Stokes’ Theorem
Learning Goal: to see the theorem and examples of it in action.
In two dimensions we had Green’s Theorem, that for a region R with boundary C and vector
field F, ∫ F ⋅ ds = ∫ (∇ × F)⋅ k dA . Note that for a region in the plane, k is the normal vector, so
C
R
k dA is really dS. So we have
∫
C
F ⋅ ds = ∫ (∇ × F)⋅ dS . It should be no surprise that this extends
R
into three dimensions without any change:
Theorem (Stokes’ Theorem): Let S be an oriented surface with compatibly oriented boundary
∂S. Assume both are nice enough to do surface/line integrals and assume F is a differentiable
vector field. Then ∫ F ⋅ ds = ∫ (∇ × F)⋅ dS .
∂S
S
What is “compatibly oriented?” It basically means right-handed, in that if you stand on the
surface, on the side given by the orientation, and you walk around the boundary, the surface will
be on your left-hand side.
The meaning is basically the same as it is in two dimensions. The curl is basically how much the
field is twisting at each point, so if you add the total amount of twist over the surface you should
get the circulation of the field around the edge of the surface. In fact, we will use the theorem in
a little bit to give a more precise idea of what curl actually means. First, though, some examples.
Example: verify Stokes’ Theorem where the surface S is the triangle with vertices (1, 0, 2), (–1,
1, 4), and (2, 2, –1) (going around in that order!) and F is the vector field zi – 2xj + yk.
First, let’s do the line integral around the boundary. We parameterize the boundary in three
pieces: σ 1(t) = (1 – 2t, t, 2 + 2t), σ 2(t) = (–1 + 3t, 1 + t, 4 – 5t), and σ 3(t) = (2 – t, 2 – 2t, –1 + 3t),
each of which runs from t = 0 to t = 1. So we get
1
1
∫ (2 + 2t,−2 + 4t,t)⋅(−2,1,2)dt + ∫ (4 − 5t,2 − 6t,1+ t)⋅(3,1,−5)dt
0
0
1
+ ∫ (−1+ 3t,2t − 4,2 − 2t)⋅(−1,−2, 3)dt
0
Evaluating, we get
1
∫ (−4 − 4t − 2 + 4t + 2t + 12 − 15t + 2 − 6t − 5 − 5t + 1− 3t − 4t + 8 + 6 − 6t )dt
0
1
= ∫ 18 − 37t dt = −1 / 2.
0
Now we try the surface integral. The curl is ∇ × F = (1, 1, –2). Since this is constant, we really
only have to multiply it by the area it “sees” looking at the triangle. Basically, we’ll take the
cross product of the sides of the triangle (to find its normal direction and area) dot it with this
curl (multiply the length of this vector by the area of the triangle and the cosine of the angle
between them to compensate for the angle between them) and divide by two (it’s a triangle, not a
parallelogram!). So, two vectors that lie along sides of the triangle are (–2, 1, 2) and (3, 1, –5)
(going in the same order around the vertices ensures our cross product will put the normal vector
on the correct “side” of the triangle) whose cross product is (–7, –4, –5). Obviously, the dot
product of this with (1, 1, –2), divided by 2, is –1/2.
Example: verify Stokes’ Theorem where F is the vector field (y,
–x, exz) and the surface S is the sphere of radius 2 centered at
(0, 0, 1) with the bottom cut off below the xy-plane.
The surface integral is going to be something of a nightmare.
First, we note that to parameterize we will take a standard sphere,
expand it by 2 and raise it up one, and then cut off the bottom
by only running ϕ to 3π/4. So here are all the details:
x = 2 sin(φ )cos(θ )
y = 2 sin(φ )sin(θ )
z = 1+ 2 cos(φ )
Tφ = ( 2 cos(φ )cos(θ ), 2 cos(φ )sin(θ ),− 2 sin(φ ))
Tθ = (− 2 sin(φ )sin(θ ), 2 sin(φ )cos(θ ),0)
N = (2sin 2 (φ )cos(θ ),2sin 2 (φ )sin(θ ),2sin(φ )cos(φ ))
F = (y,−x,e xz )
∇ × F = (0, ze xz ,−2)
(Notice that we have chosen a parameterization with outward-pointing vector, so the compatible
direction around the circle will be counterclockwise).
