (130)

=x

D

dA+2x

D

y dA

= π.

The first integral is the area of the region enclosed by the unit circle. The second integral is zero. One mayconfirm this result by computing the integral explicitly, either with Cartesian or planar polar coordinates.One may also conclude that it is zero because the functiony is an odd function –y> 0 over the region abovethex-axis andy< 0 over the region below thex-axis which is a mirror image of the region above.

You can also confirm the above result by explicitly computingthe circulation integral using the parametriza-tion r(t) = (cost,sint).

3. Consider the following vector field,

F =−y

x2+y2 i +x

x2+y2 j . (131)

The curl ofF iscurl F = ∇×F = 0, (x,y) 6= 0. (132)

Because the curl is not defined at (0,0), we can use Green’s Theorem only for simple closed curvesC thatdo not enclose the origin (0,0). (Recall that one of the assumptions in Green’s Theorem was that the partialderivatives existed atall points(x,y) ∈ D, the region enclosed byC.) In this case, all circulation integralsare zero: ∮

CF ·dr = 0. (133)

For this vector field, however, wecannotuse Green’s Theorem to conclude anything about circulationinte-grals over closed curvesC that enclose the origin. However, we can compute the circulation integral usingparametrization, as long as the curveC avoids the origin. In the case thatC is the circleCR of radiusRcentered at the origin is 2π, one can compute that the circulation is 2π (Exercise).



7 Gradient vector fields

Suppose that there exists a scalar-valued functionf with an associated vector fieldF which is such that

F = ∇ f (134)

The vector fieldF is an example of agradient vector field. For vector fields defined onR3, the scalar functionfwould generally be a function of three variables, i.e.,f (x,y,z).

In physics, the convention is to denote the scalar function as−V so that

F =−∇V. (135)

In such a case, the functionV is known as thepotentialor potential energyand the vector fieldF is known as aconservative force.

Here is another example of a gradient vector field that is important in physics – the gravitational forceF(r) exertedby a massM situated at the origin on a massm at the pointP with position vectorr :

F(r) =−GMm

r3 r , where r = |r |=√

x2+y2+z2. (136)

Note that

∇(

1r

)

=−1r3 r . (137)

If we multiply both sides byGMm, we have

∇(

GMmr

)

=−GMm

r3 r = F(r). (138)

In other words, the gravitational fieldF is a gradient field,F = ∇ f , with

f (r) =GMm

r(139)

The potential energy associated with this force is

V(r) =− f (r) =−GMm

r. (140)

As a third example, we have previously seen in our discussions of contour lines that, for a flat hot plate on whichwe define a coordinate systemx,y, on which the temperature at any(x,y) is given by a functionT(x,y), Fourier’slaw of heat conduction says that heat flow per unit area,q, in the direction of unit vectorn, along which we measuredistances, is given by

q=−λdTds

whereλ is the thermal conductivity of the material from which the plate is made.

We saw that the heat flow per unit areaq can be rewritten as the directional derivative

q= (−λ∇T) · n (141)

and we noted that the negative sign is due to the fact that heatflows from hot to cold areas, whilst the temperaturegradient∇T points in the direction of maximum temperature increase. Let us define the vector field

q =−λ∇T (142)

so that Fourier’s law of heat conduction can be stated asq = q · n. The fieldq is termed theheat flux density,and gives the rate of heat flow per unit area that is normal to the isotherms. From (142), we see that the heat fluxdensity is another example of a gradient field. This field is illustrated in Figure 20.



HOT

COLD

q=−λ∇T

T(x,y)=T4

T(x,y)=T3

T(x,y)=T2

T(x,y)=T1

q

q

Figure 20: The heat flux densityq = −λ∇T is a gradient field that gives the heat flow rate per unit area inadirection normal to the isotherms, the contour lines of the functionT(x,y).

7.1 Determining whether or not a vector field is a gradient field

We now address the following question:

Suppose that we are given a vector fieldF : Rn → Rn.

1. How do we determine whether or notF is a gradient field?

2. And if F is a is a gradient field, how do we find the scalar fieldf : Rn → R such that

F = ∇ f . (143)

We shall answer these questions by examining the dimensionsn= 1,2,3 separately.

