If is a -dimensional surface in , and if is a vector field, then Stokes’ Theorem relates the integral over of with the integral of over , the boundary of .
For Stokes’ Theorem, we will always consider a surface that is a subset of a smooth (or piecewise smooth) surface and “the boundary of †will be understood to mean “the boundary of within .†We will occasionally call this the Stokes boundary of to distinguish it from the definition of boundary in Section 1.1, which is different. When it is clear from the context that we are talking about the Stokes boundary, we will omit the word “Stokesâ€.
Example 1.
Consider shown below:
This is a subset of the larger smooth surface for some , pictured below as a blue mesh:
A point belongs to the Stokes boundary - that is, the boundary of within - if Thus the Stokes boundary of is the yellow curve at the edge of , shown below.
Here it is again, with omitted from the picture for clarity.
disclaimer Some aspects of the above discussion are a little imprecise. We could only make them precise at the expense of introducing more technical detail, and we choose not to do this here. This may be true at other points in this section.
We have discussed this issue at some length, but in practice identifying the Stokes boundary of a surface is usually very straightforward. The long discussion is to explain the difference between the Stokes boundary and our definition of boundary from Section 1.1.
Orientation
For Stokes’ Theorem, we will always Suppose that (as well as the larger smooth surface that we have Supposed contains ) is oriented by a choice of the unit normal . Informally, the orientation for may be described by saying that if someone walks around with their head pointing in the direction of the unit normal , and with on the left, then the Stokes orientation corresponds to the direction in which they travel around .
Slightly more mathematically, this means that at a point , if denotes the positively oriented normal to and if denoted the tangent to at with the Stokes orientation, then
Example 2.
In Example 1 above, one can see that if is oriented with pointed upwards, then the Stokes orientation for is counterclockwise (as seen from above.)
This is more generally true: is a graph, that is, a surface parametrized by where is a convex subset of and also (as usual) a regular region with piecewise boundary. If is oriented with pointing upwards, then again the Stokes orientation for points in the counterclockwise direction (as seen from above) around . Many concrete problems that you will see concern surfaces that can be put in the form , so it is a good idea to remember this basic fact about the orientation of .
A formula for the orientation (optional!)
In practice, you will probably never need to write down a formula that describes the Stokes orientation of some boundary . But in case you are reassured to know that such a formula exists, here it is (in the case when the surface has a nice parametrization):
Suppose that is regular region in contained in an open set , and that parametrizes the smooth surface , with its restriction to parametrizing . Also Suppose (as usual) that never vanishes, and that points in the same direction as the unit normal .
Then for every point , there exists some such that . In this situation, the positive orientation for is the direction of , where is a positively oriented (in the sense of Green’s Theorem) tangent vector to .
Example 3.
Suppose that is parametrized by for , where is the closed unit ball .
We know that at a point , the positively oriented unit tangent is given by . Thus at the pint , the Stokes orientation points in the same direction as For instance, in Example 3, points in have the form $ (u,v, u3-3uv2) $ and the Stokes orientation for the boundary at that point is given by . (If you want a unit vector, you would have to divide by the right positive constant.) However, you will probably never need a formula of this sort.
Statement of Stokes’ Theorem
Let be a piecewise smooth surface in with Stokes’ boundary as described above, that is oriented by the unit normal and that has the compatible (Stokes) orientation. Suppose also that is any vector field that is in an open set containing . Then
Sketch of proof.
Some ideas in the proof of Stokes’ Theorem are:
As in the proof of Green’s Theorem and the Divergence Theorem, first prove it for of a simple form, and then prove it for more general by dividing it into pieces of the simple form, applying the theorem on each such piece, and adding up the results.
In this case, the simple case consists of a surface that can be parametrized by a single for . More precisely, is a regular region with piecewise smooth boundary, contained in an open set , and that parametrizes a smooth or piecewise smooth surface. Then the idea of the proof is to use a change of variables to rewrite and as integrals over , then apply Green’s Theorem on .
