Rot Ac Ional

S E C T I O N 18.2 Stokes’ Theorem 1021

Further Insights and ChallengesIn Exercises 40–43, the Laplace operator ! is defined by

!" = #2"

#x2 + #2"

#y2 13

For any vector field F = !F1, F2", define the conjugate vector fieldF# = !$F2, F1".40. Show that if F = %", then curlz(F#) = !".

41. Let n be the outward-pointing unit normal vector to a simple closedcurve C. The normal derivative of a function ", denoted #"

#n , is the di-rectional derivative Dn(") = %" · n. Prove that

!

C

#"

#nds =

""

D!" dA

where D is the domain enclosed by a simple closed curve C. Hint: LetF = %". Show that #"

#n = F# · T where T is the unit tangent vector,and apply Green’s Theorem.

42. Let P = (a, b) and let Cr be the circle of radius r centered at P .The average value of a continuous function " on Cr is defined as theintegral

I"(r) = 12$

" 2$

0"(a + r cos %, b + r sin %) d%

(a) Show that

#"

#n(a + r cos %, b + r sin %)

= #"

#r(a + r cos %, b + r sin %)

(b) Use differentiation under the integral sign to prove that

d

drI"(r) = 1

2$r

"

Cr

#"

#nds

(c) Use Exercise 41 to conclude that

d

drI"(r) = 1

2$r

""

D(r)!" dA

where D(r) is the interior of Cr .

43. Prove that m(r) & I"(r) & M(r), where m(r) and M(r) are theminimum and maximum values of " on Cr . Then use the continuity of" to prove that lim

r'0I"(r) = "(P ).

In Exercises 44 and 45, let D be the region bounded by a simple closedcurve C. A function "(x, y) on D (whose second-order partial deriva-tives exist and are continuous) is called harmonic if !" = 0, where!" is the Laplace operator defined in Eq. (13).

44. Use the results of Exercises 42 and 43 to prove the mean-valueproperty of harmonic functions: If " is harmonic, then I"(r) = "(P )

for all r .

45. Show that f (x, y) = x2 $ y2 is harmonic. Verify the mean-valueproperty for f (x, y) directly [expand f (a + r cos %, b + r sin %) as afunction of % and compute I"(r)]. Show that x2 + y2 is not harmonicand does not satisfy the mean-value property.

18.2 Stokes’ TheoremStokes’ Theorem is an extension of Green’s Theorem to three dimensions in which circu-lation is related to a surface integral in R3 (rather than to a double integral in the plane).In order to state it, we introduce some definitions and terminology.

Figure 1 shows three surfaces with different types of boundaries. The boundary of asurface is denoted #S. Observe that the boundary in (A) is a single, simple closed curveand the boundary in (B) consists of three closed curves. The surface in (C) is called aclosed surface because its boundary is empty. In this case, we write #S = (.

FIGURE 1 Surfaces and their boundaries.

Recall from Section 17.5 that an orientation is a continuously varying choice ofunit normal vector at each point of a surface S . When S is oriented, we can specify anorientation of #S, called the boundary orientation. Imagine that you are a unit normal

1022 C H A P T E R 18 FUNDAMENTAL THEOREMS OF VECTOR ANALYSIS

(A) (B)

C1C1

n

nC2

S S

C2

FIGURE 2 The orientation of the boundary!S for each of the two possibleorientations of the surface S.

vector walking along the boundary curve. The boundary orientation is the direction forwhich the surface is on your left as you walk. For example, the boundary of the surface inFigure 2 consists of two curves, C1 and C2. In (A), the normal vector points to the outside.The woman (representing the normal vector) is walking along C1 and has the surface toher left, so she is walking in the positive direction. The curve C2 is oriented in the oppositedirection because she would have to walk along C2 in that direction to keep the surface onher left. The boundary orientations in (B) are reversed because the opposite normal hasbeen selected to orient the surface.

All that’s left is to define curl. The curl of the vector field F = !F1, F2, F3" is thevector field defined by the symbolic determinant

curl(F) =

!!!!!!!!!!

i j k

!

!x

!

!y

!

!z

F1 F2 F3

!!!!!!!!!!