Case 1n= 1: This case is straightforward. A general one dimensional vector field has the formF = F1(x)i. Andin one dimension, the gradient vector associated with a function f (x) is simply∇ f = f ′(x)i. Therefore, forF to beequal to∇ f , we must have

f ′(x) = F1(x). (144)

Therefore we can findf (x) by simple antidifferentiation:

f (x) =∫

F1(x) dx. (145)

For example, ifF(x) = x2i, then f (x) = 13x3.

Case 2n= 2: We now deal with fields of the form

F(x,y) = F1(x,y)i +F2(x,y)j . (146)



If F is a gradient field, then by (143),

F(x,y) =∂ f∂x

i −∂ f∂y

j . (147)

Equating components, we have

∂ f∂x

= F1(x,y) (a),∂ f∂y

= F2(x,y) (b). (148)

Now differentiate both sides of (a) with respect toy and both sides of (b) with respect tox:

∂ 2 f∂y∂x

=∂F1

∂y(c),

∂ 2 f∂x∂y

=∂F2

∂x(d). (149)

Assuming thatf is sufficiently “smooth,” i.e., its second partial derivatives are continuous, then

∂ 2 f∂y∂x

=∂ 2 f

∂x∂y. (150)

This implies that∂F1

∂y=

∂F2

∂x. (151)

This is a necessary condition onF1 andF2 for F to be a gradient field.

Example 1: The forceF = (3x2−3y2)i −6xyj . HereF1(x,y) = 3x2−3y2 andF2(x,y) =−6xy. We compute:

∂F1

∂y=−6y,

∂F2

∂x=−6y. (152)

ThereforeF is a gradient field.

Example 2: The forceF = (4x2−4y2)i +(8xy− lny)j . HereF1(x,y) = 4x2−4y2 andF2(x,y) = 8xy− lny. Wecompute:

∂F1

∂y=−8y,

∂F2

∂x= 8y. (153)

ThereforeF is not a gradient field. (It is insufficient that the above partial derivatives are equal on the liney= 0.)

Now suppose that we have determined that a fieldF : R2 → R2 is a gradient field. How can we find the associated

scalar functionf (x,y) such that (143) holds? We shall use the force in Example 1 above to illustrate the procedure.Since that force is conservative, it follows that there exists a f (x,y) such that

(3x2−3y2)i −6xyj =∂ f∂x

i +∂ f∂y

j . (154)

Equating components, we have the relations

∂ f∂x

= 3x2−3y2 (a),∂ f∂y

=−6xy (b). (155)

From (a), we are looking for a functionf (x,y) which, when differentiated partially with respect tox, gives 3x2−3y2. We can “work backwards” by antidifferentiating with respect tox, keepingy fixed:

f (x,y) =∫

3x2−3y2∂x= x3−3xy2+g(y). (c) (156)



We have used the notation∂x to emphasize this partial antidifferentiation with respect to x. Note also that the“constant of integration” is an unknown functiong(y) since any constant or function ofy will be eliminated uponpartial differentiation with respect tox. Obviously, we must determineg(y). If we differentiate (c) partially withrespect toy:

∂ f∂y

=−6xy+g′(y), (d) (157)

Note that the derivative ofg with respect toy is a normal derivative sinceg was assumed to be a functiononlyof yand notx. We now compare (d) and (b):

g′(y) = 0, (158)

which implies thatg(y) =C a constant. The final result is

f (x,y) = x3−3xy2+C. (159)

This is a one-parameter family of potential functions associated with the conservative forceF. You should alwayscheck your result:

∂V∂x

= 3x2−3y2 = F1,∂V∂y

=−6xy= F2, (160)

so our result is correct.

Note that we used both pieces of information (a) and (b) above. We could have started, however, with (b), partiallyintegrated with respect toy, etc.

