Stokes and Gauss' Divergence theorem on a closed smooth surface in $\Bbb R^3$ I have a problem understanding Stokes' Theorem and Gauss' Divergence Theorem. Suppose the following:
Let $F$ be a vector field in $\Bbb R^3$.
Let $S$ be an oriented closed smooth Surface enclosing a volume $V$ and let $C$ be a positively-oriented closed curve surrounding $S$
Stokes' Theorem says:
$$
 \int_C F·dr=\iint_S (\nabla \times F) · dS
$$
Then, by the Divergence Theorem:
$$
\iint_S (\nabla \times F)·dS = \iiint_V \nabla·(\nabla \times F)  dV
$$
But $ \nabla·(\nabla \times F)=0$. So everything is $0$
What is it that I am not seeing?
 A: While Daniel Fischer's answer probably is more direct, I find this way of showing that the circulation over a closed surface is zero (using Stokes theorem) more intuitive:

Here I split the closed surface $S$ into two surfaces $S_1$ and $S_2$ with a shared boundary, the curve $C_1$. If we apply Stokes theorem to the two surfaces separately we get:
$$\int{\int_{S_1} (\nabla \times \bar F) d \bar S} = \int_{C_1} {\bar F \cdot d \bar r}$$
and
$$\int{\int_{S_2} (\nabla \times \bar F) d \bar S} = - \int_{C_1} {\bar F \cdot d \bar r}$$
notice the negative sign before the right hand integral in the second equation. This is because the curve $C_1$ runs in the opposite direction to the normals of $S_2$ compared to $S_1$.
Combing the two surface integrals to get the integral over the entire surface,
$S = S_1 \cup S_2$:
$$
\begin{eqnarray}
\int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int{\int_{S_1} (\nabla \times \bar F) d \bar S} + \int{\int_{S_2} (\nabla \times \bar F) d \bar S} \\
\int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int_{C_1} {\bar F \cdot d \bar r} - \int_{C_1} {\bar F \cdot d \bar r} \\
&=& 0
\end{eqnarray}
$$