Examples and Consequences
Example 3.
Suppose that is a regular region with piecewise smooth boundary, and that . Suppose also that the unit normal to points upward.
Then and the orientation of coincide with the positive orientation (from Green’s Theorem) of .
Now if is a vector field of the form , then in this special case and so in this special case, Stokes’ Theorem reduces exactly to Green’s Theorem.
Example 4.
Let be the intersection of the cylinder and the surface oriented counterclockwise. Compute
Let be the portion of the surface bounded by the cylinder , with the unit normal oriented upwards. Then , and the orientation for is the correct one for Stokes’ Theorem. Thus A computation shows that , and we know from Exercises in Section 5.3, or direct computation, that Thus
Closed Surfaces
A surface is said to be closed if it has no (Stokes’) boundary.
An example of such a surface is the unit sphere
To explain what it means to “have no boundary†let’s define (both with oriented as for ) pictured below in blue and purple:
Since , it is reasonable to believe that . It is pretty clear that But since and have opposite orientations, we can suppose that they “cancel each other outâ€, resulting in “â€. This may seem unconvincing, but it is actually the underlying idea of the proof of Theorem 2 below. It is also the idea behind our definition of a closed surface:
A piecewise smooth surface in is closed if it there exist subsets with piecewise smooth boundaries, such that
,
, and
have opposite orientations (when have the same orientation as .)
In contrast to the unit sphere, the surface pictured below is not closed because there is not way of splitting it into 2 pieces whose boundaries completely cancel each other – the yellow curve at the top ( ) will always remain “uncancelledâ€.
If is a piecewise smooth closed surface in and is a vector field that is on an open set containing , then
Given a closed surface , let and be sets as in the definition of closed surface. Due to the additivity of integration and Stokes’ Theorem, because and have the opposite orientations, by assumption.
Example 5.
Let denote the unit sphere , and compute Here is a real number and is oriented outwards.
Solution. The integral equals zero, as a direct consequence of Theorem 2.
Remarks.
Let , and note that (the ordinary non-Stokes boundary.) If were everywhere in , then we could use the Divergence Theorem to compute since everywhere when is . The significance of Theorem 2 is that still equals zero even if, as in this example, is not .
One can check that if is a regular region with piecewise smooth boundary, then is a closed surface. To verify this, one can fix and define and for some small . It is geometrically clear (but not very easy to prove) that these will satisfy the conditions in the definition of “closed surfaceâ€.
Moving the Surface
It follows immediately from Stokes’ Theorem that if and are two (oriented) surfaces such that (with the same orientation), then This means that to evaluate an integral , we can move the surface to a (suitable) new surface .
Example 6.
Let be the part of the cone below the plane , with the unit normal oriented upward, and evaluate Below: , with its boundary shown in red
Let be the portion of the plane contained within the cone , with the unit normal oriented upwards, shown below, with its boundary in red.
Then (both oriented counterclockwise as seen from above). We can see that the boundaries coincides if we place both surfaces in the same picture:
So Stokes’ Theorem implies that Also, , and this equals on . We conclude that
Problems
Basic
Use Stokes’ Theorem to evaluate the integral where is the intersection of the planes and the surface , oriented counterclockwise (as seen from above).
Evaluate the integral where is the part of the plane contained in the cylinder . Suppose that is oriented counterclockwise (as seen from above).
Evaluate the integral where .
Evaluate the integral where with oriented outwards.
Let be the part of the ellipsoidal cone beneath the plane , with oriented upwards, and evaluate the integral
Evalute where is the curve formed by the intersection of the surfaces and the (oriented counterclockwise, as seen from above.)
Advanced
Find a piecewise smooth simple closed (oriented) curve that maximizes among all curves that are constrained to lie on the surface of the unit sphere . A typical such curve is pictured below.
Prove that if is a vector field and is a function, and if is a surface satisfying the assumptions of Stokes’ Theorem, then
Suppose that is a vector field such that for every such that . Prove that there does not exist any vector field such that .