THE CURL AND STOKES’ THEOREM

11. THE CURL AND STOKES’ THEOREM

We have already defined curl V = ∇ × V [see (7.3)] and have considered one application of the curl, namely, to determine whether or not a line integral between two points is independent of the path of integration (Section 8). Here is another application of the curl. Suppose a rigid body is rotating with constant angular velocity ω; this means that |ω| is the magnitude of the angular velocity and ω is

a vector along the axis of rotation (see Figure 2.6). Then we showed in Section

2 that the velocity v of a particle in the rigid body is v = ω × r, where r is

a radius vector from a point on the rotation axis to the particle. Let us calculate ∇× v = ∇×(ω×r); we can evaluate this by the method described in Section 7. We

use the formula for the triple vector product A × (B × C) = (A · C)B − (A · B)C, being careful to remember that ∇ is not an ordinary vector—it has both vector and differential-operator properties, and so must be written before variables that it differentiates. Then

∇× (ω × r) = (∇ · r)ω − (ω · ∇)r.

Since ω is constant, the first term of (11.1) means

In the second term of (11.1) we intentionally wrote ω · ∇ instead of ∇ · ω since ω is constant, and ∇ operates only on r; this term means

ω x +ω y

(ix + jy + kz) = iω x + jω y + kω z =ω ∂x

+ω z

∂y

∂z

Section 11 The Curl and Stokes’ Theorem 325

since ∂y/∂x = ∂z/∂x = 0, etc. Then

1 = (∇ × v).

(11.3) ∇× v = ∇ × (ω × r) = 2ω

or

This result gives a clue as to the name curl v (or rotation v or rot v as it is sometimes called). For this simple case curl v gave the angular velocity of rotation. In a more complicated case such as flow of fluid, the value of curl v at a point is a measure of the angular velocity of the fluid in the neighborhood of the point. When ∇× v = 0 everywhere in some region, the velocity field v is called irrotational in that region. Notice that this is the same mathematical condition as for a force F to be conservative.

Figure 11.1

Consider a vector field V (for example, V = vρ for flow of water, or V = force F). We define the circulation as the line integral curve. If V is a force F, then this integral is equal to the work done by the force. For flow of water, we can get a physical picture of the meaning of the circulation in the following way. Think of placing a tiny paddle-wheel probe (Figure 11.1c) in any of the flow patterns pictured in Figure 11.1. If the velocity of the fluid is greater on one side of the wheel than on the other, for example, as in (c), then the wheel will turn. Suppose we calculate the circulation wheel along a closed curve in a plane perpendicular to the axis (plane of the paper in Figure 11.1). If V = vρ is larger on one side of the wheel than the other, then the circulation is different from zero, but if [as in (b)] V is the same on both sides, then the circulation is zero. We shall show that the component of curl V along the

326 Vector Analysis Chapter 6

axis of the paddle wheel equals

dσ→0 dσ

where dσ is the area inclosed by the curve along which we calculate the circulation. The paddle wheel then acts as a “curl meter” to measure curl V; if it does not

lines are parallel. In (d), it is possible to have curl V = 0 even though the stream lines go around a corner; in fact, for the flow of water around a corner, curl V = 0. What you should realize is that the value of curl V at a point depends upon the circulation in the neighborhood of the point and not on the overall flow pattern.

We want to show the relation between the circulation point P and a direction n, let us find the component of

curl V in the direction n at P . Draw a plane through P perpendicular to n and choose axes so that it is the (x, y) plane with n parallel to k. Find the circulation around an element of area dσ centered on P . (See Figures 9.5 and 11.2.) By (9.18) with area A replaced by the element

Figure 11.2 of area dσ, and with n = k

V · dr =

(curl V) · k dx dy =

(curl V) · n dσ

around dσ

dσ

dσ

Note that, since we proved (9.7) and so (9.18) for non-rectangular areas A (see Section 9), dσ here may be more general than dx dy, say with curved or slanted sides.

