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CLP-4 Vector Calculus

Section 4.4 Stokes’ Theorem

Our last variant of the fundamental theorem of calculus is Stokes’
 1 
Sir George Gabriel Stokes (1819–1903) was an Irish physicist and mathematician. In addition to Stokes’ theorem, he is known for the Navier-Stokes equations of fluid dynamics and for his work on the wave theory of light. He gave evidence to the Royal Commission on the Use of Iron in Railway Structures after the Dee bridge disaster of 1847.
theorem, which is like Green’s theorem, but in three dimensions. It relates an integral over a finite surface in R3 with an integral over the curve bounding the surface.
Note that
  • in Stokes’ theorem, S must be an oriented surface. In particular, S may not be a Möbius strip. (See Example 3.5.3.)
  • If S is part of the xy-plane, then Stokes’ theorem reduces to Green’s theorem. Our proof of Stokes’ theorem will consist of rewriting the integrals so as to allow an application of Green’s theorem.
  • If S is a simple closed curve and
    • when you look at S from high on the z-axis, it is oriented counterclockwise (look at the figure in Theorem 4.4.1), then
    • n^ is upward pointing, i.e. has positive z-component, at least near S.

Proof.

Write F=F1ıı^+F2ȷȷ^+F3k^. Both integrals involve F1 terms and F2 terms and F3 terms. We shall show that the F1 terms in the two integrals agree. In other words, we shall assume that F=F1ıı^. The proofs that the F2 and F3 terms also agree are similar. For simplicity, we’ll assume
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Otherwise, decompose S into simpler pieces, analogously to what we did in the proof of the divergence theorem.
that the boundary of S consists of just a single curve, and that we can
  • pick a parametrization of S with
    S={ r(u,v)=(x(u,v),y(u,v),z(u,v)) | (u,v) in RR2 }
    and with r(u,v) orientation preserving in the sense that n^dS=+ru×rvdudv. Also
  • pick a parametrization of the curve, R, bounding R as (u(t),v(t)), atb, in such a way that when you walk along R in the direction of increasing t, then R is on your left.
Then the curve S bounding S can be parametrized as R(t)=r(u(t),v(t)), atb.
The orientation of R(t):
We’ll now verify that the direction of increasing t for the parametrization R(t) of S is the direction of the arrow on S in the figure on the left above.
By continuity, it suffices to check the orientation at a single point.
Find a point (u0,v0) on R where the forward pointing tangent vector is a positive multiple of ıı^. The horizontal arrow on R in the figure on the left below is at such a point. Suppose that t=t0 at this point — in other words, suppose that (u0,v0)=(u(t0),v(t0)). Because the forward pointing tangent vector to R at (u0,v0), namely (u(t0),v(t0)), is a positive multiple of ıı^, we have u(t0)>0 and v(t0)=0. The tangent vector to S at R(t0)=r(u0,v0), pointing in the direction of increasing t, is
R(t0)=ddtr(u(t),v(t))|t=t0=u(t0)ru(u0,v0)+v(t0)rv(u0,v0)=u(t0)ru(u0,v0)
and so is a positive multiple of ru(u0,v0). See the figure on the right below.
If we now walk along a path in the uv-plane which starts at (u0,v0), holds u fixed at u0 and increases v, we move into the interior of R starting at (u0,v0). Correspondingly, if we walk along the path, r(u0,v), in R3 with v starting at v0 and increasing, we move into the interior of S. The forward tangent to this new path, rv(u0,v0), points from r(u0,v0) into the interior of S. It’s the blue arrow in the figure on the right below.
Now imagine that you are walking along S in the direction of increasing t. At time t0 you are at R(t0). You point your right arm straight ahead of you. So it is pointing in the direction ru(u0,v0). You point your left arm out sideways into the interior of S. It is pointing in the direction rv(u0,v0). If the direction of increasing t is the same as the forward direction of the orientation of S, then the vector from our feet to our head, which is ru(u0,v0)×rv(u0,v0), should be pointing in the same direction as n^. And since n^dS=+ru×rvdudv, it is.
Now, with our parametrization and orientation sorted out, we can examine the integrals.
The surface integral:
Since F=F1ıı^, so that
×F=det[ıı^ȷȷ^k^xyzF100]=(0,F1z,F1y)
and
n^dS=ru×rv dudv=det[ıı^ȷȷ^k^xuyuzuxvyvzv]=(yuzvzuyv)ıı^+(zuxvxuzv)ȷȷ^+(xuyvyuxv)k^
and
S×Fn^dS=R(0,F1z,F1y)ru×rv dudv=R{F1z(zuxvxuzv)F1y(xuyvyuxv)}dudv
Now we examine the line integral and show that it equals this one.
The line integral:
SFdr=abF(r(u(t),v(t)))ddtr(u(t),v(t)) dt=abF(r(u(t),v(t)))[ru(u(t),v(t))dudt(t)+rv(u(t),v(t))dvdt(t)] dt
We can write this as the line integral
RM(u,v) du+N(u,v) dv=ab[M(u(t),v(t)) dudt(t)+N(u(t),v(t)) dvdt(t)] dt
around R, if we choose
M(u,v)=F(r(u,v))ru(u,v)=F1(x(u,v),y(u,v),z(u,v))xu(u,v)N(u,v)=F(r(u,v))rv(u,v)=F1(x(u,v),y(u,v),z(u,v))xv(u,v)
Finally, we show that the surface integral equals the line integral:
By Green’s Theorem, we have
SFdr=RM(u,v) du+N(u,v) dv=R{NuMv} dudv=R{u[F1(x(u,v),y(u,v),z(u,v))]xv+F12xuvv[F1(x(u,v),y(u,v),z(u,v))]xuF12xvu} dudv=R{(F1xxu+F1yyu+F1zzu)xv+F12xuv(F1xxv+F1yyv+F1zzv)xuF12xvu} dudv=R{(F1yyu+F1zzu)xv(F1yyv+F1zzv)xu}dudv=S×Fn^dS
which is the conclusion that we wanted.
Before we move on to some examples, here are a couple of remarks.
  • Stokes’ theorem says that CFdr=S×Fn^ dS for any (suitably oriented) surface whose boundary is C. So if S1 and S2 are two different (suitably oriented) surfaces having the same boundary curve C, then
    S1×Fn^ dS=S2×Fn^ dS
    For example, if C is the unit circle
    C={ (x,y,z) | x2+y2=1, z=0 }
    oriented counterclockwise when viewed from above, then both
    S1={ (x,y,z) | x2+y21, z=0 }S2={ (x,y,z) | z0, x2+y2+z2=1 }
    with upward pointing unit normal vectors, have boundary C. So Stokes’ tells us that S1×Fn^ dS=S2×Fn^ dS.
    It should not be a surprise that S1×Fn^ dS=S2×Fn^ dS, for the following reason. Let
    V={ (x,y,z) | x2+y2+z21, z0 }
    be the solid between S1 and S2. The boundary V of V is the union of S1 and S2.
    But beware that the outward pointing normal to V (call it N^) is +n^ on S2 and n^ on S1. So the divergence theorem gives
    S2×Fn^ dSS1×Fn^ dS=S2×FN^ dS+S1×FN^ dS=V×FN^ dS=V(×F) dVby the divergence theorem=0
    by the vector identity Theorem 4.1.7.a.
  • As a second remark, suppose that the vector field F obeys ×F=0 everywhere. Then Stokes’ theorem forces CFdr=0 are around all closed curves C, which implies that F is conservative, by Theorem 2.4.7. So Stokes’ theorem provides another proof of Theorem 2.4.8.
Here is an easy example which shows that Stokes’ can be very useful when ×F simplifies.

