TITLE: Proving that $\iint_S (\nabla \times F) \cdot \hat{n} dS =0$
QUESTION [8 upvotes]: I have the following question:

Prove that $$\iint_S (\nabla \times \vec{F}) \cdot \hat{n} dS =0$$  for any closed surface $S$ and twice differentiable vector field $\vec F:\mathbb{R^3} \to \mathbb{R^3} $ . 

I need to prove this using Stokes' theorem. 
The only thing I want to verify is whether or not for every closed surface $S$, we have:
 $$\iint_S (\nabla \times \vec{F}) \cdot \hat{n} dS =\int_C \vec F \cdot d\vec r$$ and the last term is trivially zero, because $C=\emptyset $ ($S$ is a closed surface).
Is this correct?
Thanks in advance

REPLY [6 votes]: Yes you can use Stokes theorem but as well you can use Gauss(divergence) theorem
$$\iint_S (\nabla \times \vec{F}) \cdot \hat{n} dS =\iiint_{\text{Interior}(S)} 
\nabla \cdot ( \nabla \times \vec{F})  dV$$
But divergence of curl is identically zero ie
$$ \nabla \cdot ( \nabla \times \vec{F}) = 0$$