Prove that $\oint_{\partial S} \psi \; d\ell = \iint_S (\hat{\mathbf{n}} \times \nabla \psi) \; dS$

In the Wikipedia article on vector calculus identities, we have the following

$$\oint_{\partial S} \psi \; d\ell = \iint_S (\hat{\mathbf{n}} \times \nabla \psi) \; dS$$

How do I prove this? I tried Stokes' theorem, to no avail. Perhaps there are some identities for exterior derivatives that I'm not aware of which may be useful.

• Err... That is the (Kelvin-)Stokes theorem... en.wikipedia.org/wiki/… – Eric Towers Jun 15 '14 at 4:30
• Kelvin-Stokes states that the surface integral of the curl is equal to the line integral on the boundary. I don't see why this reduces to that? – user908123 Jun 15 '14 at 5:03

Hint: Pick an arbitrary constant vector $e$, and take the inner product with it on both sides. Then use Stokes theorem to prove that $$\int_{\partial S}\psi\,(e,dl) = \int_S(e,dS\times \nabla \psi).$$
• Question: what does the volume form $dS$ mean here? It's certainly not the flux integral of $n \times \nabla \psi$. Where can I find a good reference for this? (I haven't seen this notation anywhere else) – user908123 Jun 15 '14 at 13:14
• @user908123 $dS$ is a vector pointing in the direction of the normal to surface, with its length equal to the area of the infinitesimal surface element. This is the same as $\mathbf{n}\,dS$ in another notation. – Kirill Jun 15 '14 at 14:16