We assume that the components of V have continuous first derivatives; then curl V is continuous. Thus the value of (curl V) · n over dσ is nearly the same as (curl V) · n at P , so the double integral in (11.5) is approximately the value of (curl V) · n at P multiplied by dσ. If we divide (11.5) by dσ and take the limit as dσ → 0, we have an exact equation

(11.6) (∇ × V) · n = lim

V · dr.

dσ→0 dσ

around dσ

This equation can be used as a definition of curl V; then the discussion above shows that [see equation (9.16)] the components of curl V are those given in our previous definition (7.3).

In evaluating the line integral we must go around the area element dσ as in Figure 11.2 keeping the area to our left. Another way of saying this is that we go around dσ in the direction indicated by n and the right-hand rule; that is, if the thumb of your right hand points in the direction n, your fingers curve in the direction you must go around the boundary of dσ in evaluating the line integral. (See Figure 11.2 with n = k.)

Section 11 The Curl and Stokes’ Theorem 327

Stokes’ Theorem This theorem relates an integral over an open surface to the line integral around the curve bounding the surface (Figure 11.3). A butterfly net is a good example of what we are talking about; the net is the surface and the supporting rim is the curve bounding the surface. The surfaces we consider here (and which arise in applications) will be surfaces which could be obtained by deforming a hemisphere (or the butterfly net of Figure 11.3). In particular, the surfaces we consider must be two-sided. You can easily construct a one-sided surface by taking a long strip of paper, giving it a half twist, and joining the ends (Figure 11.4). A belt of this shape is sometimes used for driving machinery. This surface is called a Moebius strip, and you can verify that it has only one side by tracing your finger around it or imagining trying to paint one side. Stokes’ theorem does not apply to such surfaces because we cannot define the sense of the normal vector n to such a surface. We require the bounding curve to be simple (that is, it must not cross itself) and closed.

Figure 11.4 Figure 11.3

Consider the kind of surface we have described and imagine it divided into area elements dσ by a network of curves as in Figure 11.5. Draw a unit vector n perpendicular to each area element; n, of course, varies from element to element, but all n’s must be on the same side of the two-sided surface. Each area element

Figure 11.5

is approximately an element of the tangent plane to the surface at a point in dσ. Then, as in (11.5), we have

around dσ

dσ

for each element. Recall from Section 9 and the comment just after equation (11.5), that dσ includes area elements such as those along the edges in Figure 11.5. Then if we sum the equations in (11.7) for all the area elements of the whole surface area,

328 Vector Analysis Chapter 6

From Figure 11.5 we see that all the interior line integrals cancel because along a border between two dσ’s the two integrals are in opposite directions. Then the left side of (11.8) becomes simply the line integral around the outside curve bounding the surface. Thus we have Stokes’ theorem as stated in (9.19):

Stokes’ theorem

curve bounding σ

surface σ

You should have it clearly in mind that this is for an open surface bounded by a simple closed curve. Recall the example of a butterfly net. Notice that Stokes’ the- orem says that the line integral over any surface of which the curve is a boundary; in other words, you don’t change the value of the integral by deforming the butterfly net! An easy way to determine the direction of integration for the line integral is to imagine collapsing the surface and its bounding curve into a plane; then the “surface” is just the plane area inside the curve and n is normal to the plane. The direction of integration is then given by the right-hand rule as discussed just after equation (11.6).

Example 1. Given V = 4yi + xj + 2zk, find (∇ × V) · n dσ over the hemisphere x 2 + y 2 +z 2 =a 2 , z ≥ 0. Using (7.3), we find that ∇ × V = −3k. There are several ways we could do the problem: (a) integrate the expression as it stands; (b) use Stokes’ theorem and evaluate

2 +y 2 =a 2 in the (x, y) plane; (c) use Stokes’ theorem to say that the integral is the same over any surface bounded by this circle,

for example, the plane area inside the circle! Since this plane area is in the (x, y) plane, we have

n = k, (∇ × V) · n = −3k · k = −3, so the integral is

2 2 −3 dσ = −3 · πa = −3πa .