So we integral (∇ × F)⋅N, plugging in the appropriate formulas for x, y, and z, to obtain
3π /4
2π
0
0
∫ ∫
2(1+ 2 cos(φ ))e
awful, but the first part is
left with
3π /4
2π
0
0
∫ ∫
2 sin(φ )cos(θ )(1+ 2 cos(φ ))
∫
2π
0
sin 2 (φ )sin(θ ) − 4 sin(φ )cos(φ )dθ dφ . Now that looks
aeb cos(θ ) sin(θ )dθ which we can do, and which is zero. So we are
−4 sin(φ )cos(φ )dθ dφ = ∫
3π /4
0
3π /4
−8π sin(φ )cos(φ )dφ = −4π sin 2 (φ ) 0
= −2π .
Let’s try the line integral. Note that we have cut off the sphere along the xy-plane where they
intersect in…the unit circle! So we parameterize the boundary (counterclockwise!) as (cos(θ),
sin(θ), 0). So
∫
C
2π
2π
0
0
F ⋅ ds = ∫ (sin(θ ),− cos(θ ),1)⋅(− sin(θ ),cos(θ ),0)dθ = ∫ −1dθ = −2π . Hmm,
that was a lot easier.
Example: let F = ∇f. Verification of Stokes’ Theorem is really easy now. For integrating ∇f
around the boundary—which must be one or more closed curves—must give zero (use the
fundamental theorem of line integrals, and note that the end point and start point are the same!).
On the other hand, the surface integral easily gives zero, because the curl of a gradient is always
zero.
We can use Stokes’ Theorem to get a better understanding of what the curl is. Let S be a tiny
square parallel to the xy-plane. Then for that square, ∫ (∇ × F)⋅ dS = ∫ F ⋅ ds . We know the
S
∂S
latter integral is the circulation of F around the boundary of the square. On the other hand, if the
square is small enough, we can assume that the curl is essentially constant. dS = N dx dy, and
since the square is parallel to the xy-plane, N = (area)i. So the surface integral is essentially just
the x-component of the curl times the area of the square. Dividing both sides by the area, and
taking limits as this area goes to zero (to make sure the curl really is constant!) we find that the xcomponent of the curl is just the horizontal circulation per unit (horizontal) area.
The same applies in the yz- and xz- planes, so we can say that the curl is the net circulation per
unit area, tilted to be normal to the plane of that circulation.
Curvy considerations
It should be fairly clear that every orientable surface has a boundary consisting of a number of
simple closed curves. Non-orientable surfaces have such boundaries, too, but we don’t need to
worry about them right now since we’re doing surface vector integrals. Of course, there are
closed surfaces like a sphere or a torus whose boundaries are empty (what does Stokes say in this
situation?).
An interesting question is the reverse: is every simple closed
curve (or collection of such) the boundary of an orientable
surface? The answer might seem like it should obviously be
“yes,” until you start to consider that curves might be knotted!
Does the trefoil knot at right bound a nice orientable surface?
The answer is still “yes” only now it is not so obvious!
The technique is to start with a projection of the curve onto a
plane, with little gaps to indicate which strand of the curve
“crosses over” the other:
Since we need everything to be oriented anyway, draw little arrows all along your diagram to
indicate the orientation. Then at every crossing, change the diagram to a non-crossing, keeping
the strands oriented in the same direction:
�������
Now the diagram falls apart into a number of unlinked oriented loops. Now for each loop we
can easily make a surface which spans it. Orient this surface upward if the loop runs around it
counterclockwise, and downward if clockwise. Connect the little surface pieces together by halftwisted bands which run along the former crossings.
It takes a little doing but it is ultimately not too hard to prove that this arrangement makes all the
orientations compatible with each other (think: if they’re not, we can walk along the boundary
and the orientation would have to change, but each piece is oriented strictly compatibly with its
boundary). So we’ve created what’s known as a “Seifert surface” which is an oriented surface
whose boundary is our original curve (or collection of curves).
(Sorry—I can’t draw the pictures for these electronically.)