="

!F3

!y# !F2

!z

#i #

"!F3

!x# !F1

!z

#j +

"!F2

!x# !F1

!y

#k

In more compact form, the curl is the symbolic cross product

curl(F) = $ % F

where $ is the del “operator” (also called “nabla”):

$ =$

!

!x,

!

!y,

!

!z

%

In terms of components, curl(F) is the vector field

curl(F) =$!F3

!y# !F2

!z,!F1

!z# !F3

!x,!F2

!x# !F1

!y

%

It is straightforward to check that curl obeys the linearity rules:

curl(F + G) = curl(F) + curl(G)

curl(cF) = c curl(F) (c any constant)


EXAMPLE 1 Calculating the Curl Calculate the curl of F =!xy, ex, y + z

".

Solution We compute the curl as a symbolic determinant:

curl(F) =

#########

i j k

!

!x

!

!y

!

!z

xy ex y + z

#########

=$

!

!y(y + z) ! !

!zex

%i !

$!

!x(y + z) ! !

!zxy

%j +

$!

!xex ! !

!yxy

%k

= i + (ex ! x)k

EXAMPLE 2 Conservative Vector Fields Have Zero Curl Verify:

If F = "V , then curl(F) = 0. That is, curl("V ) = 0. 1

Solution The curl of a vector field is zero if

!F3

!y! !F2

!z= 0,

!F1

!z! !F3

!x= 0,

!F2

!x! !F1

!y= 0

But these equations are equivalent to the cross-partials condition that is satisfied by everyconservative vector field F = "V .

In the next theorem, we assume that S is an oriented surface with parametrizationG : D # S, where D is a domain in the plane bounded by smooth, simple closed curves,and G is one-to-one and regular, except possibly on the boundary of D. More generally,S may be a finite union of surfaces of this type. The surfaces in applications we consider,such as spheres, cubes, and graphs of functions, satisfy these conditions.

THEOREM 1 Stokes’ Theorem For surfaces S as described above,

&

!SF · ds =

''

Scurl(F) · dS 2

The integral on the left is defined relative to the boundary orientation of !S. If S isclosed (that is, !S is empty), then the surface integral on the right is zero.

Proof Each side of Eq. (2) is equal to a sum over the components of F:

The curl measures the extent to which Ffails to be conservative. If F isconservative, then curl(F) = 0 and Stokes’Theorem merely confirms what we alreadyknow: The circulation of a conservativevector field around a closed path is zero.

&

CF · ds =

&

CF1 dx + F2 dy + F3 dz

''

Scurl(F) · dS =

''

Scurl(F1i) · dS +

''

Scurl(F2j) · dS +

''

Scurl(F3k) · dS

The proof consists of showing that the F1-, F2-, and F3-terms are separately equal.Because a complete proof is quite technical, we will prove it under the simplifying

assumption that S is the graph of a function z = f (x, y) lying over a domain D in thexy-plane. Furthermore, we will carry the details only for the F1-terms. The calculation


for F2-components is similar, and we leave as an exercise the equality of the F3-terms(Exercise 31). Thus, we shall prove that

!

CF1 dx =

""

Scurl(F1(x, y, z)i) · dS 3

Orient S with upward-pointing normal as in Figure 3 and let C = !S be the boundarycurve. Let C0 be the boundary of D in the xy-plane, and let c0(t) = (x(t), y(t)) (for

z

yx

C0

(x, y)

(x, y, f (x, y))

D

S

C

n

FIGURE 3

a ! t ! b) be a counterclockwise parametrization of C0 as in Figure 3. The boundarycurve C projects onto C0, so C has parametrization

c(t) =#x(t), y(t), f (x(t), y(t))

$

and thus!

CF1(x, y, z) dx =

" b

aF1

#x(t), y(t), f (x(t), y(t))

$dx

dtdt

The integral on the right is precisely the integral we obtain by integratingF1

#x, y, f (x, y)

$dx over the curve C0 in the plane R2. In other words,

!

CF1(x, y, z) dx =

"

C0

F1#x, y, f (x, y)

$dx

By Green’s Theorem applied to the integral on the right,!

CF1(x, y, z) dx = "

""

D

!