This is the easiest way to do the problem; however, for this simple case it is not too hard by the other methods. We shall leave (b) for you to do and do (a). Since the surface is a sphere with center at the origin, r is normal to it (but for any surface we could get the normal from the gradient). Then on the surface

ix + jy + kz

n=

|r|

(∇ × V) · n = −3k · = −3 .

Section 11 The Curl and Stokes’ Theorem 329

We want to evaluate −3(z/a) dσ over the hemisphere. In spherical coordinates (see Chapter 5, Section 4) we have

z = r cos θ, dσ = r 2 sin θ dθ dφ.

For our surface r = a. Then the integral is 2π π/2

2 1 2 = −3a · 2π · = −3πa

(as before). Amp` ere’s Law Stokes’ theorem is of interest in electromagnetic theory. (Com-

pare the use of the divergence theorem in connection with Gauss’s law in Section 10.) Amp`ere’s circuital law (in SI units) says that

H · dr = I,

where H = B/µ 0 , B is the magnetic field, µ 0 is a constant (called the permeability of free space), C is a closed curve, and I is the current “linking” C, that is crossing any surface area bounded by C. The surface area and the curve C are related just as in Stokes’ theorem (butterfly net and its rim). If we think of a bundle of wires linking a closed curve C (Figure 11.6) and then spreading out, we can see that the same current crosses any surface whose bounding curve is C.

Just as Gauss’s law (10.23) is useful in computing electric fields, so Amp`ere’s law is useful in computing magnetic fields. Consider, for example, a long straight wire carrying a current I (Figure 11.7). At a distance r from the wire, H is tangent

Figure 11.6 Figure 11.7 to a circle of radius r in a plane perpendicular to the wire. By symmetry, |H| same

at all points of the circle. We can then find |H| by Amp`ere’s law. Taking C to be the circle of radius r, we have

H · dr =

|H|r dθ = |H|r · 2π = I

or

|H| =

2πr

330 Vector Analysis Chapter 6

If, in Figure 11.6, J is the current density (current crossing unit area perpendic- ular to J), then J · n dσ is the current across a surface element dσ [compare (10.4)] and σ J · n dσ, over any surface σ bounded by C, is the total current I linking C. Then by Amp`ere’s law

H · dr =

J · n dσ.

By Stokes’ theorem

H · dr =

(∇ × H) · n dσ,

so we have

(∇ × H) · n dσ =

J · n dσ.

Since this is true for any σ, we have ∇ × H = J, which is one of the Maxwell equa- tions. Alternatively, we could start with the Maxwell equation and apply Stokes’ theorem to get Amp`ere’s law.

Conservative Fields We next want to state carefully, and use Stokes’ the- orem to prove, under what conditions

a given field F is conservative (see Sec- tion 8). First, recall that in physical problems we are often interested only in a particular region of space, and our formulas (say for F) may very well be correct only in that region. For exam- ple, the gravitational pull of the earth

on an object is proportional to 1/r 2 for

r ≥ earths’ radius R, but this is not a correct formula for r < R (see Prob- lem 8.21). The electric field in the re-

Figure 11.8 gion between the plates of a cylindrical capacitor is proportional to 1/r (problem 10.12), but only in this region is this formula correct. We must, then, consider the kind of region in which a given field

F is defined. Consider the shaded regions in Figure 11.8. We say that a region is simply connected if any simple † closed curve in the region can be shrunk to a point without encountering any points not in the region. You can see in Figure 11.8c that the dotted curve surrounds the “hole” and so cannot be shrunk to a point in the region; this region is then not simply connected. The “hole” is sometimes only a single point, but this is enough to make the region not simply connected. In three dimensions the region between cylindrical capacitor plates (infinitely long) is not simply connected since a loop of string around the inner cylinder (see cross section, Figure 11.8c) cannot be drawn up to a knot. Similarly, the interior of an inner tube is not simply connected. The region between two concentric spheres is simply connected, however. You should see this by realizing that you could pull up into a

A simple curve does not cross itself; for example, a figure eight is not a simple curve.