Example 4.4.2.

Evaluate CFdr where F=[2z+sin(x146)]ıı^5zȷȷ^5yk^ and the curve C is the circle x2+y2=4, z=1, oriented counterclockwise when viewed from above.
Solution.
The x146 in F will probably make a direct evaluation of the integral difficult. So we’ll use Stokes’ theorem. To do so we need a surface S with S=C. The simplest is just the flat disk
S={ (x,y,z) | x2+y24,  z=1 }
Since
×F=det[ıı^ȷȷ^k^xyz2z+sin(x146)5z5y]=ıı^det[yz5z5y]ȷȷ^det[xz2z+sin(x146)5y]+k^det[xy2z+sin(x146)5z]=2ȷȷ^
and the normal to S is k^, Stokes’ theorem gives
CFdr=S×Fn^dS=S(2ȷȷ^)k^dS=0
Now we’ll repeat the last example with a harder curve.

Example 4.4.3.

Evaluate CFdr where F=[2z+sin(x146)]ıı^5zȷȷ^5yk^ and the curve C is the intersection of x2+y2+z2=4 and z=y, oriented counterclockwise when viewed from above.
Solution.
The surface x2+y2+z2=4 is the sphere of radius 2 centred on the origin and z=y is a plane which contains the origin. So C, being the intersection of a sphere with a plane through the centre of the sphere, is a circle, with centre (0,0,0) and radius 2. The part of the circle in the first octant is sketched on the left below.
The x146 in F will probably make a direct evaluation of the integral difficult. So we’ll use Stokes’ theorem. To do so we need a surface S with S=C. The simplest is the flat disk
S={ (x,y,z) | x2+y2+z24,  z=y }
The first octant of S is shown in the figure on the right above. We saw in the last Example 4.4.2 that
×F=2ȷȷ^
So Stokes’ theorem gives
CFdr=S×Fn^dS=2Sȷȷ^n^dS
We’ll evaluate the integral 2Sȷȷ^n^dS in two ways. The first way is more efficient, but also requires more insight. Since (zy)=k^ȷȷ^, the upward unit normal to the plane zy=0, and hence to S, is n^=12(k^ȷȷ^). Consequently the integrand
ȷȷ^n^=ȷȷ^(ȷȷ^+k^2)=12
is a constant and we do not need a formula for n^dS:
CFdr=2Sȷȷ^n^dS=2SdS=2Area(S)=2π22=42π
Alternatively, we can evaluate the integral Sȷȷ^n^dS using our normal protocol. As S is part of the plane z=f(x,y)=y,
n^dS=±(fxıı^fyȷȷ^+k^)dxdy=±(ȷȷ^+k^)dxdy
To get the upward pointing normal pointing normal, we take the + sign so that n^dS=(ȷȷ^+k^)dxdy. As (x,y,z) runs over
S={ (x,y,z) | x2+y2+z24,  z=y }={ (x,y,z) | x2+2y24,  z=y }={ (x,y,z) | x24+y221,  z=y }
(x,y) runs over the elliptical disk R={ (x,y) | x24+y221 }. The part of this ellipse in the first octant is the shaded region in the figure below.
This ellipse has semiaxes a=2 and b=2 and hence area πab=22π. So
CFdr=2Sȷȷ^n^dS=2Rȷȷ^(ȷȷ^+k^)dxdy=2Rdxdy=2Area(R)=42π

Example 4.4.4.