!yF1(x, y, f (x, y)) dA

By the Chain Rule,

!

!yF1

#x, y, f (x, y)

$= F1y

#x, y, f (x, y)

$+ F1z

#x, y, f (x, y)

$fy(x, y)

so finally we obtain!

CF1 dx = "

""

D

%F1y

#x, y, f (x, y)

$+ F1z

#x, y, f (x, y)

$fy(x, y)

&dA 4

To finish the proof, we compute the surface integral of curl(F1i) using the parametriza-REMINDER Calculating a surfaceintegral:""

SF · dS =

""

DF(u, v) · n(u, v) du dv

If S is a graph z = f (x, y), parametrizedby G(x, y) = (x, y, f (x, y)), then

n(x, y) ='"fx(x, y),"fy(x, y), 1

(

tion G(x, y) = (x, y, f (x, y)) of S:

n ='"fx(x, y),"fy(x, y), 1

((upward-pointing normal)

curl(F1i) · n ='0, F1z, "F1y

(·'"fx(x, y),"fy(x, y), 1

(

= "F1z

#x, y, f (x, y)

$fy(x, y) " F1y

#x, y, f (x, y)

$

""

Scurl(F1i) · dS = "

""

D

%F1z(x, y, z)fy(x, y) + F1y

#x, y, f (x, y)

$&dA 5

The right-hand sides of Eq. (4) and Eq. (5) are equal. This proves Eq. (3).

EXAMPLE 3 Verifying Stokes’ Theorem Verify Stokes’ Theorem for

F = #"y, 2x, x + z$

and the upper hemisphere with outward-pointing normal vectors (Figure 4):

S = {(x, y, z) : x2 + y2 + z2 = 1, z % 0}


Solution We will show that both the line integral and the surface integral in Stokes’

FIGURE 4 Upper hemisphere with orientedboundary.

Theorem are equal to 3! .

Step 1. Compute the line integral around the boundary curve.The boundary of S is the unit circle oriented in the counterclockwise direction with

REMINDER In Eq. (6), we use! 2!

0cos2 t dt =

! 2!

0

1 + cos 2t

2dt = !

parametrization c(t) = (cos t, sin t, 0). Thus,

c!(t) = "# sin t, cos t, 0$F(c(t)) = "# sin t, 2 cos t, cos t$

F(c(t)) · c!(t) = "# sin t, 2 cos t, cos t$ · "# sin t, cos t, 0$

= sin2 t + 2 cos2 t = 1 + cos2 t

"

"SF · ds =

! 2!

0(1 + cos2 t) dt = 2! + ! = 3! 6

Step 2. Compute the curl.

curl(F) =

#########

i j k

"

"x

"

"y

"

"z

#y 2x x + z

#########

=$

"

"y(x + z) # "

"z2x

%i #

$"

"x(x + z) # "

"z(#y)

%j

+$

"

"x2x # "

"y(#y)

%k

= "0, #1, 3$

Step 3. Compute the surface integral of the curl.We parametrize the hemisphere using spherical coordinates:REMINDER Stokes’ Theorem states

"

"SF · ds =

!!

Scurl(F) · dS G(#, $) = (cos # sin $, sin # sin $, cos $)

By Eq. (2) of Section 17.4, the outward-pointing normal vector is

n = sin $ "cos # sin $, sin # sin $, cos $$

Therefore,

curl(F) · n = sin $ "0, #1, 3$ · "cos # sin $, sin # sin $, cos $$

= # sin # sin2 $ + 3 cos $ sin $

The upper hemisphere S corresponds to 0 % $ % !2 , so

!!

Scurl(F) · dS =

! !/2

$=0

! 2!

#=0(# sin # sin2 $ + 3 cos $ sin $) d# d$

= 0 + 2!

! !/2

$=03 cos $ sin $ d$ = 2!

$32

sin2 $

% ####!/2

$=0

= 3!


