Stokes’ theorem is the second major integral theorem of vector calculus and is a relationship between certain types of line and surface integrals. We have previously seen the relationship
(∇×u)·ˆnδA= I
C
u·dr (118)
which, for a vector field u says that the line integral around a small closed loop is equal to the component of the curl normal to the elemental area enclosed by the loop. We can think of this later quantity as the ‘flux’ of the vector field formed by the curl∇×u through the elemental areaδA.
Stokes’ theorem extends this relationship to integral form, relating the line integral of u around a finite closed loop C to the flux of∇×u through any surface spanning the loop. The concept of a surface spanning a loop C is illustrated in Figure 16, where the surfaces S1and S2span the loop C. Note that a curve C in the three-dimensional space can be spanned by an infinite number of surfaces.
C S
1S
2u
u
Figure 16: Two surfaces, S1and S2, spanning the curve C.
The statement of Stokes’ theorem is as follows: if S is a surface that spans a closed curve C then I
C
u·dr= Z
S(∇×u)·dA
where the direction in which C is traversed and the direction of the normal to the surface S are related by the right-hand rule, as illustrated in Figure 17 (for example, if the normal to S points upwards, the curve C is traversed counterclockwise.
Stokes’ theorem can be interpreted as follows. Imagine the surface S spanning the closed curve C to be divided into a large number of surface elementsδA=δA ˆn, as illustrated in Figure 18, left. Suppose that each elemental area spans a small closed loop Cδ. As with (118), for each of these elements, the relation
(∇×u)·ˆnδA= I
Cδ
u·dr (119)
holds. If we calculate the total flux of the curl∇×u through the entire surface S, then, from (119), this is equivalent to summing the circulation along the boundary of each elemental area making up the surface S. If we sum all the circulations of u along the boundary of each element of area, then all line integrals along the common boundary lines of adjoining surface elements that do not lie on the curve C will cancel (as in Figure 18, right), leaving only the line integrals along the curve C. This is because, during the line integration procedure on all the elemental areas, each boundary is traversed just once in each direction, and it is only along the boundary line C, at the edge
n
S C
ˆ
Figure 17: Orientation of the curve C with respect to the normal to the surface S.
of the surface S, that the contributions remain to give the total circulation along the curve C. In other words, calculating the total flux of∇×u through the entire surface S is equal to taking the circulation along C, the curve that is spanned by S, which is the statement of Stokes’ theorem.
It is important to note that Stokes’ theorem applies to any surface S spanning the loop C. If we imagine multiple surfaces S1,S2,· · ·spanning the same loop C (as in Figure 16), then the flux of∇×u through each of these surfaces is the same.
C
u
u
δA δA 1 δ A 2
δ A 3 δA 4 Line integrals along common
boundaries of elemental areas cancel out S
Figure 18: Stokes’ theorem.
Example 3. Suppose F=−yi+xj+2zk and B=∇×F. Find the outward flux of B through the northern hemi-sphere|r|=R (illustrated in Figure 19) using
1. Stokes’ theorem 2. Gauss’ Theorem
3. Direct integration of the flux of B through the hemisphere.
R
S2 (the hemispherical 'shell')
S1 (the base of the hemisphere)
n ˆ
1n ˆ
2Curve C
Figure 19: The northern hemisphere S2and the base of the hemisphere S1.
1. Let C be the curve that encloses the base of the hemisphere S2. Note that C is a circle of radius R centered on the origin. We wish to find x
S2
B·ˆn2dS=x
S2
(∇×F)·ˆn2dS
(note that, since we want the flux coming out of the hemisphere, we take ˆn2pointing out of the hemispherical shell S2). By Stokes’ theorem, the right hand side of this equation is equal to
x
S2
(∇×F)·ˆn2dS= I
C
F·dr
With respect to the curve C, the normal ˆn2point ‘upwards’, and therefore when using Stokes’ theorem we must calculate the line integral around C in the direction that obeys the right hand rule with respect to ˆn2. This direction is the counterclockwise direction looking at C from ‘above’, and a parametrization of the curve C that follows this direction is
r(t) =x(t)i+y(t)j+z(t)k
=R cos(t)i+R sin(t)j+0k (120)
Note that if we wanted to integrate in the opposite direction, the parametrization would have been r(t) =R cos(t)i−R sin(t)j+0k
Now with the parametrization (120)
dr(t) = [−R sin(t)i+R cos(t)j+0k]dt On the curve C,
F=−y(t)i+x(t)j+2z(t)k
=−R sin(t)i+R cos(t)j+0k This yields F·dr=R2dt and therefore
x
S2
B·ˆn2dS=x
S2
(∇×F)·ˆn2dS= I
C
F·dr= Z 2π
t=0
R2dt=2πR2
Another way of applying Stokes’ theorem would have been to use the fact that for any surface spanning C
such as S1or S2 x
S2
(∇×F)·ˆn2dS=x
S1
(∇×F)·ˆn1dS
Note that ˆn1=1k. The vector ˆn1points in the positive z direction, because with respect the curve C it must be orientated in the same direction as the vector ˆn2.
Now∇×F=2k, and therefore x
S1
(∇×F)·ˆn1dS=x
S1
2 dS=2πR2 .
2. We wish to finds
S2B·ˆn2dS. Let S be the surface composed of both S1and S2in Figure 19, enclosing the hemispherical volume V .