Evaluate CFdr where F=(x+y)ıı^+2(xz)ȷȷ^+(y2+z)k^ and C is the oriented curve obtained by going from (2,0,0) to (0,3,0) to (0,0,6) and back to (2,0,0) along straight line segments.
Solution 1.
In this first solution, we’ll evaluate the integral directly. The first line segment (C1 in the figure above) may be parametrized as
r(t)=(2,0,0)+t{(0,3,0)(2,0,0)}=(22t,3t,0)0t1
So the integral along this segment is
01F(r(t))drdt dt=01(2+t,2(22t),(3t)2)(2,3,0) dt=01(814t) dt=[8t7t2]01=1
The second line segment (C2 in the figure above) may be parametrized as
r(t)=(0,3,0)+t{(0,0,6)(0,3,0)}=(0,33t,6t)0t1.
So the integral along this segment is
01F(r(t))drdt dt=01(3(1t),12t,9(1t)2+6t)(0,3,6) dt=01[36t+54(1t)2+36t] dt=[18t218(1t)3+18t2]01=54
The final line segment (C3 in the figure above) may be parametrized as
r(t)=(0,0,6)+t{(2,0,0)(0,0,6)}=(2t,0,66t)0t1
So the line integral along this segment is
01F(r(t))drdt dt=01(2t,4t12(1t),6(1t))(2,0,6) dt=01[4t36(1t)] dt=[2t2+18(1t)2]01=16
The full line integral is
CFdr=1+5416=39
Solution 2.
This time we shall apply Stokes’ Theorem. The curl of F is
×F=det[ıı^ȷȷ^k^xyzx+y2(xz)y2+z]=(2y+2)ıı^(00)ȷȷ^+(21)k^=2(y+1)ıı^+k^
The curve C is a triangle and so is contained in a plane. Any plane has an equation of the form Ax+By+Cz=D. Our plane does not pass through the origin (look at the figure above) so the D must be nonzero. Consequently we may divide Ax+By+Cz=D through by D giving an equation of the form ax+by+cz=1.
  • Because (2,0,0) lies on the plane, a=12.
  • Because (0,3,0) lies on the plane, b=13.
  • Because (0,0,6) lies on the plane, c=16.
So the triangle is contained in the plane x2+y3+z6=1. It is the boundary of the surface S that consists of the portion of the plane x2+y3+z6=1 that obeys x0, y0 and z0. Rewrite the equation of the plane as z=63x2y. For this surface
n^ dS=(3ıı^+2ȷȷ^+k^)dxdy
by 3.3.2, and we can write
S={ (x,y,z) | x0, y0, z0, z=63x2y }={ (x,y,z) | x0,y0, 63x2y0, z=63x2y }
As (x,y,z) runs over S, (x,y) runs over the triangle
R={ (x,y,z) | x0, y0, 3x+2y6 }={ (x,y,z) | x0, 0y32(2x) }
Using horizontal strips as in the figure on the left below,
CFdr=S×Fn^dS=R[2(y+1)ıı^+k^][3ıı^+2ȷȷ^+k^] dxdy=R[6y+7] dxdy=03dy013(62y)dx [6y+7]=03dy 13[6y+7][62y]=1303dy [12y2+22y+42]=13[4y3+11y2+42y]03=[4×9+11×3+42]=39
Alternatively, using vertical strips as in the figure on the right above,
CFdr=R[6y+7] dxdy=02dx032(2x)dy [6y+7]=02dx [33222(2x)2+732(2x)]=[27413(2x)321212(2x)2]02=948+2144=39

Example 4.4.5.

Evaluate CFdr where F=(cosx+y+z)ıı^+(x+z)ȷȷ^+(x+y)k^ and C is the intersection of the surfaces
x2+y22+z23=1andz=x2+2y2
oriented counterclockwise when viewed from above.
Solution.
First, let’s sketch the curve. x2+y22+z23=1 is an ellipsoid centred on the origin and z=x2+2y2 is an upward opening paraboloid that passes through the origin. They are sketched in the figure below. The paraboloid is red.
Their intersection, the curve C, is the blue curve in the figure. It looks like a deformed
 3 
By Salvador Dali?
circle.
One could imagine parametrizing C. For example, substituting x2=z2y2 into the equation of the ellipsoid gives 32y2+13(z+32)2=74. This can be solved to give y as a function of z and then x2=z2y2 also gives x as a function of z. However this would clearly yield, at best, a really messy integral. So let’s try Stokes’ theorem.
In fact, since
×F=det[ıı^ȷȷ^k^xyzcosx+y+zx+zx+y]=ıı^(11)ȷȷ^(11)+k^(11)=0
This F is conservative! (In fact F=(sinx+xy+xz+yz).) As C is a closed curve, CFdr=0.

Example 4.4.6.

Evaluate SGn^dS where G=(2x)ıı^+(2z2x)ȷȷ^+(2x2z)k^ and
S={ (x,y,z) | z=(1x2y2)(1y3)cosx ey, x2+y21 }
with upward pointing normal
Solution 1.
The surface S is sketched below. It is a pretty weird surface. About the
only simple thing about it is that its boundary, S, is the circle x2+y2=1, z=0. It is clear that we should not try to evaluate the integral directly
 4 
That way lies pain.
. In this solution we will combine the divergence theorem with the observation that
G=x(2x)+y(2z2x)+z(2x2z)=0
to avoid ever having work with the surface S. Here is an outline of what we will do.
  • We first select a simple surface S whose boundary S is also the circle x2+y2=1, z=0. A nice simple choice of S, and the surface that we will use, is the disk
    S={ (x,y,z) | x2+y2=1, z=0 }
  • Then we define V to be the solid whose top surface is S and whose bottom surface is S. So the boundary of V is the union of S and S.
  • For S, we will use the upward pointing normal n^=k^, which is minus the outward pointing normal to V on S. So the divergence theorem says that
    VGdV=SGn^dSSGn^dS
    The left hand side is zero because, as we have already seen, G=0. So
    SGn^dS=SGn^dS
  • Finally, we compute SGn^dS.
We saw an argument like this (with G=×F) in the first remark following the proof of Theorem 4.4.1.
So all that we have to do now is compute
SGn^dS=SGn^dS=SGk^dS=x2+y21z=0(2x2z)dxdy=x2+y21z=0(2x)dxdy=0
simply because the integrand is odd under xx.
Solution 2.
In this second solution we’ll use Stokes’ theorem instead of the divergence theorem. To do so, we have to express G in the form ×F. So the first thing to do is to check if G passes the screening test, Theorem 4.1.12, for the existence of vector potentials. That is, to check if G=0. It is. We saw this in Solution 1 above.
Next, we have to find a vector potential. In fact, we have already found, in Example 4.1.15, that
F=(z22xz)ıı^+(x22xz)ȷȷ^
is a vector potential for G, which we can quickly check.
Parametrizing C by r(t)=costıı^+sintȷȷ^, 0t2π, Stokes’ theorem gives (recalling that z=0 on C so that F(r(t))=x2ȷȷ^|x=cost=cos2t)
SGn^dS=S×Fn^dS=CFdr=02πF(r(t))drdt dt=02π(cos2t)(cost) dt
Of course this integral can be evaluated by using that one antiderivative of the integrand cos3t=(1sin2t)cost is sint13sin3t and that this antiderivative is zero at t=0 and at t=2π. But it is easier to observe that the integral of any odd power of sint or cost over any full period is zero. Look, for example, at the graphs of sin3x and cos3x, below.
Either way
SGn^dS=0