Notice that curl(F) contains the partial derivatives!F1

!yand

!F1

!zbut not the partial

!F1

!x. So if F1 = F1(x) is a function of x alone, then

!F1

!y= !F1

!z= 0, and F1 does not

contribute to the curl. The same holds for the other components. In summary, if each ofF1, F2, and F3 depends only on its corresponding variable x, y, or z, then

curl!"F1(x), F2(y), F3(z)

#$= 0

EXAMPLE 4 Use Stokes’ Theorem to show that%

CF · ds = 0, where

F ="sin(x2), ey2 + x2, z4 + 2x2#

and C is the boundary of the triangle in Figure 5 with the indicated orientation.

FIGURE 5

Solution We apply Stokes’ Theorem

%

CF · ds =

&&

Scurl(F) · dS

and show that the integral on the right is zero.By the preceding remark, the first component sin(x2) does not contribute to the curl

since it depends only on x. Similarly, ey2and z4 drop out of the curl, and we have

curl'"

sin x2, ey2 + x2, z4 + 2x2#(

=

Automatically zero) *+ ,curl

'"sin x2, ey2

, z4#(

+ curl'"

0, x2, 2x2#(

=-0, ! !

!x2x2,

!

!xx2

.= "0, !4x, 2x#

Now, it turns out (by the author’s design) that we can show the surface integral iszero without actually computing it. Referring to Figure 5, we see that C is the boundaryof the triangular surface S contained in the plane

x

3+ y

2+ z = 1

Therefore, u =" 1

3 , 12 , 1

#is a normal vector to this plane. But u and curl(F) are orthogonal:

curl(F) · u = "0, !4x, 2x# ·-

13,

12, 1

.= !2x + 2x = 0

In other words, the normal component of curl(F) along S is zero. Since the surface integralof a vector field is equal to the surface integral of the normal component, we conclude

that&&

Scurl(F) · dS = 0.


CONCEPTUAL INSIGHT Recall that if F is conservative—that is, F = !V —then for anytwo paths C1 and C2 from P to Q (Figure 6),

!

C1

F · ds =!

C2

F · ds = V (Q) " V (P )

In other words, the line integral is path independent. In particular,"

CF · ds is zero if C

is closed (P = Q).Analogous facts are true for surface integrals when F = curl(A). The vector field

A is called a vector potential for F. Stokes’ Theorem tells us that for any two surfacesS1 and S2 with the same oriented boundary C (Figure 7),

!!

S1

F · dS =!!

S2

F · dS ="

CA · ds

In other words, the surface integral of a vector field with vector potential A is surfaceindependent, just as a vector field with a potential function V is path independent.

If the surface is closed, then the boundary curve is empty and the surface integralis zero:

!!

SF · dS = 0 if F = curl(A) and S is closed

C1

C2

! P

+ Q

FIGURE 6 Two paths with the sameboundary Q " P .

C

S1

S2

FIGURE 7 Surfaces S1 and S2 have thesame oriented boundary.

Vector potentials are not unique: IfF = curl(A), then F = curl(A + B) forany vector field B such that curl(B) = 0.

THEOREM 2 Surface Independence for Curl Vector FieldsIf F = curl(A), then the flux of F through a surface S depends only on the orientedboundary !S and not on the surface itself:

!!

SF · dS =

"

!SA · ds 7

In particular, if S is closed (that is, !S is empty), then!!

SF · dS = 0.

EXAMPLE 5 Let F = curl(A), where A =#y + z, sin(xy), exyz

$. Find the flux of FREMINDER By the flux of a vector field

through a surface, we mean the surfaceintegral of the vector field.

through the surfaces S1 and S2 in Figure 8 whose common boundary C is the unit circlein the xz-plane.

Solution With C oriented in the direction of the arrow, S1 lies to the left, and by Eq. (7),

FIGURE 8

!!

S1

F · dS ="

CA · ds

We shall compute the line integral on the right. The parametrization c(t) = (cos t, 0, sin t)

traces C in the direction of the arrow because it begins at c(0) = (1, 0, 0) and moves inthe direction of c

%"2

&= (0, 0, 1). We have

A(c(t)) =#0 + sin t, sin(0), e0$ = #sin t, 0, 1$

A(c(t)) · c%(t) = #sin t, 0, 1$ · #" sin t, 0, cos t$ = " sin2 t + cos t

"

CA · ds =

! 2"

0(" sin2 t + cos t) dt = ""


We conclude that!!