Sincev
SB·dS gives the flux out of the hemisphere, {
S
B·dS=x
S1
B·(−ˆn1)dS+x
S2
B·ˆn2dS (121)
Note that we have taken the normal to the surface S1to be−ˆn1since that is the normal to S1that points out of the volume V . At the same time, we have used ˆn2as the normal to S2since that is the normal to S2that points out of the volume V .
By Gauss’ theorem
y
V
∇·B dV={
S
B·dS
=x
S1
B·(−ˆn1)dS+x
S2
B·ˆn2dS
(122)
Note that for any vector field F, we have the identity
∇·(∇×F) =0
Therefore∇·B and the left hand side of (122) both equal zero. From (122), we then have x
S2
B·ˆn2dS=x
S1
B·ˆn1dS
The right hand side of this equation is easy to compute: B=2k and therefore B·ˆn1=2 and x
S1
B·ˆn1dS=x
S1
2dS=2πR2
3. We wish to finds
S2B·ˆn2dS. The position vector r always points normal to a spherical surface centered on the origin. Therefore on the surface S2ˆn2=|r|r =Rr. Since B=2k, we have B·ˆn2=2z. Hence,
x
S2
B·ˆn2dS=x
S2
2 Rz dS
In spherical polar coordinates z=R cosφ, whilst on the surface S2, dS=R2sinφdθdφ. This gives x
S2
B·ˆn2dS=x
S2
2 Rz dS=
Z π2 0
Z θ=2π θ=0
2
R(R cosφ)(R2sinφdθdφ)
= Z π
2
0
Z θ=2π
θ=0 R2sin(2φ)dθdφ=2πR2Z
π 2
0 sin(2φ)dφ=2πR2
6.1 Green’s theorem: Stokes’ theorem in the plane
As with the divergence theorem in the plane, a special case of Stokes’ theorem can be stated when the surface of interest is flattened to give a subset D of a plane. When the flattened surface D is bounded by a closed curve C, then Stokes’ theorem relates circulation of a vector field F around C to the total flux of∇×F through D. This is known as Green’s theorem.
More specifically, let F(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 region D⊂R. Also assume that the partial derivatives∂F2
∂x and∂F1
∂y exists at all points(x,y)∈D.
I C
F·dr=x
D
∂F2
∂x −∂F1
∂y
dA, (123)
where the line integration is performed over C in a counterclockwise direction, with D lying to the left of the path.
Note that:
1. The integral on the left is a line integral – the circulation of the vector field F over the closed curve C.
2. The integral on the right is a double integral over the region D enclosed by C.
Special case: If
∂F2
∂x =
∂F1
∂y , (124)
for all points(x,y)∈D, then the circulation integralHCF·dr is zero. But recall that the above equality condition for the partial derivatives is the condition for F to be conservative. In fact, the integrand in Green’s Theorem, Eq.
(123) is the k component of the curl of F. 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 inR2is a vector that points in the z-direction. This is related to the convention of assigning a velocity vector that points along an axis of rotation using the right-hand rule. With the above result, we can rewrite Green’s Theorem as
I C
F·dr=x
D
[curl F]kdA, (125)
where[v]kdenotes the k-component of vector v. In this case, the k-component of curl F is its only component.
Examples:
1. The vector field F=−Kyi+Kxj. (This is the velocity vector field of a thin plate on the xy-plane that is rotating about the z-axis with angular speed dθ
dt =K.) Now let CRdenote the circle of radius R>0 centered at the origin. In the previous lecture, we computed the circulation integral of F over CRto be
I
F·dr=2πKR2. (126)
This was done by a direct calculation of the line integral using the parametrization of CRas r(t) = (R cost,R sint).
Let us now compute this result using Green’s Theorem. Here F1=−Ky and F2=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
∂y dA=2K x
D
dA=2KA(D) =2KπR2, (128)
where A(D)denotes the area of region D (area of a circular region, radius R).
It is sometimes misleading to present this example because people may associate the curl of F, 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, I
C
F·dr=2KA(D), (129)
where A(D)denotes the area of the region D enclosed by C.
2. Compute the circulation of the vector field F=−y2i+xj around the circle of radius 1 centered at the origin.
Here, F1=−y2and F2=x. Then x
D
∂F2
∂x −
∂F1
∂y dA = 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 may confirm this result by computing the integral explicitly, either with Cartesian or planar polar coordinates.
One may also conclude that it is zero because the function y is an odd function – y>0 over the region above the x-axis and y<0 over the region below the x-axis which is a mirror image of the region above.
You can also confirm the above result by explicitly computing the circulation integral using the parametriza-tion r(t) = (cost,sint).
3. Consider the following vector field,
F=− y
x2+y2i+ x
x2+y2j. (131)
The curl of F is
curl 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 curves C that do not enclose the origin (0,0). (Recall that one of the assumptions in Green’s Theorem was that the partial derivatives existed at all points(x,y)∈D, the region enclosed by C.) In this case, all circulation integrals
are zero: I
C
F·dr=0. (133)
For this vector field, however, we cannot use Green’s Theorem to conclude anything about circulation inte-grals over closed curves C that enclose the origin. However, we can compute the circulation integral using parametrization, as long as the curve C avoids the origin. In the case that C is the circle CRof radius R centered at the origin is 2π, one can compute that the circulation is 2π(Exercise).