# Improper use of Stokes and Divergence Theorem. Find the problem

Could someone point out what is wrong with this equality? Assume that $\mathbf{F}$ is continuous (and hence, its partial derivatives).

\begin{align} \oint \mathbf{F}\cdot d\mathbf{s} & =^\text{by Stokes} \iint_S \nabla \times \mathbf{F} \cdot d\mathbf{S} \\ &=^\text{by Div} \iiint_V \nabla\cdot( \nabla \times \mathbf{F} ) \, dV \\ &=\iiint_V 0 \,dV \\ &=0\\ &\implies \oint \mathbf{F}\cdot d\mathbf{s}= 0 \; \forall \mathbf{F} \end{align}

Since we assumed $\mathbf{F}$ and its partials are all continuous. But obviously this is wrong if $\mathbf{F}$ is non-conservative. But everything seems to agree. What went wrong?

EDIT. For a refinement of the problem. Let me specifically state that $S$ is a closed surface with a boundary curve that is also closed. So $V$ here is the volume of that surface and since $S$ is closed it has a volume

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$S$ is not the boundary of a volume. Edit: Or, as Schmitty says, $\partial S$ is empty. –  Neal Aug 15 '12 at 21:28
Isn't $V$ is the volume over the entire enclosed surface? –  Hawk Aug 15 '12 at 21:30
You need to be more explicit about what curve/surface/volume you are integrating over. Perhaps take a simple example, say a circle, and tell us what you think the domains of the integrals are. –  Rahul Aug 15 '12 at 22:02
@RahulNarain, take a circle to be the boundary. Then I attach a hemisphere to that boundary to make my surface. My volume integral will integrate the volume of that hemisphere –  Hawk Aug 15 '12 at 22:21
@jak If you attach a hemisphere without base, you cannot use the divergence theorem. If you attach a hemisphere with base, then your surface is closed, hence the boundary is empty, not the circle you started with. –  Tunococ Aug 15 '12 at 22:53

Actually nothing is wrong with that. You start with a vector field integrated over a closed curve. Your first equality which does use Stokes's Theorem goes to an integral over a surface S for which your original curve must be the boundary. Your next equality uses the divergence theorem and goes to an integral over a volume for which your surface S must be the boundary implying S is a closed surface. Since your assumptions indicate that S is a closed surface S doesn't have a boundary- or rather, the boundary of S is the empty set. So the integral you started with is over the empty set----> hence it's zero.

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But like Tunococ said, my surface is actually open –  Hawk Aug 16 '12 at 0:01
As Tunococ said for an open surface the divergence theorem doesn't apply which gives you the problem you were looking for. In the case of a closed surface the argument is valid. Whatever F is if you integrate it over the empty set you're going to get zero. –  Schmitty Aug 17 '12 at 16:47
How could you integrate over the empty set (the boundary)? if it is closed? I am confused –  Hawk Aug 23 '12 at 21:58

This is one of my favorite phenomena in multivariable calculus. I remember noticing this when I was first learning the subject, and spent many hours wondering how this could be.

The explanation for this phenomenon lies in the following geometric principle:**

The boundary of a boundary is empty.

This geometric fact is in some sense "dual" to the fact that $\text{div}(\text{curl}\,\mathbf{F}) = 0$ for all $\mathbf{F}$.

In particular, if you have a volume $V$ that bounds a surface $S$, then the surface $S$ cannot have a boundary curve. Said another way, the boundary curve $C = \partial S$ is the empty set, so integrating anything over it is zero.

Example: In the comments, you consider a solid hemisphere $V$. The boundary of $V$ will then be the surface of the hemisphere and also the disc base. This closed surface (consisting of both the hemisphere surface and the disc base) does not have a boundary curve.

On the other hand, the surface which is just the hemisphere (without the base) does have a boundary curve: namely, the circle. However, this surface cannot be said to enclose any volume.

Note 1: Typically when one talks about a "closed" surface, one specifically means a surface which does not have a boundary curve. This is an unfortunate piece of terminology since the term "closed" can also refer to being a closed subset of $\mathbb{R}^3$, and these two definitions are not equivalent.

For instance, the hemisphere together with its boundary curve (but not including the disc base) is a closed as a subset of $\mathbb{R}^3$, but is generally not called a "closed surface." However, the hemisphere together with the disc base is a closed surface (and is also closed as a subset of $\mathbb{R}^3$).

** Note 2: This principle is somewhat vague as stated. In order to make it precise, one needs to rigorously define the notion of "boundary." This can be done in a couple of ways; some definitions will satisfy this principle, while others won't. For now, let's not get into these details.

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This closed surface (consisting of both the hemisphere surface and the disc base) does not have a boundary curve. Isn't it a circle? –  Hawk Dec 23 '12 at 3:49
No. I understand your confusion, though a simple explanation escapes me... All I can say is: topologically speaking, the hemisphere together with the disc base is like a sphere that has been squashed and then pinched. Certainly you wouldn't say that a sphere has a boundary curve, right? Same thing here. –  Jesse Madnick Dec 23 '12 at 17:34
As another example, a (circular) cylinder together with both caps does not have a boundary curve. However, a cylinder with no caps has a boundary consisting of 2 (disjoint) circles. Still different is a cylinder together with one cap, which has boundary consisting of 1 circle (namely the circle that isn't capped off). –  Jesse Madnick Dec 23 '12 at 17:36
Isn't the cap bounded by a circle anyways? –  Hawk Dec 25 '12 at 1:05
Yes, the boundary of the cap is a circle. But the boundary of a cylinder together with one cap consists of 1 circle -- namely, the circle that is the edge of the uncapped side. Is all of this clear? –  Jesse Madnick Dec 25 '12 at 1:11