S1

F · dS = !! . On the other hand, S2 lies on the right as you traverse

C. Therefore S2 has oriented boundary !C, and!!

S2

F · dS ="

!CA · ds = !

"

CA · ds = !

CONCEPTUAL INSIGHT Interpretation of the Curl In Section 18.1, we showed that the

quantity"F2

"x! "F1

"yin Green’s Theorem is the “circulation per unit of enclosed area.”

A similar interpretation is valid in R3.Consider a plane through a point P with unit normal vector en and let D be a small

domain containing P with boundary curve C (Figure 9). By Stokes’ Theorem,"

CF · ds "

!!

D(curl(F) · en) dS 8

The vector field curl(F) is continuous (its components are derivatives of the componentsof F), so its value does not change much on D if D is sufficiently small. To a firstapproximation, we can replace curl(F) by the constant value curl(F)(P ), giving us theapproximation

!!

D(curl(F) · en) dS "

!!

D(curl(F)(P ) · en) dS

9" (curl(F)(P ) · en)Area(D)

Furthermore, curl(F)(P ) · en = #curl(F)(P )# cos # , where # is the angle betweencurl(F) and en. Together, Eq. (8) and Eq. (9) give us

"

CF · ds " #curl(F)(P )#(cos #)Area(D) 10

This is a remarkable result. It tells us that curl(F) encodes the the circulation per unitof enclosed area in every plane through P in a simple way—namely, as the dot productcurl(F)(P ) · en. In particular, the circulation rate varies (to a first-order approximation)as the cosine of the angle # between curl(F)(P ) and en.

We can also argue (as in Section 18.1 for vector fields in the plane) that if F is thevelocity field of a fluid, then a small paddle wheel with normal en will rotate with anangular velocity of approximately 1

2 curl(F)(P ) · en (see Figure 10).

yC

z

x

Dcurl(F )

P

en

FIGURE 9 The curve C around P lies in theplane through P with normal vector en.

C

r

curl(F )

P

encurl(F )

en

FIGURE 10 The paddle wheel can beoriented in different ways, as specified bythe normal vector en.

EXAMPLE 6 Vector Potential for a Solenoid An electric current flowing through asolenoid (a tightly wound spiral of wire; see Figure 11) creates a magnetic field B. If weassume that the solenoid is infinitely long, with radius R and the z-axis as central axis,then

B =

#$

%0 if r > R

Bk if r < R


x

y

I

z

B

I

R

x

y

A

S

z

r

B

FIGURE 11 The magnetic field of a longsolenoid is nearly uniform inside and weakoutside. In practice, we treat the solenoidas “infinitely long” if it is very long incomparison with its radius.

where r = (x2 + y2)1/2 and B is a constant that depends on the current strength and thespacing of the turns of wire.

(a) Show that a vector potential for B is

A =

!""#

""$

12R2B

%! y

r2 ,x

r2 , 0&

if r > R

12B "!y, x, 0# if r < R

(b) Calculate the flux of B through the surface S (with upward-pointing normal) in Fig-ure 11 whose boundary is a circle of radius r where r > R.

Solution

(a) For any functions f and g,

curl("f, g, 0#) ='!gz, fz, gx ! fy

(

Applying this to A for r < R, we obtain

curl(A) = 12B

)0, 0,

!

!xx ! !

!y(!y)

*= "0, 0, B# = Bk = B

We leave it as an exercise [Exercise 29] to show that curl(A) = B = 0 for r > R.(b) The boundary circle of S with counterclockwise parametrization c(t) =(r cos t, r sin t, 0), so

The vector potential A is continuous butnot differentiable on the cylinder r = R,that is, on the solenoid itself (Figure 12).The magnetic field B = curl(A) has ajump discontinuity where r = R. We takefor granted the fact that Stokes’ Theoremremains valid in this setting.

rR

BR

!A!

12

FIGURE 12 The magnitude $A$ of thevector potential as a function of distance r

to the z-axis.

c%(t) = "!r sin t, r cos t, 0#

A(c(t)) = 12R2Br!1 "! sin t, cos t, 0#

A(c(t)) · c%(t) = 12R2B

+(! sin t)2 + cos2 t

,= 1

2R2B

By Stokes’ Theorem, the flux of B through S is equal to

--

SB · dS =

.

