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
Best Answer
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.