Section 11 The Curl and Stokes’ Theorem 331

knot, a loop of string placed anywhere in this region. We shall now state and prove our theorem.

(11.10) If the components of F and their first partial derivatives are continuous in a

simply connected region, then any one of the following five conditions implies all the others.

(a) curl F = 0 at every point of the region. (b)

(c) F is conservative, that is B

A F · dr is independent of the path of integration from A to B. (The path must, of course, lie entirely in the region.)

(d) F · dr is an exact differential of a single valued function. (e) F = grad W , W single-valued.

We shall show that each of these conditions implies the one following it. We can use Stokes’ theorem to prove (b) assuming (a). First select any simple closed curve and let it be the bounding curve for the surface in Stokes’ theorem. Since the region is simply connected we can think of shrinking the curve to a point in the region; as it shrinks it traces out a surface which we use as the Stokes’ theorem surface. Assuming (a), we have curl F = 0 at every point of the region and so also at every point of the surface. Thus the surface integral in Stokes’ theorem is zero and therefore the line integral around the closed curve equals zero. This gives (b).

To show that (b) implies (c), consider any two paths I and II from A to B (Figure 11.9). From (b) we have

F · dr +

F · dr = 0.

Figure 11.9 Since an integral from A to B is the negative of an integral from B to A, we have

path I

path II

F · dr −

F · dr = 0.

path I

path II

which is (c). To show that (c) implies (d), select some reference point O in the region and calculate

F · dr from the reference point to every other point of the region. For each point P we find a single value of the integral no matter what path of integration we choose from O to P . Let this value be the value of the function W at the point P . We then have a single-valued function W such that

F · dr = W (P ).

0 to P

332 Vector Analysis Chapter 6

Then (since F is continuous), dW = F · dr, that is, F · dr is the differential of a single-valued function W . Since dW = ∇W · dr = F · dr for arbitrary dr, we have

F = ∇W which is (e). Finally, (e) implies (a) as we proved in Section 8. (The continuity of the com- ponents of F and their partial derivatives makes the second-order mixed partial derivatives of W equal.) Thus we have shown that any one of the five conditions (a) to (e) implies the others under the conditions of the theorem. It is worth observ- ing carefully the requirement that F and its partial derivatives must be continuous in a simply connected region. A simple example makes this clear. Look at Example

2 in Section 8; you can easily compute curl F and find that it is zero everywhere except at the origin (where it is undefined). You might then be tempted to assume that integral of dθ along a circle with center at the origin is 2π. What is wrong? The trouble is that F does not have continuous partial derivatives at the origin, and any simply connected region containing the circle of integration must contain the origin. Then curl F is not zero at every point inside the integration curve. Notice also that F · dr = dθ is an exact differential, but not of a single-valued function; θ increases by 2π every time we go around the origin.

A vector field V is called irrotational (or conservative or lamellar ) if curl V = 0; in this case V = grad W , where W (or its negative) is called the scalar potential. If div V = 0, the vector field is called solenoidal ; in this case V = curl A, where A is a vector function called the vector potential. It is easy to prove (Problem 7.17d) that if V = ∇ × A, then div V = 0. It is also possible to construct an A (actually an infinite number of A’s) so that V = curl A if we know that ∇ · V = 0.

Example 2. Given V = i(x 2 − yz) − j2yz + k(z 2 − 2zx), find A such that V = ∇ × A. We find

div V = (x 2 (z − yz) + 2 (−2yz) + − 2zx)

= 2x − 2z + 2z − 2x = 0. Thus V is solenoidal and we proceed to find A. We are looking for an A such that

V = curl A = = i(x 2 − yz) − j2yz + k(z 2 − 2zx).