Example 4.4.7.

In this example we compute, in three different ways, CFdr where
F=(zy)ıı^(x+z)ȷȷ^(x+y)k^
and C is the curve x2+y2+z2=4, z=y oriented counterclockwise when viewed from above.
Solution 1.
Direct Computation:
In this first computation, we parametrize the curve C and compute CFdr directly. The plane z=y passes through the origin, which is the centre of the sphere x2+y2+z2=4. So C is a circle which, like the sphere, has radius 2 and centre (0,0,0). We use a parametrization of the form
r(t)=c+ρcostıı^+ρsintȷȷ^0t2π
where
  • c=(0,0,0) is the centre of C,
  • ρ=2 is the radius of C and
  • ıı^ and ȷȷ^ are two vectors that
    1. are unit vectors,
    2. are parallel to the plane z=y and
    3. are mutually perpendicular.
The trickiest part is finding suitable vectors ıı^ and ȷȷ^:
  • The point (2,0,0) satisfies both x2+y2+z2=4 and z=y and so is on C. We may choose ıı^ to be the unit vector in the direction from the centre (0,0,0) of the circle towards (2,0,0). Namely ıı^=(1,0,0).
  • Since the plane of the circle is zy=0, the vector (zy)=(0,1,1) is perpendicular to the plane of C. So k^=12(0,1,1) is a unit vector normal to z=y. Then ȷȷ^=k^×ıı^=12(0,1,1)×(1,0,0)=12(0,1,1) is a unit vector that is perpendicular to ıı^ and k^. Since ȷȷ^ is perpendicular to k^, it is parallel to z=y.
Substituting in c=(0,0,0), ρ=2, ıı^=(1,0,0) and ȷȷ^=12(0,1,1) gives
r(t)=2cost(1,0,0)+2sint12(0,1,1)=2(cost,sint2,sint2)0t2π
To check that this parametrization is correct, note that x=2cost, y=2sint, z=2sint satisfies both x2+y2+z2=4 and z=y.
At t=0, r(0)=(2,0,0). As t increases, z(t)=2sint increases and r(t) moves upwards towards r(π2)=(0,2,2). This is the desired counterclockwise direction (when viewed from above). Now that we have a parametrization, we can set up the integral.
r(t)=(2cost,2sint,2sint)r(t)=(2sint,2cost,2cost)F(r(t))=(z(t)y(t),x(t)z(t),x(t)y(t))=(2sint2sint,2cost2sint,2cost2sint)=(0,2cost+2sint,2cost+2sint)F(r(t))r(t)=[42cos2t+4costsint]=[22cos(2t)+22+2sin(2t)]
by the double angle formulae sin(2t)=2sintcost and cos(2t)=2cos2t1. So
CFdr=02πF(r(t))r(t) dt=02π[22cos(2t)+22+2sin(2t)] dt=[2sin(2t)+22tcos(2t)]02π=42π
Oof! Let’s do it an easier way.
Solution 2.
Stokes’ Theorem:
To apply Stokes’ theorem we need to express C as the boundary S of a surface S. As
C={ (x,y,z) | x2+y2+z2=4, z=y }
is a closed curve, this is possible. In fact there are many possible choices of S with S=C. Three possible S’s (sketched below) are
S={ (x,y,z) | x2+y2+z24, z=y }S={ (x,y,z) | x2+y2+z2=4, zy }S={ (x,y,z) | x2+y2+z2=4, zy }
The first of these, which is part of a plane, is likely to lead to simpler computations than the last two, which are parts of a sphere. So we choose what looks like the simpler way.
In preparation for application of Stokes’ theorem, we compute ×F and n^dS. For the latter, we apply the formula n^dS=±(fx,fy,1)dxdy (of Equation 3.3.2) to the surface z=f(x,y)=y. We use the + sign to give the normal a positive k^ component.
×F=det[ıı^ȷȷ^k^xyzzyxzxy]=ıı^(1(1))ȷȷ^(11)+k^(1(1))=2ȷȷ^n^dS=(0,1,1)dxdy×Fn^dS=(0,2,0)(0,1,1)dxdy=2dxdy
The integration variables are x and y and, by definition, the domain of integration is
R={ (x,y) | (x,y,z) is in S for some z }
To determine precisely what this domain of integration is, we observe that since z=y on S, x2+y2+z24 is the same as x2+2y24 on S,
S={ (x,y,z) | x2+2y24, z=y }R={ (x,y) | x2+2y24 }
So the domain of integration is an ellipse with semimajor axis a=2, semiminor axis b=2 and area πab=22π. The integral is then
CFdr=S×Fn^dS=R(2)dxdy=2 Area(R)=42π
Remark (Limits of integration):
If the integrand were more complicated, we would have to evaluate the integral over R by expressing it as an iterated integrals with the correct limits of integration. First suppose that we slice up R using thin vertical slices. On each such slice, x is essentially constant and y runs from (4x2)/2 to (4x2)/2. The leftmost such slice would have x=2 and the rightmost such slice would have x=2. So the correct limits with this slicing are
Rf(x,y)dxdy=22dx(4x2)/2(4x2)/2dy f(x,y)
If, instead, we slice up R using thin horizontal slices, then, on each such slice, y is essentially constant and x runs from 42y2 to 42y2. The bottom such slice would have y=2 and the top such slice would have y=2. So the correct limits with this slicing are
Rf(x,y)dxdy=22dy42y242y2dx f(x,y)
Note that the integral with limits
22dy22dx f(x,y)
corresponds to a slicing with x running from 2 to 2 on {\bf every} slice. This corresponds to a rectangular domain of integration, not what we have here.
Stokes’ Theorem, Again:
Since the integrand is just a constant (after Stoking — not the original integrand) and S is so simple (because we chose it wisely), we can evaluate the integral S×Fn^dS without ever determining dS explicitly and without ever setting up any limits of integration. We already know that ×F=2ȷȷ^. Since S is the level surface zy=0, the gradient (zy)=ȷȷ^+k^ is normal to S. So n^=12(ȷȷ^+k^) and
CFdr=S×Fn^dS=S(2ȷȷ^)12(ȷȷ^+k^)dS=S2dS=2 Area(S)
As S is a circle of radius 2, CFdr=42π, yet again.

