1
$\begingroup$

I'm trying to figure out what $\iint_R \nabla\times\vec{F}\cdot d\textbf{S}$ means. I have a feeling that it has something to do with the classical Stokes' theorem. The Stokes' theorem that I have says $$ \int\limits_C W_{\vec{F}} = \iint\limits_S \Phi_{\nabla\times\vec{F}} $$ where $\vec{F}$ is a vector field, $W_{\vec{F}}$ is the work form of $\vec{F}$, and $\Phi_{\nabla\times\vec{F}}$ is the flux form of the curl of $\vec{F}$. Is the notation in question the same as the RHS of the above equation?

$\endgroup$
  • $\begingroup$ Yes, that's the integral of a curl $\nabla \times \mathbf F$ of a vector field $\mathbf F$ over the surface $R$ (in your integral) or $S$ in your alternative notation, and is mostly used in Stoke's theorem. $\endgroup$ – Kaster Jun 7 '13 at 10:04
1
$\begingroup$

It seems to me that the integrals $$\int\limits_C W_{\vec{F}}~~~~\text{and}~~~~\oint_{\mathfrak{C}}\vec{F}\cdot d\textbf{r}$$ have the same meanings. I don't know the notation $ \Phi_{\nabla\times\vec{F}}$, but if it means as $$\textbf{curl F}\cdot \hat{\textbf{N}} ~dS$$ then your answer is Yes.

$\endgroup$
  • $\begingroup$ I am wondering about your latter speculation. The notation $\Phi_{\nabla\times\vec{F}}$ would be $x\,dy\wedge dz - y\,dx\wedge dz + z\,dx\wedge dy$ if $\vec{F} = [x,y,z]^T$. $\endgroup$ – user59083 Jun 7 '13 at 10:28
  • 1
    $\begingroup$ @5space: Thank you for remarking me that. Honestly, I don't know that notation but I see now that is related to $n-$forms. :) $\endgroup$ – mrs Jun 7 '13 at 10:30

Your Answer

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