!SA · ds =

- 2"

0A(c(t)) · c%(t) dt = 1

2R2B

- 2"

0dt = "R2B


CONCEPTUAL INSIGHT There is an interesting difference between scalar and vector po-tentials. If F = !V , then the scalar potential V is constant in regions where the fieldF is zero (since a function with zero gradient is constant). This is not true for vectorpotentials. As we saw in Example 6, the magnetic field B produced by a solenoid is zeroeverywhere outside the solenoid, but the vector potential A is not constant outside thesolenoid. In fact, A is proportional to

!" y

r2 ,x

r2 , 0". This is related to an intriguing phe-

nomenon in physics called the Aharonov-Bohm (AB) effect, first proposed on theoreticalgrounds in the 1940s.

According to electromagnetic theory, a magnetic field B exerts a force on a movingelectron, causing a deflection in the electron’s path. We do not expect any deflection whenan electron moves past a solenoid because B is zero outside the solenoid (in practice,the field is not actually zero, but it is very small—we ignore this difficulty). However,according to quantum mechanics, electrons have both particle and wave properties. Ina double-slit experiment, a stream of electrons passing through two small slits creates awavelike interference pattern on a detection screen (Figure 13). The AB effect predictsthat if we place a small solenoid between the slits as in the figure (the solenoid is sosmall that the electrons never pass through it), then the interference pattern will shiftslightly. It is as if the electrons are “aware” of the magnetic field inside the solenoid,even though they never encounter the field directly.

The AB effect was hotly debated until it was confirmed definitively in 1985, inexperiments carried out by a team of Japanese physicists led by Akira Tonomura. TheAB effect appeared to contradict “classical” electromagnetic theory, according to whichthe trajectory of an electron is determined by B alone. There is no such contradiction inquantum mechanics, because the behavior of the electrons is governed not by B but bya “wave function” derived from the nonconstant vector potential A.

Solenoid

Electron stream

B

Detection screen

FIGURE 13 A stream of electrons passingthrough a double slit produces aninterference pattern on the detectionscreen. The pattern shifts slightly when anelectric current flows through the solenoid.

18.2 SUMMARY

• The boundary of a surface S is denoted !S. We say that S is closed if !S is empty.• Suppose that S is oriented (a continuously varying unit normal is specified at eachpoint of S). The boundary orientation of !S is defined as follows: If you walk along theboundary in the positive direction with your head pointing in the normal direction, thenthe surface is on your left.•

curl(F) =

#########

i j k

!

!x

!

!y

!

!z

F1 F2 F3

#########

=$

!F3

!y" !F2

!z

%i "

$!F3

!x" !F1

!z

%j +

$!F2

!x" !F1

!y

%k

Symbolically, curl(F) = ! # F where ! is the del operator

! =&

!

!x,

!

!y,

!

!z

'

• Stokes’ Theorem relates the circulation around the boundary to the surface integral ofthe curl:


!

!SF · ds =

""

Scurl(F) · dS

• If F = !V , then curl(F) = 0.• Surface Independence: If F = curl(A), then the flux of F through a surface S dependsonly on the oriented boundary !S and not on the surface itself:

""

SF · dS =

!

!SA · ds

In particular, if S is closed (that is, !S is empty) and F = curl(A), then""

SF · dS = 0.

• The curl is interpreted as a vector that encodes circulation per unit area: If P is any pointand en is a unit normal vector, then

"

CF · ds " (curl(F)(P ) · en)Area(D)

where C is a small, simple closed curve around P in the plane through P with normalvector en, and D is the enclosed region.

18.2 EXERCISES

Preliminary Questions1. Indicate with an arrow the boundary orientation of the boundary

curves of the surfaces in Figure 14, oriented by the outward-pointingnormal vectors.

FIGURE 14

2. Let F = curl(A). Which of the following are related by Stokes’Theorem?(a) The circulation of A and flux of F.(b) The circulation of F and flux of A.