Example 4.4.8.

In Warning 4.1.17, we stated that if a vector field fails to pass the screening test B=0 at even a single point, for example because the vector field is not defined at that point, then B can fail to have a vector potential. An example is the point source
B(x,y,z)=r^(x,y,z)r(x,y,z)2
of Example 3.4.2. Here, as usual,
r(x,y,z)=x2+y2+z2r^(x,y,z)=xıı^+yȷȷ^+zk^x2+y2+z2
This vector field is defined on all of R3, except for the origin, and its divergence
B=x(x(x2+y2+z2)3/2)+y(y(x2+y2+z2)3/2)+z(z(x2+y2+z2)3/2)=(1(x2+y2+z2)3/23x2(x2+y2+z2)5/2)+(1(x2+y2+z2)3/23y2(x2+y2+z2)5/2)+(1(x2+y2+z2)3/23z2(x2+y2+z2)5/2)=3(x2+y2+z2)3/23(x2+y2+z2)(x2+y2+z2)5/2
is zero everywhere except at the origin, where it is not defined.
This vector field cannot have a vector potential on its domain of definition, i.e. on R3{(0,0,0)}={ (x,y,z) | (x,y,z)(0,0,0) }. To see this, suppose to the contrary that it did have a vector potential A. Then its flux through any closed surface
 5 
If you are uncomfortable with the surface not having a boundary, poke a very small hole in the surface, giving it a very small boundary. Then take the limit as the hole tends to zero.
(i.e. surface without a boundary) S would be
SBn^dS=S×An^dS=SAdr=0
by Stokes’ theorem, since S is empty. But we found in Example 3.4.2, with m=1, that the flux of B through any sphere centred on the origin is 4π.

Subsection 4.4.1 The Interpretation of Div and Curl Revisited

In sections 4.1.4 and 4.1.5 we derived interpretations of the divergence and of the curl. Now that we have the divergence theorem and Stokes’ theorem, we can simplify those derivations a lot.

Subsubsection 4.4.1.1 Divergence

Let ε>0 be a tiny positive number, and then let
Bε(x0,y0,z0)={ (x,y,z) | (xx0)2+(yy0)2+(zz0)2<ε2 }
be a tiny ball of radius ε centred on the point (x0,y0,z0). Denote by
Sε(x0,y0,z0)={ (x,y,z) | (xx0)2+(yy0)2+(zz0)2=ε2 }
its surface. Because Bε(x0,y0,z0) is really small, v is essentially constant in Bε(x0,y0,z0) and we essentially have
Bε(x0,y0,z0)v dV=v(x0,y0,z0) Vol(Bε(x0,y0,z0))
Of course we are really making an approximation here, based on the assumption that v(x,y,z) is continuous and so takes values very close to v(x0,y0,z0) everywhere on the domain of integration. The approximation gets better and better as ε0 and a more precise statement is
v(x0,y0,z0)=limε0Bε(x0,y0,z0)v dVVol(Bε(x0,y0,z0))
By the divergence theorem, we also have
Bε(x0,y0,z0)v dV=Sε(x0,y0,z0)vn^ dS
Think of the vector field v as the velocity of a moving fluid which has density one. We have already seen, in §3.4, that the flux integral for a velocity field has the interpretation
Sε(x0,y0,z0)vn^ dS={the volume of fluid leaving Bε(x0,y0,z0)through Sε(x0,y0,z0) per unit time
We conclude that, as we said in 4.1.19,
v(x0,y0,z0)=limε0the rate at which fluid is exiting Bε(x0,y0,z0)Vol(Bε(x0,y0,z0))={rate at which fluid is exiting an infinitesimal sphere centred at (x0,y0,z0)per unit time, per unit volume=strength of the source at (x0,y0,z0)
If our world is filled with an incompressible fluid, a fluid whose density is constant and so never expands or compresses, we will have v=0.