Case 3n= 3: We now deal with forces of the form

F(x,y,z) = F1(x,y,z)i +F2(x,y,z)j +F3(x,y,z)k. (161)

If F is a gradient field, then by (143),

F(x,y,z) =∂ f∂x

i +∂ f∂y

j +∂ f∂z

k. (162)

Equating components, we have

∂ f∂x

= F1(x,y,z) (a),∂ f∂y

= F2(x,y,z) (b),∂ f∂z

= F3(x,y,z) (c). (163)

As we did in the case ofn= 2, we differentiate both sides of (a) with respect toy and both sides of (b) with respectto x:

∂ 2 f∂y∂x

=∂F1

∂y(d),

∂ 2 f∂x∂y

=∂F2

∂x(e). (164)

Because of the equality of the two mixed derivatives, it follows that

∂F1

∂y=

∂F2

∂x. (165)

But this is not enough, we have to look at the other two possible pairings within (a)-(c). Now differentiate bothsides of (a) with respect tozand both sides of (c) with respect tox:

∂ 2 f∂z∂x

=∂F1

∂z(f),

∂ 2 f∂x∂z

=∂F3

∂x(g). (166)




∂F1

∂z=

∂F3

∂x. (167)

Finally, differentiate both sides of (b) with respect toz and both sides of (c) with respect toy:

∂ 2 f∂z∂y

=∂F2

∂z(h),

∂ 2 f∂y∂z

=∂F3

∂y(i). (168)


∂F2

∂z=

∂F3

∂y. (169)

To summarize, the relations that must be satisfied by the three components of a conservative forceF in R3 are:

∂F1

∂y=

∂F2

∂x,

∂F1

∂z=

∂F3

∂x,

∂F2

∂z=

∂F3

∂y. (170)

These relations, which must holdsimultaneously, look quite complicated. In fact, if we go back to the definitionof the curl of a vector fieldF

∇×F =

(∂∂x

i +∂∂y

i +∂∂z

k)

× (F1i +F2j +F3k) (171)

=

∣∣∣∣∣∣

i j k∂∂x

∂∂y

∂∂z

F1 F2 F3

∣∣∣∣∣∣

=

(∂F3

∂y−

∂F2

∂z

)

i +(

∂F1

∂z−

∂F3

∂x

)

j +(

∂F2

∂x−

∂F1

∂y

)

k.

we notice that the if the three relations in (170) are satisfied, then the three entries of the vector∇×F are zero. Inother words,

∇×F = 0 implies thatF is a gradient field.

In fact, the above relation goes both ways so that “implies that” can be replaced by “if and only if” but the abovewill be sufficient for our needs. To check whether or not a vector field is conservative, you should examine its curl.

The above result may also be expressed in the following way. For any scalar fieldf : R3 → R,

∇× (∇ f ) = 0. (172)

In other words, the curl of the gradient fieldF = ∇ f is zero, and because the curl gives a measure of the rotationaleffects of a vector field,gradient or conservative fields are said to beirrotational. Of course, in all cases, we areassuming that these results hold true at all points for whichall necessary derivatives exist.

Finally, how would one determine the scalar functionf (x,y,z) associated with a gradient fieldF in R3? The

answer: By tedious, systematic integration along the linesof what was done forR2 earlier. We shall not pursuethis topic.

7.2 Line integrals of gradient vector fields and the Generalized Fundamental Theoremof Calculus

Let us consider the line integral∫

C F ·dsof the vector fieldF = 2xi+4yj +zk over a couple of curves that have thesame starting and ending points:



1. The curver(t) = (cost,sint, t), with 0≤ t ≤ 2π. This is a helical curve that starts at(1,0,0) and ends at(1,0,2π).

Step 1: Evaluate the velocity vector:dr(t)dt = (−sint,cost,1).

Step 2: EvaluateF at points on the curve:

F(r(t)) = (2x(t),4y(t),z(t)) = (2cost,4sint, t) (173)


F(r(t)) ·dr(t)

dt= (2cost,4sint, t) · (−sint,cost,1) = 2cost sint + t (174)

Now evaluate the line integral:

∫

CF ·ds =

∫ 2π

0(2cost sint + t) dt (175)

=

[

sin2 t +12

t2]2π

0

= 2π2.

2. The straight line from(1,0,0) to (1,0,2π). Since the value of the line integral is independent of theparametrization used, we’ll use the simplest one,r(t) = (1,0, t), with 0≤ t ≤ 2π.

Step 1: Evaluate the velocity vector:dr(t)dt = (0,0,1).