∂x ∂y ∂z

There are many A’s satisfying this equation; we shall show first how to find one of them and then a general formula for all. It is possible to find an A with one zero component; let us take A x = 0. Then the y and z components of curl A each involve just one component of A. From (11.11), the y and z components of curl A are

∂A z

z 2 ∂A − 2zx = y .

(11.12) −2yz = −

∂x

∂x

Section 11 The Curl and Stokes’ Theorem 333

If we integrate (11.12) partially with respect to x (that is, with y and z constant), we find A y and A z except for possible functions of y and z which could be added without changing (11.12):

A 2 y 2 =z x − zx +f 1 (y, z),

A z = 2xyz + f 2 (y, z).

Substituting (11.13) into the x component of (11.11), we get

∂f 2 2 ∂f 1 (11.14)

∂z We now select f 1 and f 2 to satisfy (11.14). There is much leeway here and this

can easily be done by inspection. We could take f 2 = 0, f 1 = 1 2 yz 2 , or f 1 = 0,

f 2 =− 1 2 y 2 z, and so forth. Using the second choice, we have (11.15)

A = j(z 2 2 1 x − zx 2 ) + k(2xyz − y z).

You may wonder why this process works and what div V = 0 has to do with it. We can answer both these questions by following the above process with a general

V rather than a special example. Given that div V = 0, we want an A such that

V = curl A. We try to find one with A x = 0. Then the y and z components of

V = curl A are

Then we have

V y dx + g(y, z). The x component of V = curl A is

dx + h(y, z). ∂y

Since div V = 0, we can put

into (11.18), getting

∂V x

dx + h(y, z).

∂x

This is correct with proper choice of h(y, z). When we know one A, for which a given V is equal to curl A, all others are of the form

A + ∇u,

where u is any scalar function. For (see Problem 7.17b), ∇ × ∇u = 0, so the addition of ∇u to A does not affect V. Also we can show that all possible A’s are

334 Vector Analysis Chapter 6

of the form (11.20). For if V = curl A 1 and V = curl A 2 , then curl(A 1 −A 2 ) = 0, so A 1 −A 2 is the gradient of some scalar function.

A careful statement and proof that div V = 0 is a necessary and sufficient con- dition for V = curl A requires that V have continuous partial derivatives at every point of a region which is simply connected in the sense that every closed surface (rather than closed curve) can be shrunk to a point in the region (for example, the region between two concentric spheres is not simply connected in this sense).

PROBLEMS, SECTION 11

1. Do case (b) of Example 1 above.

2. Given the vector A = (x 2 −y 2 )i + 2xyj.

(a) Find ∇ × A. (b)

Evaluate RR (∇ × A) · dσ over a rectangle in the (x, y) plane bounded by the lines x = 0, x = a, y = 0, y = b.

(c) Evaluate H A · dr around the boundary of the rectangle and thus verify Stokes’ theorem for this case.

Use either Stokes’ theorem or the divergence theorem to evaluate each of the following integrals in the easiest possible way.

3. RR

2 j−y 2 k)·n dσ, where σ is the part of the surface z = 4−x −y above the (x, y) plane. 4. RR curl(yi + 2j ) · n dσ, where σ is the surface in the first

surface σ curl(x 2 i+z 2 2

octant made up of part of the plane 2x + 3y + 4z = 12, and triangles in the (x, z) and (y, z) planes, as indicated in the figure.

5. RR r · n dσ over the surface in Problem 4, where r = ix + jy + kz. Hint: See Problem 10.9.

6. RR V · n dσ over the closed surface of the tin can bounded by x 2 +y 2 = 9, z = 0, z = 5, if

V = 2xyi − y 2 j + (z + xy)k.