3. What is the definition of a vector potential?

4. Which of the following statements is correct?(a) The flux of curl(A) through every oriented surface is zero.(b) The flux of curl(A) through every closed, oriented surface is zero.

5. Which condition on F guarantees that the flux through S1 is equalto the flux through S2 for any two oriented surfaces S1 and S2 with thesame oriented boundary?

ExercisesIn Exercises 1–4, calculate curl(F).

1. F =#z # y2, x + z3, y + x2$

2. F =%y

x,y

z,

z

x

&3. F =

#ey, sin x, cos x

$

4. F =%

x

x2 + y2 ,y

x2 + y2 , 0&

In Exercises 5–8, verify Stokes’ Theorem for the given vector field andsurface, oriented with an upward-pointing normal.

5. F = $2xy, x, y + z%, the surface z = 1 # x2 # y2 forx2 + y2 & 1

6. F = $yz, 0, x%, the portion of the planex

2+ y

3+ z = 1 where

x, y, z ' 0

7. F =#ey#z, 0, 0

$, the square with vertices (1, 0, 1), (1, 1, 1),

(0, 1, 1), and (0, 0, 1)

8. F ='y, x, x2 + y2

(, the upper hemisphere

x2 + y2 + z2 = 1, z ' 0

In Exercises 9 and 10, calculate curl(F) and then use Stokes’Theorem tocompute the flux of curl(F) through the given surface as a line integral.

9. F ='ez2 # y, ez3 + x, cos(xz)

(, the upper hemisphere

x2 + y2 + z2 = 1, z ' 0 with outward-pointing normal

10. F ='x + y, z2 # 4, x

)y2 + 1)

(, surface of the wedge-shaped

box in Figure 15 (bottom included, top excluded) with outward-point-ing normal


y

x + y = 1

z

2

x

(0, 1, 2)

(1, 0, 2)

1

1

FIGURE 15

11. Let S be the surface of the cylinder (not including the top and bot-tom) of radius 2 for 1 ! z ! 6, oriented with outward-pointing normal(Figure 16).(a) Indicate with an arrow the orientation of !S (the top and bottomcircles).(b) Verify Stokes’ Theorem for S and F =

!yz2, 0, 0

".

FIGURE 16

R

!R

FIGURE 17

12. Let S be the portion of the plane z = x contained in the half-cylinder of radius R depicted in Figure 17. Use Stokes’ Theorem tocalculate the circulation of F = "z, x, y + 2z# around the boundary ofS (a half-ellipse) in the counterclockwise direction when viewed fromabove. Hint: Show that curl(F) is orthogonal to the normal vector tothe plane.

13. Let I be the flux of F =!ey, 2xex2

, z2"through the upper hemi-

sphere S of the unit sphere.

(a) Let G =!ey, 2xex2

, 0". Find a vector field A such that

curl(A) = G.(b) Use Stokes’ Theorem to show that the flux of G through S is zero.Hint: Calculate the circulation of A around !S .(c) Calculate I . Hint: Use (b) to show that I is equal to the flux of!0, 0, z2"

through S.

14. Let F = "0, $z, 1#. Let S be the spherical cap x2 + y2 + z2 ! 1,

where z % 12 . Evaluate

##

SF · dS directly as a surface integral. Then

verify that F = curl(A), where A = (0, x, xz) and evaluate the surfaceintegral again using Stokes’ Theorem.

15. Let A be the vector potential and B the magnetic field of the infinitesolenoid of radius R in Example 6. Use Stokes’ Theorem to compute:

(a) The flux of B through a circle in the xy-plane of radius r < R

(b) The circulation of A around the boundary C of a surface lyingoutside the solenoid

16. The magnetic field B due to a small current loop (which weplace at the origin) is called a magnetic dipole (Figure 18). Let" = (x2 + y2 + z2)1/2. For " large, B = curl(A), where

A =$$ y

"3 ,x

"3 , 0%

(a) Let C be a horizontal circle of radius R with center (0, 0, c), wherec is large. Show that A is tangent to C.(b) Use Stokes’ Theorem to calculate the flux of B through C.

