5.6 Stokes’ Theorem

5.6 Stokes’ Theorem

  1. Statement of Stokes’ Theorem
  2. Examples and Consequences
  3. Problems

    

Statement of Stokes’ Theorem

The Stokes Boundary

If S is a 2-dimensional surface in R3, and if F is a C1 vector field, then Stokes’ Theorem relates the integral over S of curlF with the integral of F over S, the boundary of S.

For Stokes’ Theorem, we will always consider a surface S that is a subset of a smooth (or piecewise smooth) surface S0 and “the boundary of S” will be understood to mean “the boundary of S within S0.” We will occasionally call this the Stokes boundary of S 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 “Stokes”.

Example 1.

Consider S={G(u): uR2,|u|1}, for G(u)=(u,v,u33uv2), shown below:

S0S1
This is a subset of the larger smooth surface S0={G(u): uR2,|u|<1+δ} for some δ>0, pictured below as a blue mesh:
S0S1

A point x belongs to the Stokes boundary - that is, the boundary of S within S0 - if r>0,B(r,x)S   and   B(r,x)(S0S) are both nonempty. Thus the Stokes boundary of S is the yellow curve at the edge of S, shown below.

S0S1

Here it is again, with S0 omitted from the picture for clarity.

S0S1
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 S 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 S (as well as the larger smooth surface S0 that we have Supposed contains S) is oriented by a choice of the unit normal n. Informally, the orientation for S may be described by saying that if someone walks around S with their head pointing in the direction of the unit normal n, and with S on the left, then the Stokes orientation corresponds to the direction in which they travel around S.

Slightly more mathematically, this means that at a point xS, if n denotes the positively oriented normal to S0 and if t denoted the tangent to S at x with the Stokes orientation, then n×t points into S.

Example 2.

In Example 1 above, one can see that if S is oriented with n pointed upwards, then the Stokes orientation for S is counterclockwise (as seen from above.)

This is more generally true: S is a graph, that is, a surface parametrized by (1)G(u,v)  of the form  (u,v,ϕ(u,v)),  (u,v)RR2, where R is a convex subset of R2 and also (as usual) a regular region with piecewise C1 boundary. If S is oriented with n pointing upwards, then again the Stokes orientation for S points in the counterclockwise direction (as seen from above) around S.
Many concrete problems that you will see concern surfaces S that can be put in the form , so it is a good idea to remember this basic fact about the orientation of S.

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 S. But in case you are reassured to know that such a formula exists, here it is (in the case when the surface S has a nice parametrization):

Suppose that R is regular region in R2 contained in an open set R0, and that G:R0R3 parametrizes the smooth surface S0, with its restriction to R parametrizing S. Also Suppose (as usual) that uG×vG never vanishes, and that uG×vG points in the same direction as the unit normal n.

Then for every point xS, there exists some uR such that x=G(u). In this situation, the positive orientation for S is the direction of DG(u)T, where T is a positively oriented (in the sense of Green’s Theorem) tangent vector to R.

Example 3.

Suppose that S is parametrized by G(u)=(u,v,ϕ(u,v)) for uR, where R is the closed unit ball {uR2:|u|1}.

We know that at a point (u,v)R, the positively oriented unit tangent is given by T=(v,u). Thus at the pint G(u,v), the Stokes orientation points in the same direction as DG(u)(v,u)=(1001uϕ(u)vϕ(u))(vu)=(vuvuϕ(u,v)+uvϕ(u,v)). For instance, in Example 3, points in S have the form $ (u,v, u3-3uv2) $ and the Stokes orientation for the boundary at that point is given by (v,u,3(v33u2v). (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 S be a piecewise smooth surface in R3 with Stokes’ boundary S as described above, that S is oriented by the unit normal n and that S has the compatible (Stokes) orientation. Suppose also that F is any vector field that is C1 in an open set containing S. Then ScurlFndA=SFdx.

Sketch of proof.

Some ideas in the proof of Stokes’ Theorem are:

  1. As in the proof of Green’s Theorem and the Divergence Theorem, first prove it for S of a simple form, and then prove it for more general S by dividing it into pieces of the simple form, applying the theorem on each such piece, and adding up the results.

  2. In this case, the simple case consists of a surface S that can be parametrized by a single G:RR3 for RR2. More precisely, R is a regular region with piecewise smooth boundary, contained in an open set R0, and that G:R0R3 parametrizes a smooth or piecewise smooth surface. Then the idea of the proof is to use a change of variables to rewrite ScurlFndA and
    SFdx as integrals over R, then apply Green’s Theorem on R.

Examples and Consequences

Example 3.

Suppose that RR2 is a regular region with piecewise smooth boundary, and that S={(x,y,0):(x,y)R}. Suppose also that the unit normal n to S points upward.

Then S={(x,y,0):(x,y)R} and the orientation of S coincide with the positive orientation (from Green’s Theorem) of R.

Now if F is a vector field of the form F(x,y,z)=(F1(x,y),F2(x,y),0), then in this special case ScurlFndA =R(F2xF1y)dA and SFdx=R(F1(x,y),F2(x,y))dx, so in this special case, Stokes’ Theorem reduces exactly to Green’s Theorem.

Example 4.

Let C be the intersection of the cylinder x2+y2=1 and the surface z=ex2ysin(yx2), oriented counterclockwise. Compute CFdx for F(x,y,z)=(sinyz,x+xzcosyz,z+xycosyz).

