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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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    If I take my hemisphere to be my surface (which is closed), then I say for my base (which is a circle), I'll attach a circle that is the boundary for the open surface of the open hemisphere or the disk itself. Then doesn't the Div Thrm still hold?2012-08-18

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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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    How could you integrate over the empty set (the boundary)? if it is closed? I am confused2012-08-23
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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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    Intuitively, yes, I suppose you could say that. I think it's more accurate, however, to say that the Divergence Theorem simply doesn't apply to the hemisphere (without cap), meaning that you shouldn't have gotten $0$ at all: the calculation should have stopped at Stokes' Theorem. Said another way, the hemisphere doesn't have a well-defined volume. And yes, I mentioned in my post that there are essentially 2 different definitions of "boundary" -- one for vector calculus (and differential geometry) and one for point-set topology.2013-01-01