# 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?

• If $S$ encloses $V$, then there is no curve $C$ surrounding $S$, since $S$ is a closed surface (or a "sum" of several such). – Daniel Fischer Feb 23 '14 at 21:28
• Thanks, I know understand how Stockes' theorem works, I was quite confused! – Francisco Feb 24 '14 at 13:05

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, $S = S_1 \cup S_2$:

$$\begin{eqnarray} \int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int{\int_{S_1} (\nabla \times \bar F) d \bar S} + \int{\int_{S_2} (\nabla \times \bar F) d \bar S} \\ \int{\int_{S} (\nabla \times \bar F) d \bar S} &=& \int_{C_1} {\bar F \cdot d \bar r} - \int_{C_1} {\bar F \cdot d \bar r} \\ &=& 0 \end{eqnarray}$$