Rc

Current loop

A

z

y

x

FIGURE 18

17. Auniform magnetic field B has constant strengthb in the z-direction[that is, B = "0, 0, b#].(a) Verify that A = 1

2 B & r is a vector potential for B, where r ="x, y, 0#.(b) Calculate the flux of B through the rectangle with vertices A, B,C, and D in Figure 19.

18. Let F =!$x2y, x, 0

". Referring to Figure 19, let C be the closed

path ABCD. Use Stokes’ Theorem to evaluate#

CF · ds in two ways.

First, regard C as the boundary of the rectangle with vertices A, B, C,and D. Then treat C as the boundary of the wedge-shaped box withopen top.

FIGURE 19

19. Let F =!y2, 2z + x, 2y2"

. Use Stokes’ Theorem to find a planewith equation ax + by + cz = 0 (where a, b, c are not all zero) such

that&

CF · ds = 0 for every closed C lying in the plane. Hint: Choose

a, b, c so that curl(F) lies in the plane.

20. Let F =!$z2, 2zx, 4y $ x2"

and let C be a simple closed curve inthe plane x + y + z = 4 that encloses a region of area 16 (Figure 20).


Calculate!

CF · ds, where C is oriented in the counterclockwise direc-

tion (when viewed from above the plane).

FIGURE 20

21. Let F ="y2, x2, z2#

. Show that$

C1

F · ds =$

C2

F · ds

for any two closed curves lying on a cylinder whose central axis is thez-axis (Figure 21).

FIGURE 21

22. The curl of a vector field F at the origin is v0 = !3, 1, 4". Estimatethe circulation around the small parallelogram spanned by the vectorsA =

"0, 1

2 , 12#

and B ="0, 0, 1

3#.

23. You know two things about a vector field F:(i) F has a vector potential A (but A is unknown).

(ii) The circulation of A around the unit circle (oriented counterclock-wise) is 25.Determine the flux of F through the surface S in Figure 22, orientedwith upward pointing normal.

S

y

x

z

1

Unit circle

FIGURE 22 Surface S whose boundary is the unit circle.

24. Suppose that F has a vector potential and that F(x, y, 0) = k. Findthe flux of F through the surface S in Figure 22, oriented with upwardpointing normal.

25. Prove that curl(f a) = #f $ a, where f is a differentiable func-tion and a is a constant vector.

26. Show that curl(F) = 0 if F is radial, meaning that F =f (!) !x, y, z" for some function f (!), where ! =

%x2 + y2 + z2.

Hint: It is enough to show that one component of curl(F) is zero, be-cause it will then follow for the other two components by symmetry.

27. Prove the following Product Rule:

curl(f F) = f curl(F) + #f $ F

28. Assume that f and g have continuous partial derivatives of order 2.Prove that

!

"Sf #(g) · ds =

$$

S#(f ) $ #(g) · ds

29. Verify that B = curl(A) for r > R in the setting of Example 6.

30. Explain carefully why Green’s Theorem is a special caseof Stokes’ Theorem.

Further Insights and Challenges31. In this exercise, we use the notation of the proof of Theorem 1 andprove

!

CF3(x, y, z)k · ds =

$$

Scurl(F3(x, y, z)k) · dS 11

In particular, S is the graph of z = f (x, y) over a domain D, and C isthe boundary of S with parametrization (x(t), y(t), f (x(t), y(t))).(a) Use the Chain Rule to show that

F3(x, y, z)k · ds = F3(x(t), y(t), f (x(t), y(t))&fx(x(t), y(t))x%(t) + fy(x(t), y(t))y%(t)

'dt

and verify that

!

CF3(x, y, z)k · ds =

!

C0

"F3(x, y, z)fx(x, y), F3(x, y, z)fy(x, y)

#· ds

where C0 has parametrization (x(t), y(t)).

(b) Apply Green’s Theorem to the line integral over C0 and show thatthe result is equal to the right-hand side of Eq. (11).

32. Let F be a continuously differentiable vector field in R3, Q a point,and S a plane containing Q with unit normal vector e. Let Cr be a circleof radius r centered at Q in S, and let Sr be the disk enclosed by Cr .Assume Sr is oriented with unit normal vector e.

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