Let S be the portion of the surface z=ex2ysin(yx2) bounded by the cylinder x2+y2=1, with the unit normal n oriented upwards. Then C=S, and the orientation for C is the correct one for Stokes’ Theorem. Thus CFdx=ScurlFndA. A computation shows that curlF=(0,0,1), and we know from Exercises in Section 5.3, or direct computation, that S(0,0,1)ndA = π. Thus CFdx=π

Closed Surfaces

A surface SR3 is said to be closed if it has no (Stokes’) boundary.

An example of such a surface is the unit sphere S={(x,y,z)R3:x2+y2+z2=1},n oriented outwards.

To explain what it means to “have no boundary” let’s define S1={(x,y,z)R3:x2+y2+z2=1,x0},S2={(x,y,z)R3:x2+y2+z2=1,x0}, (both with n oriented as for S) pictured below in blue and purple:
nobdy

Since S=S1S2, it is reasonable to believe that S=S1+S2. It is pretty clear that S1=S2={(x,y,z)S:x=0}. But since S1 and S2 have opposite orientations, we can suppose that they “cancel each other out”, resulting in “S=0”. 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 S in R3 is closed if it there exist subsets S1,S2 with piecewise smooth boundaries, such that

In contrast to the unit sphere, the surface S 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 ( S) will always remain “uncancelled”.

nobdy
If S is a piecewise smooth closed surface in R3 and F is a vector field that is C1 on an open set containing S, then ScurlFndA=0.

Given a closed surface S, let S1 and S2 be sets as in the definition of closed surface. Due to the additivity of integration and Stokes’ Theorem, ScurlFndA=S1curlFndA+S2curlFndA=S1Fdx+S2Fdx=0 because S1 and S2 have the opposite orientations, by assumption.

Example 5.

Let S denote the unit sphere S={(x,y,z)R3:x2+y2+z2=1}, and compute ScurlFndA for F=(yx2+y2+αz2,xx2+y2+αz2,0). Here α>0 is a real number and n is oriented outwards.

Solution. The integral equals zero, as a direct consequence of Theorem 2.

Remarks.

  1. Let R={xR3:|x|1}, and note that S=R (the ordinary non-Stokes boundary.) If F were C2 everywhere in R, then we could use the Divergence Theorem to compute ScurlFndA=RcurlFndA=RdivcurlFdV=0 since divcurlF=0 everywhere when F is C2. The significance of Theorem 2 is that ScurlFndA still equals zero even if, as in this example, F is not C2.

  2. One can check that if RR3 is a regular region with piecewise smooth boundary, then S=R is a closed surface. To verify this, one can fix pS and define S1={xS:|xp|ε} and S2={xS:|xp|ε} for some small ε>0. It is geometrically clear (but not very easy to prove) that these will satisfy the conditions in the definition of “closed surface”.

Moving the Surface

It follows immediately from Stokes’ Theorem that if S and S are two (oriented) surfaces such that S=S (with the same orientation), then ScurlFndA=SFdx=SFdx=ScurlFndA. This means that to evaluate an integral ScurlFndA, we can move the surface to a (suitable) new surface S.

Example 6.

Let S be the part of the cone z=2x2+y2 below the plane x+z=1, with the unit normal oriented upward, and evaluate ScurlFndA for F=(ex,1,(x+z1)2). Below: S, with its boundary shown in red

nobdy

Let S be the portion of the plane x+z=1 contained within the cone z=2x2+y2, with the unit normal oriented upwards, shown below, with its boundary in red.

nobdy
Then S=S={(x,y,z)R3:x+z=1, z=2x2+y2} (both oriented counterclockwise as seen from above). We can see that the boundaries coincides if we place both surfaces in the same picture:
nobdy
So Stokes’ Theorem implies that ScurlFndA=ScurlFndA. Also, curlF=(0,2(x+z1),0), and this equals 0 on S. We conclude that
ScurlFndA=0.

Problems

Basic

  1. Use Stokes’ Theorem to evaluate the integral CFdx ,F=(ex+y,eyx,sinz) where C is the intersection of the planes x=0,x=1,y=0,y=1 and the surface z=x+2y, oriented counterclockwise (as seen from above).

  2. Evaluate the integral CFdx ,F=(3x2yz,x3z,(x3+x)y) where C is the part of the plane z=x contained in the cylinder x2+y2=1. Suppose that C is oriented counterclockwise (as seen from above).

  3. Evaluate the integral SFdx ,F=(ycosxyyz,xcosxy+z,y) where S={(x,y,z)R3:0yx1,z=ex2}.

  4. Evaluate the integral ScurlFndA,F=(ecosxy,sinyzx2+y2+z4,xcosy) where S={(x,y,z)R3:x2+2y2+3z2=4} with n oriented outwards.

  5. Let S be the part of the ellipsoidal cone z=x2+(y/3)2 beneath the plane 4z+y=2, with n oriented upwards, and evaluate the integral ScurlFndA,F=(y,x,eyz).

  6. Evalute CFdx,F=(y+sin2x,xz+x,ez) where C is the curve formed by the intersection of the surfaces xz=1 and the (x5)2+(y+ey)2=4 (oriented counterclockwise, as seen from above.)

Advanced

  1. Find a piecewise smooth simple closed (oriented) curve C that maximizes CFdx,F=(y(z+1),x(z+1),0) among all curves that are constrained to lie on the surface of the unit sphere x2+y2+z2=1. A typical such curve is pictured below.

    curveonsphere

  2. Prove that if F is a C1 vector field and g is a C1 function, and if S is a surface satisfying the assumptions of Stokes’ Theorem, then SgcurlFndA=S(g×F)ndA+SgFdx.

  3. Suppose that G:R3R3 is a vector field such that G(x)x>0 for every x such that |x|=1. Prove that there does not exist any C1 vector field F such that G=×F.

    

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