Step 2: EvaluateF at points on the curve, using the parametrization. Here,

F(r(t)) = (2x(t),4y(t),z(t)) = (2,0, t). (176)


F(r(t)) ·dr(t)

dt= (2,0, t) · (0,0,1) = t (177)

Now evaluate the line integral:∫

CF ·ds=

∫ 2π

0t dt = 2π2. (178)

Note that the results of Examples 1 and 2 are identical. This could be a coincidence but if you tried other pathswith the same endpoints, you will obtain 2π2. We’ll show very shortly that for vector fieldF = (x,2y,4z), the lineintegral

∫

CF ·ds= 2π2 (179)

for anysufficiently continuous path that starts at(1,0,0) and ends at(1,0,2π). In other words,the line integral isindependent of path.

The reason for this independence of the path is the fact thatF is agradient vector field, i.e. there exists a scalarfunction f (x,y,z) such thatF = ∇ f . In physics, whereF normally represents a force,F is said to be aconservativeforce or conservative vector field. In physics, a conservative force is one for which there exists a scalar-valuedfunctionV(x,y,z) so thatF =−∇V.

In the discussion that follows, we shall use the term “gradient vector field”. However, when we shall discussapplications to physics, we’ll use the term “conservative force”. The two ideas are equivalent, since the scalar-valued functions are related as follows,

f (x,y,z) =−V(x,y,z). (180)

As you’ll see later, the minus sign in front ofV is very convenient in physics. It allows the total mechanical energyof a particle to be given by the sum ofEK , the kinetic energy, andV, thepotential energy.



Now returning to the vector fieldF used above, note that

F = 2xi +4yj +zk = ∇ f , (181)

where

f (x,y,z) = x2+2y2+12

z2. (182)

Here is our claim:

Let C be a curver(t), a≤ t ≤ b, andF = ∇ f a gradient vector field. Then∫

CF ·ds=

∫

C∇ f ·ds= f (r(b))− f (r(a)). (183)

This result is known as theGeneralized Fundamental Theorem of Calculusfor functions of several variables.Notethat the line integral depends only on the endpoints and not on the path taken.

Why “Generalized Fundamental Theorem”? Let’s go back to the Fundamental Theorem of Calculus for functionsof a single variable, which implies that

∫ b

af ′(x) dx= f (b)− f (a), (184)

since f (x) is an antiderivative off ′(x). Comparing (184) and (183) it appears that the gradient off , ∇ f , is thenatural “derivative” of a functionf of several variables. (Of course, for the single variable case, they are the same:∇ f = f ′(x)i.)

We now prove the GFTC in (183):

∫

C∇ f ·ds =

∫ b

a∇ f (r(t)) ·

dr(t)dt

dt (185)

=∫ b

a

(∂ f∂x

,∂ f∂y

,∂ f∂z

)

·

(dxdt

,dydt

,dzdt

)

dt

=

∫ b

a

(∂ f∂x

dxdt

+∂ f∂y

dydt

+∂ f∂z

dzdt

)

dt

=∫ b

a

ddt

f (r(t)) dt

= f (r(b))− f (r(a)).

Revisiting Examples 1 and 2 above:Let us now return to the vector fieldF used in Examples 1 and 2, and theknowledge that it is a gradient field, as shown in Eq. (182). From the GFTC, the line integral will simply be thedifference of the functionf evaluated at the endpoints:

∫

CF ·ds =

∫

C∇ f ·ds (186)

=

[

x2+2y2+12

z2](1,0,2π)

(1,0,0)

= 2π2.



7.3 Conservative vector fields and line integrals around closed curves

If we go back to the Generalized Fundamental Theorem of Calculus, which states that ifF is a gradient or conser-vative vector field andC is a curve that travels from pointA to pointB, then

∫

CF ·dr =

∫

C∇ f ·dr (187)

= f (B)− f (A)

We now consider the case of the line integral of a vector field around a closed curve, which, as we have seen, isdenoted ∮

CF ·dr (188)

In other words, the curveC starts atA, travels around for a while and then ends back atA, so thatA= B. Supposethe regionD in the domain off andF which is enclosed by the curveC is simply connected, meaning that withinthe regionD there are no ‘holes’ (as illustrated in Figure 21) or singularities of f ,F or the partial derivatives of thecomponents ofF. Then, sincef (A) = f (B), we can conclude that for conservative fieldsF,

∮

CF ·dr = 0 (189)

Region D

y

x

Curve CRegion D

y

x

Curve C

Figure 21: Left: regionD in thexyplane is simply connected. Right: regionD not simply connected.