Subsubsection 4.4.1.2 Curl

Again let ε>0 be a tiny positive number and let Dε(x0,y0,z0) be a tiny flat circular disk of radius ε centred on the point (x0,y0,z0) and denote by Cε(x0,y0,z0) its boundary circle. Let n^ be a unit normal vector to Dε. It tells us the orientation of Dε. Give the circle Cε the corresponding orientation using the right hand rule. That is, if the fingers of your right hand are pointing in the corresponding direction of motion along Cε and your palm is facing Dε, then your thumb is pointing in the direction n^.
Because Dε(x0,y0,z0) is really small, ×v is essentially constant on Dε(x0,y0,z0) and we essentially have
Dε(x0,y0,z0)×vn^ dS=×v(x0,y0,z0)n^ Area(Dε(x0,y0,z0))=πε2 ×v(x0,y0,z0)n^
Again, this is really an approximate statement which gets better and better as ε0. A more precise statement is
×v(x0,y0,z0)n^=limε0Dε(x0,y0,z0)×vn^ dSπε2
By Stokes’ theorem, we also have
Dε(x0,y0,z0)×vn^ dS=Cε(x0,y0,z0)vdr
Again, think of the vector field v as the velocity of a moving fluid. Then Cεvdr is called the circulation of v around Cε.
To measure the circulation experimentally, place a small paddle wheel in the fluid, with the axle of the paddle wheel pointing along n^ and each of the paddles perpendicular to Cε and centred on Cε.
Each paddle moves tangentially to Cε. It would like to move with the same speed as the tangential speed vt^ (where t^ is the forward pointing unit tangent vector to Cε at the location of the paddle) of the fluid at its location. But all paddles are rigidly fixed to the axle of the paddle wheel and so must all move with the same speed. That common speed will be the average value of vt^ around Cε. If ds represents an element of arc length of Cε, the average value of vt^ around Cε is
vT=12πεCεvt^ ds=12πεCεvdr
since dr has direction t^ and length ds so that dr=t^ds, and since 2πε is the circumference of Cε. If the paddle wheel rotates at Ω radians per unit time, each paddle travels a distance Ωε per unit time (remember that ε is the radius of Cε). That is, vT=Ωε. Combining all this information,
×v(x0,y0,z0)n^=limε0Dε(x0,y0,z0)×vn^ dSπε2=limε0Cεvdrπε2=limε02πε vTπε2=limε02πε (Ωε)πε2=2Ω
so that
Ω=12×v(x0,y0,z0)n^
The component of ×v(x0,y0,z0) in any direction n^ is twice the rate at which the paddle wheel turns when it is put into the fluid at (x0,y0,z0) with its axle pointing in the direction n^. The direction of ×v(x0,y0,z0) is the axle direction which gives maximum rate of rotation and the magnitude of ×v(x0,y0,z0) is twice that maximum rate of rotation. For this reason, ×v is called the “vorticity”.

Subsection 4.4.2 Optional — An Application of Stokes’ Theorem — Faraday’s Law

Magnetic induction refers to a physical process whereby an electric voltage is created (“induced”) by a time varying magnetic field. This process is exploited in many applications, including electric generators, induction motors, induction cooking, induction welding and inductive charging. Michael Faraday
 6 
Michael Faraday (1791–1867) was an English physicist and chemist. He ended up being an extremely influential scientist despite having only the most basic of formal educations.
is generally credited with the discovery of magnetic induction. Faraday’s law is the following. Let S be an oriented surface with boundary C. Let E and B be the (time dependent) electric and magnetic fields and define
CEdr=voltage around CSBn^dS=magnetic flux through S
Then the voltage around C is the negative of the rate of change of the magnetic flux through S. As an equation, Faraday’s Law is
CEdr=tSBn^dS
We can reformulate this as a partial differential equation. By Stokes’ Theorem,
CEdr=S(×E)n^dS
so Faraday’s law becomes
S(×E+Bt)n^dS=0
This is true for all surfaces S. So the integrand, assuming that it is continuous, must be zero.
To see this, let G=(×E+Bt). Suppose that G(x0)0. Pick a unit vector n^ in the direction of G(x0). Let S be a very small flat disk centered on x0 with normal n^ (the vector we picked). Then G(x0)n^>0 and, by continuity, G(x)n^>0 for all x on S, if we have picked S small enough. Then S(×E+Bt)n^dS>0, which is a contradiction. So G=0 everywhere and we conclude that
×E+Bt=0
This is one of Maxwell’s electromagnetic field equations
 7 
For the others, see Example 4.1.2
.

Exercises 4.4.3 Exercises

Exercise Group.

Exercises — Stage 1
1.
Each of the figures below contains a sketch of a surface S and its boundary S. Stokes’ theorem says that SFdr=S×Fn^dS if n^ is a correctly oriented unit normal vector to S. Add to each sketch a typical such normal vector.
(a)
(b)
(c)
2.
Let
  • R be a finite region in the xy-plane,
  • the boundary, C, of R consist of a single piecewise smooth, simple closed curve
    • that is oriented (i.e. an arrow is put on C) consistently with R in the sense that if you walk along C in the direction of the arrow, then R is on your left
  • F1(x,y) and F2(x,y) have continuous first partial derivatives at every point of R.
Use Stokes’ theorem to show that
C[F1(x,y)dx+F2(x,y)dy]=R(F2xF1y) dxdy
i.e. to show Green’s theorem.
3.
Verify the identity  Cϕψdr=Cψϕdr  for any continuously differentiable scalar fields ϕ and ψ and curve C that is the boundary of a piecewise smooth surface.

Exercise Group.

