Related Stokes' Theorem

Question

How would you prove the following using Stokes's Theorem? $$ \int_{S}(d\vec{S}\times\vec{\nabla})\times\vec{v}=\oint_{C}d\vec{l}\times\vec{v} $$ I know you pretty much should use a substitution that involves $\vec{a}=\vec{v}\times\vec{c}$ where $\vec{c}$ is a constant vector. Hence doing Stokes' Theorem on $\vec{a}$ you should get the above statement. I get stuck and don't know what cross-product property I should use, any suggestions?

Answer 1

Assuming $$ \int\limits_S \mbox{rot } F \cdot dS = \int\limits_{\partial S} F \cdot dx $$ is the Stokes theorem you have in mind.

We could now try to use the identities for the vector product (anticommutativity, bac-cab rule), but because of the nabla operator involved, which is not commutative in general, I prefer to check this identity component-wise: \begin{align} e_1 \cdot \int\limits_{\partial S} dx \times v &= \int\limits_{\partial S} \epsilon_{1jk} dx_j v_k \\ &= \int\limits_{\partial S} dx_2 v_3 - dx_3 v_2 \\ &= \int\limits_S \mbox{rot } (0, v_3, -v_2)^T \cdot dS \\ &= \int\limits_S dS_1 (\partial_2 (-v_2) - \partial_3 v_3) + dS_2(-\partial_1 (-v_2)) + dS_3(\partial_1 v_3) \\ &= \int\limits_S (dS_3\partial_1-dS_1\partial_3)v_3 + (dS_2\partial_1-dS_1\partial_2)v_2 \\ &= \int\limits_S (dS\times \nabla)_2 v_3 - (dS\times \nabla)_3 v_2 \\ &= e_1 \cdot \int\limits_S (dS\times \nabla) \times v \end{align} We could do the other two components as well, but I am too lazy for this.

Answer 2

Hint: you should use the following Jacobi Identity: $$ \vec x\times(\vec y\times \vec z)+\vec y\times(\vec z\times \vec x)+\vec z\times(\vec x\times \vec y)=0. $$ and the fact that $\nabla\times\vec c=0$ for a constant vector function $\vec c$.