7.4 Gradient fields summary

We can now summarize the main results relating to gradient vector fields as follows:

Let D be a simply connected domain (no holes, no singularities). Then the following are equivalent:

1. curl(F) = ∇×F = 0 onD (F is an irrotational field);

2. There exists a functionf such thatF = ∇ f onD (F is a gradient field);

3.∫

CF ·dr is independent of the path for every curveC onD (F is a conservative field).



7.5 Conservative force fields and energy considerations

Conservative fields theory is particularly useful for analyzing physics problems involving energy and work. First,we formally re-state some important definitions.

Definition: In physics, a force fieldF : Rn → Rn is conservative (i.e. gradient) if there exists a scalar-valued

functionV : Rn → R such thatF =−∇V. (190)

The scalar-valued functionV(r) is known as thepotential energyfunction.

Definition: In mathematics, a vector fieldF : Rn → Rn is agradient vector field(i.e. conservative) if there exists

a scalar-valued functionf : Rn → R such thatF = ∇ f . (191)

Of course,f andV are related byV = − f . In this section, where definite applications to physics areof concern,we shall be using the definition of conservative forces.

Let us consider some examples that apply these definitions.

7.5.1 Conservative forces in one-dimension

The simplest situation occurs if the forceF is one-dimensional, i.e., a function only of one spatial variable:

F = F(x)i. (192)

You have studied this situation in first-year physics. The potential energyV(x) is a function only ofx. The relationF =−∇V implies that

F(x) =−V ′(x) or V ′(x) =−F(x). (193)

This implies that

V(x) =−∫

F(x)+C (194)

Alternatively, let us integrate both sides of the expressionV ′(x) =−F(x) from fromx0 to a generalx value, wherex0 is a reference point:

V(x)−V(x0) =−∫ x

x0

F(s) ds. (195)

Typically, the reference pointx0 is chosen so thatV(x0) = 0. The physical interpretation of this definition is thatthepotential energy V(x) is the work done against the force f(x) in moving the mass m from x0 to x. Very soon,we’ll see the higher-dimensional analogy of this property.

Example 1: Near the surface of the earth, the gravitational force on a small object of massm is well approximatedby F(x) =−mgi, wherex denotes the height of the object above the surface of the earth and the unit vectori pointsdirectly upward. For convenience, we letx= 0 be the surface of the earth so that the potential energy of the massbecomes

V(x) =−∫ x

0−mg ds= mgx. (196)

Then for an object moving under the influence of gravity near the surface of the earth (for example, a ball that hasbeen thrown upward, or dropped from a heighth> 0), its total mechanical energy is given by

E(t) =12

mv(t)2+mgx(t). (197)



Example 2: Recall Hooke’s Law for the force exerted by a spring on a mass that is allowed to move only in onedimension:F(x) = −kxi, wherek is the spring constant andx is the displacement from the equilibrium positionx= 0. Once again we letx= 0 be reference point so that the potential energy function isdefined as

V(x) =−∫ x

0−ks ds=

12

kx2. (198)

7.5.2 Conservative forces inR2

Consider the two-dimensional mass-spring system sketchedbelow. (We are assuming that the mass is lying on africtionless floor.)

m

x

y

O

k1

k2

For small displacements from the equilibrium point(0,0), the force exerted by the springs on the massm is givenby

F(x,y) = F1(x,y)i +F2(x,y)j =−k1xi −k2yj . (199)

This force is conservative because the necessary relation,

∂F2

∂y=

∂F1

∂x, (200)

is satisfied: In this case, both terms are zero.

It is not too difficult to find a scalar-valued functionV(x,y) such thatF =−∇V, i.e.,

∂V∂x

=−k1x,∂V∂y

=−k2y. (201)

From partial integration, we find that

V(x,y) =12

k1x2+12

k2y2+C, (202)

which you can easily verify by differentiation. The reference conditionV(0,0) = 0 implies thatC = 0. Note thatV(x,y) is simply the sum of the “Hookean” potential energies of the two springs.

