0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Err... That is the (Kelvin-)Stokes theorem... en.wikipedia.org/wiki/… $\endgroup$ – Eric Towers Jun 15 '14 at 4:30
  • $\begingroup$ 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? $\endgroup$ – user908123 Jun 15 '14 at 5:03
1
$\begingroup$

Hint: Pick an arbitrary constant vector $e$, and take the inner product with it on both sides. Then use Stokes theorem to prove that $$ \int_{\partial S}\psi\,(e,dl) = \int_S(e,dS\times \nabla \psi). $$

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – user908123 Jun 15 '14 at 13:14
  • $\begingroup$ @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. $\endgroup$ – Kirill Jun 15 '14 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.