Exercises — Stage 2
4.
Let C be the curve of intersection of the cylinder x2+y2=1 and the surface z=y2 oriented in the counterclockwise direction as seen from (0,0,100). Let F=(x2y,y2+x,1). Calculate CFdr
  1. by direct evaluation
  2. by using Stokes’ Theorem.
5.
Evaluate CFdr where F=yexıı^+(x+ex)ȷȷ^+z2k^ and C is the curve
r(t)=(1+cost)ıı^+(1+sint)ȷȷ^+(1sintcost)k^0t2π
6. (✳).
Find the value of S×Fn^dS where F=(zy,x,x) and S is the hemisphere
{ (x,y,z)R3 | x2+y2+z2=4, z0 }
oriented so the surface normals point away from the centre of the hemisphere.
7. (✳).
Let S be the part of the surface z=16(x2+y2)2 which lies above the xy-plane. Let F be the vector field
F=xln(1+z)ıı^+x(3+y)ȷȷ^+ycoszk^
Calculate
S×Fn^dS
where n^ is the upward normal on S.
8. (✳).
Let C be the intersection of the paraboloid z=4x2y2 with the cylinder x2+(y1)2=1, oriented counterclockwise when viewed from high on the z-axis. Let F=xzıı^+xȷȷ^+yzk^. Find CFdr.
9.
Let F=yezıı^+x3coszȷȷ^+zsin(xy)k^, and let S be the part of the surface z=(1x2)(1y2) that lies above the square 1x1, 1y1 in the xy-plane. Find the flux of ×F upward through S.
10.
Evaluate the integral CFdr, in which F=(ex2yz,sinyyz,xz+2y) and C is the triangular path from (1,0,0) to (0,1,0) to (0,0,1) to (1,0,0).
11. (✳).
Let F(x,y,z)=zıı^+xȷȷ^+yk^ be a vector field. Use Stokes’ theorem to evaluate the line integral CFdr where C is the intersection of the plane z=y and the ellipsoid x24+y22+z22=1, oriented counter-clockwise when viewed from high on the z-axis.
12. (✳).
Consider the vector field F(x,y,z)=z2ıı^+x2ȷȷ^+y2k^ in R3.
  1. Compute the line integral I1=C1Fdr where C1 is the curve consisting of three line segments, L1 from (2,0,0) to (0,2,0), then L2 from (0,2,0) to (0,0,2), finally L3 from (0,0,2) to (2,0,0).
  2. A simple closed curve C2 lies on the plane E:x+y+z=2, enclosing a region R on the plane of area 3, and oriented in a counterclockwise direction as observed from the positive x-axis. Compute the line integral I2=C2Fdr.
13. (✳).
Let C=C1+C2+C3 be the curve given by the union of the three parameterized curves
r1(t)=(2cost,2sint,0),0tπ/2r2(t)=(0,2cost,2sint),0tπ/2r3(t)=(2sint,0,2cost),0tπ/2
  1. Draw a picture of C. Clearly mark each of the curves C1, C2, and C3 and indicate the orientations given by the parameterizations.
  2. Find and parameterize an oriented surface S whose boundary is C (with the given orientations).
  3. Compute the line integral CFdr where
    F=(y+sin(x2),z3x+ln(1+y2),y+ez2)
14. (✳).
We consider the cone with equation z=x2+y2. Note that its tip, or vertex, is located at the origin (0,0,0). The cone is oriented in such a way that the normal vectors point downwards (and away from the z axis). In the parts below, both S1 and S2 are oriented this way.
Let F=(zy,zx,xycos(yz)).
  1. Let S1 be the part of the cone that lies between the planes z=0 and z=4. Note that S1 does not include any part of the plane z=4. Use Stokes’ theorem to determine the value of
    S1×Fn^dS
    Make a sketch indicating the orientations of S1 and of the contour(s) of integration.
  2. Let S2 be the part of the cone that lies below the plane z=4 and above z=1. Note that S2 does not include any part of the planes z=1 and z=4. Determine the flux of ×F across S2. Justify your answer, including a sketch indicating the orientations of S2 and of the contour(s) of integration.
15. (✳).
Consider the curve C that is the intersection of the plane z=x+4 and the cylinder x2+y2=4, and suppose C is oriented so that it is traversed clockwise as seen from above.
Let F(x,y,z)=(x3+2y,sin(y)+z,x+sin(z2)).
Use Stokes’ Theorem to evaluate the line integral CFdr.
16. (✳).
  1. Consider the vector field F(x,y,z)=(z2,x2,y2) in R3. Compute the line integral CFdr, where C is the curve consisting of the three line segments, L1 from (2,0,0) to (0,2,0), then L2 from (0,2,0) to (0,0,2), and finally L3 from (0,0,2) to (2,0,0).
  2. A simple closed curve C lies in the plane x+y+z=2. The surface this curve C surrounds inside the plane x+y+z=2 has area 3. The curve C is oriented in a counterclockwise direction as observed from the positive x-axis. Compute the line integral CFdr , where F is as in (a).
17. (✳).
Evaluate the line integral
C(z+11+z)dx+xzdy+(3xyx(z+1)2)dz
where C is the curve parameterized by
r(t)=(cost,sint,1cos2tsint)0t2π
18. (✳).
A simple closed curve C lies in the plane x+y+z=1. The surface this curve C surrounds inside the plane x+y+z=1 has area 5. The curve C is oriented in a clockwise direction as observed from the positive z-axis looking down at the plane x+y+z=1.
Compute the line integral of F(x,y,z)=(z2,x2,y2) around C.
19. (✳).
Let C be the oriented curve consisting of the 5 line segments which form the paths from (0,0,0) to (0,1,1), from (0,1,1) to (0,1,2), from (0,1,2) to (0,2,0), from (0,2,0) to (2,2,0), and from (2,2,0) to (0,0,0). Let
F=(y+exsinx)ıı^+y4ȷȷ^+ztanzk^
Evaluate the integral CFdr.
20. (✳).
Suppose the curve C is the intersection of the cylinder x2+y2=1 with the surface z=xy2, traversed clockwise if viewed from the positive z-axis, i.e. viewed “from above”. Evaluate the line integral
C(z+sinz)dx+(x3x2y)dy+(xcoszy)dz
21. (✳).
Evaluate S×Fn^dS where S is that part of the sphere x2+y2+z2=2 above the plane z=1, n^ is the upward unit normal, and
F(x,y,z)=y2ıı^+x3ȷȷ^+(ex+ey+z)k^
22. (✳).
Let
F=xsinyıı^ysinxȷȷ^+(xy)z2k^
Use Stokes’ theorem to evaluate
CFdr
along the path consisting of the straight line segments successively joining the points P0=(0,0,0) to P1=(π/2,0,0) to P2=(π/2,0,1) to P3=(0,0,1) to P4=(0,π/2,1) to P5=(0,π/2,0), and back to (0,0,0).
23. (✳).
Let
F=(2z1+y+sin(x2),3z1+x+sin(y2),5(x+1)(y+2))
Let C be the oriented curve consisting of four line segments from (0,0,0) to (2,0,0), from (2,0,0) to (0,0,2), from (0,0,2) to (0,3,0), and from (0,3,0) to (0,0,0).
  1. Draw a picture of C. Clearly indicate the orientation on each line segment.
  2. Compute the work integral CFdr.
24. (✳).
Evaluate S×Fn^dS where F=yıı^+2zȷȷ^+3xk^ and S is the surface z=1x2y2, z0 and n^ is a unit normal to S obeying n^k^0.
25. (✳).
Let S be the curved surface below, oriented by the outward normal:
x2+y2+2(z1)2=6,z0.
(E.g., at the high point of the surface, the unit normal is k^.)
Define
G=×F,whereF=(xzy3cosz)ıı^+x3ezȷȷ^+xyzex2+y2+z2k^.
Find SGn^dS.
26. (✳).
Let C be a circle of radius R lying in the plane x+y+z=3. Use Stokes’ Theorem to calculate the value of
CFdr
where F=z2ıı^+x2ȷȷ^+y2k^. (You may use either orientation of the circle.)
27.
Let S be the oriented surface consisting of the top and four sides of the cube whose vertices are (±1,±1,±1), oriented outward. If F(x,y,z)=(xyz,xy2,x2yz), find the flux of ×F through S.
28.
Let S denote the part of the spiral ramp (that is helicoidal surface) parametrized by
x=ucosv, y=usinv, z=v0u1, 0v2π
Let C denote the boundary of S with orientation specified by the upward pointing normal on S. Find
Cydxxdy+xydz