7. RR (curl V) · n dσ over any surface whose bounding curve is in the (x, y) plane, where V = (x − x 2 z)i + (yz 3 −y 2 )j + (x 2 y − xz)k.

8. RR curl(x 2 yi − xzk) · n dσ over the closed surface of the ellipsoid

Warning: Stokes’ theorem applies only to an open surface. Hints: Could you cut the given surface into two halves? Also see (d) in the table of vector identities (page 339).

9. RR V · n dσ over the entire surface of the volume in the first octant bounded by

x 2 +y 2 +z 2 = 16 and the coordinate planes, where

V = (x + x 2 2 −y 2 )i + (2xyz − 2xy)j − xz k. 10.

2 RR (curl V) · n dσ over the part of the surface z = 9 − x 2 − 9y above the (x, y) plane,

if V = 2xyi + (x 2 − 2x)j − x 2 z 2 k.

Section 11 The Curl and Stokes’ Theorem 335

11. RR V · n dσ over the entire surface of a cube in the first octant with edges of length 2 along the coordinate axes, where

V = (x 2 −y 2 )i + 3yj − 2xzk.

12. H V · dr around the circle (x − 2) 2 + (y − 3) 2 = 9, z = 0, where V = (x 2 + yz 2 )i + (2x − y 3 )j.

13. RR (2xi − 2yj + 5k) · n dσ over the surface of a sphere of radius 2 and center at the origin.

14. H (yi − xj + zk) · dr around the circumference of the circle of radius 2, center at the origin, in the (x, y) plane. 15. H

c y dx + z dy + x dz, where C is the curve of intersection of the surfaces whose

equations are x + y = 2 and x 2 +y 2 +z 2 = 2(x + y).

16. What is wrong with the following “proof” that there are no magnetic fields? By electromagnetic theory, ∇ · B = 0, and B = ∇ × A. (The error is not in these equations.) Using them, we find

(by the divergence theorem)

ZZ

A · dr (by Stokes’ theorem). Since R A· dr = 0, A is conservative, or A = ∇ψ. Then B = ∇×A = ∇×∇ψ = 0,

(∇ × A) · n dσ =

so B = 0. 17. Derive the following vector integral theorems.

inclosing τ surface

Hint: In the divergence theorem (10.17), substitute V = φC, where C is an arbitrary constant vector, to obtain C · R ∇φ dτ = C · H φn dσ. Since C is arbitrary, let C = i to show that the x components of the two integrals are equal; similarly, let C = j and C = k to show that the y components are equal and the z components are equal.

inclosing τ surface

Hint: Replace V in the divergence theorem by V × C, where C is an arbitrary constant vector. Follow the last part of the hint in (a).

(c) φ dr =

(n × ∇φ)dσ.

bounding σ curve

curve bounding σ

surface σ

Hints for (c) and (d): Use the substitutions suggested in (a) and (b) but in Stokes’ theorem (11.9) instead of the divergence theorem.

Hint: Integrate (7.6) over volume τ and use the divergence theorem.

336 Vector Analysis Chapter 6 Z

inclosing τ surface

Hint: Integrate (h) in the Table of Vector Identities (page 339) and use the divergence theorem.

I (g)

φ(∇ × V) · n dσ = (∇ × ∇φ) · n dσ + φ V · dr.

surface of σ

surface of σ

curve bounding σ

Hint: Integrate (g) in the Table of Vector Identities (page 339) and use Stokes’ Theorem.

Find vector fields A such that V = curl A for each given V. 18. V = (x 2 − yz + y)i + (x − 2yz)j + (z 2 − 2zx + x + y)k

2 2 19. 2 V = i(x − 2xz) + j(y − 2xy) + k(z − 2yz + xy) 20. V = i(ze zy + x sin zx) + jx cos xz − kz sin zx

21. V = −k

22. V = (y + z)i + (x − z)j + (x 2 +y 2 )k