7.5.3 Some important conservative forces inR3

Let us return to the following result inR3 derived in a previous lecture

∇1r=−

1r3 r , r = xi +yj +zk, (203)



wherer =√

x2+y2+z2. Now note that for any constantK,

∇(

Kr

)

=−Kr3 r . (204)

The term on the right side resembles the form of vector force fields that are associated with either a point chargeor mass situated at the origin. Recall that the electrostatic field (that is, the force per unit charge)E(r) that isgenerated by the presence of a chargeQ at the origin is given by

E(r) =Q

4πε0r3 r . (205)

Note that if we set

K =−Q

4πε0, (206)

then Eq. (204) becomes

∇(

−Q

4πε0r

)

=Q

4πε0r3 r = E(r). (207)

This implies that the electrostatic fieldE is conservative, i.e.

E =−∇Φ = ∇(−Φ), where Φ(r) =Q

4πε0r. (208)

Φ(r) is theelectrostatic potentialassociated with the electrostatic fieldE(r). In Cartesian coordinates, it is givenby

Φ(x,y,z) =Q

4πε0

1√

x2+y2+z2. (209)

The level sets ofΦ are spheres inR3 that are centered at the origin(0,0,0). The value of the functionΦ(r)decreases as we move away from the origin. Once again, this isreflected in the fact that its gradient vectors pointinward. And sinceE is thenegativeof these gradient vectors,E points outward.

You may be more familiar with theelectrostatic forceF(r) exerted by a chargeQ at (0,0,0) on a test chargeq thatis situated atr . It will be given by

F(r) =Qq

4πε0r3 r = qE(r). (210)

Thepotential energyof the chargeq at r is then

V(r) =Qq

4πε0r= qΦ(r). (211)

These quantities are related, respectively, to the field quantititesE andΦ by the test charge constantq so that

F(r) =−∇V(r). (212)

In other words, the electrostatic force fieldF is conservative.

The above method is easily applied to the gravitational fieldG(r) due to a massM situated at the origin(0,0,0)with a proper selection of the constantK. We’ll leave this as an exercise for the reader. The most important pointis that the gravitational forceF exerted by massM on another massm is conservative.

7.5.4 Conservative forces and the conservation of energy

Suppose that a conservative force acts upon a massm while it is moving from pointA to pointB in R3. Then the

total work done by the force is given by

W =

∫

CAB

F ·dr (213)

= −

∫

CAB

∇V ·dr (F =−∇V)

= −[V(B)−V(A)] (by GFTC)

= V(A)−V(B),



whereV(A) andV(B) denote the potential energies of the mass atA andB, respectively. Note that we did not haveto know the curveCAB along which the mass travelled. The above result would have been valid foranycurve thatstarted atA and ended atB.

Let us now recall a result from a previous lecture, namely, that the work done by the force (whether or not it isconservative) is equal to the difference in kinetic energies of the mass, i.e.

W = K(B)−K(A). (214)

We combine the above two results to give

V(A)−V(B) = K(B)−K(A) (215)

orK(A)+V(A) = K(B)+V(B). (216)

The sum of kinetic and potential energies of a particle at a point is the total mechanical energy E of the particle atthat point. The above states thatE(A) = E(B), i.e. energy is conserved.

This result is true for any two pointsA andB on the trajectory of the mass particle as it moves under the influenceof the forceF according to Newton’s lawF = ma. As a result, we can state that the total mechanical energy oftheparticle remains unchanged along its trajectory.

We consider the total mechanical energy of the mass as a function of time as it moves along the trajectoryr(t):

E(t) =12

m‖ v(t) ‖2 +V(r(t)). (217)

It is useful to rewrite the total energy function as follows:

E(t) =12

mv(t) ·v(t)+V(x1(t),x2(t), · · · ,xn(t)). (218)

We now differentiate both sides with respect to timet:

E′(t) =12

mddt

v(t) ·v(t)+ddt

V(x1,x2(t), · · · ,xn(t)). (219)

After a little work involving the time derivatives of the dotproduct and the Chain Rule, one finds that

E′(t) = 0. (220)

This implies thatE(t) is constant along the trajectoryr(t).


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