Exercise Group.

Exercises — Stage 3
29.
Let C be the intersection of x+2yz=7 and x22x+4y2=15. The curve C is oriented counterclockwise when viewed from high on the z-axis. Let
F=(ex2+yz)ıı^+(cos(y2)x2)ȷȷ^+(sin(z2)+xy)k^
Evaluate CFdr.
30. (✳).
  1. Find the curl of the vector field F=(2+x2+z,0,3+x2z).
  2. Let C be the curve in R3 from the point (0,0,0) to the point (2,0,0), consisting of three consecutive line segments connecting the points (0,0,0) to (0,0,3), (0,0,3) to (0,1,0), and (0,1,0) to (2,0,0). Evaluate the line integral
    CFdr
    where F is the vector field from (a).
31. (✳).
  1. Let S be the bucket shaped surface consisting of the cylindrical surface y2+z2=9 between x=0 and x=5, and the disc inside the yz-plane of radius 3 centered at the origin. (The bucket S has a bottom, but no lid.) Orient S in such a way that the unit normal points outward. Compute the flux of the vector field ×G through S, where G=(x,z,y).
  2. Compute the flux of the vector field F=(2+z,xz2,xcosy) through S, where S is as in (a).
32. (✳).
Let
F(x,y,z)=(yx+x1+x2,x2y1+y2,cos5(lnz))
  1. Write down the domain D of F.
  2. Circle the correct statement(s):
    1. D is connected.
    2. D is simply connected.
    3. D is disconnected.
  3. Compute ×F.
  4. Let C be the square with corners (3±1,3±1) in the plane z=2, oriented clockwise (viewed from above, i.e. down z-axis). Compute
    CFdr
  5. Is F conservative?
33. (✳).
A physicist studies a vector field F(x,y,z). From experiments, it is known that F is of the form
F(x,y,z)=xzıı^+(axeyz+byz)ȷȷ^+(y2xeyz2)k^
for some real numbers a and b. It is further known that F=×G for some differentiable vector field G.
  1. Determine a and b.
  2. Evaluate the surface integral
    SFn^dS
    where S is the part of the ellipsoid x2+y2+14z2=1 for which z0, oriented so that its normal vector has a positive z-component.
34. (✳).
Let C be the curve in the xy-plane from the point (0,0) to the point (5,5) consisting of the ten line segments consecutively connecting the points (0,0), (0,1), (1,1), (1,2), (2,2), (2,3), (3,3), (3,4), (4,4), (4,5), (5,5). Evaluate the line integral
CFdr
where
F=yıı^+(2x10)ȷȷ^
35. (✳).
Let F=(sinx2,xz,z2). Evaluate CFdr around the curve C of intersection of the cylinder x2+y2=4 with the surface z=x2, traversed counter clockwise as viewed from high on the z-axis.
36. (✳).
Explain how one deduces the differential form
×E=1cHt
of Faraday’s law from its integral form
CEdr=1c ddtSHn^dS
37. (✳).
Let C be the curve given by the parametric equations:
x=cost, y=2sint, z=cost, 0t2π
and let
F=zıı^+xȷȷ^+y3z3k^
Use Stokes’ theorem to evaluate
CFdr
38. (✳).
Use Stokes’ theorem to evaluate
Czdx+xdyydz
where C is the closed curve which is the intersection of the plane x+y+z=1 with the sphere x2+y2+z2=1. Assume that C is oriented clockwise as viewed from the origin.
39. (✳).
Let S be the part of the half cone
z=x2+y2,y0,
that lies below the plane z=1.
  1. Find a parametrization for S.
  2. Calculate the flux of the velocity field
    v=xıı^+yȷȷ^2zk^
    downward through S.
  3. A vector field F has curl ×F=xıı^+yȷȷ^2zk^. On the xz-plane, the vector field F is constant with F(x,0,z)=ȷȷ^. Given this information, calculate
    CFdr,
    where C is the half circle
    x2+y2=1, z=1, y0
    oriented from (1,0,1) to (1,0,1).
40.
Consider S(×F)n^dS where S is the portion of the sphere x2+y2+z2=1 that obeys x+y+z1, n^ is the upward pointing normal to the sphere and F=(yz)ıı^+(zx)ȷȷ^+(xy)k^. Find another surface S with the property that S(×F)n^dS=S(×F)n^dS and evaluate S(×F)n^dS.