Volume to surface integral of $R\times \nabla \times B$

Question

I need to transform the following integral into a surface integral (if that's possible): $$\int\int\int_\Omega R\times (\nabla \times A) dv = \int\int_{\partial \Omega} ? . {\bf n} da, $$ where $R = (x,y,z)$. Any idea?

Answer 1

Partial answer:

First, we know from our vector calculus identities that $$ \mathbf{r}\times(\nabla\times\mathbf{A}) = \nabla(\mathbf{r}\cdot\mathbf{A}) - \mathbf{r}\cdot\nabla\mathbf A - \mathbf{A}\cdot\nabla\mathbf{r}-\mathbf{A}\times(\nabla\times \mathbf{r}) = \nabla(\mathbf{r}\cdot\mathbf{A}) - \mathbf{r}\cdot\nabla\mathbf A - \mathbf{A}, $$ where we have used $\nabla\times\mathbf{r} = 0$ and $\nabla \mathbf{r} = I$. Now, we can mostly put this into divergence form using $\nabla\cdot (\mathbf{A}\mathbf{r}) = \mathbf{A}\nabla\cdot \mathbf{r} + \mathbf{r}\cdot\nabla\mathbf{A} = 3\mathbf{A} + \mathbf{r}\cdot\nabla\mathbf{A}$ and $\nabla f = \nabla\cdot(fI)$, giving $$ \mathbf{r}\times(\nabla\times\mathbf{A}) = \nabla \cdot[(\mathbf{r}\cdot\mathbf{A})I - \mathbf{A}\mathbf{r}] + 2\mathbf{A}. $$ Now we can apply the divergence theorem to get $$ \iiint_\Omega\mathbf{r}\times(\nabla\times\mathbf{A})dV = \iint_{\partial \Omega}\left[(\mathbf{r}\cdot\mathbf{A})\hat{n} - (\mathbf{r}\cdot\hat{n})\mathbf{A}\right]dS + 2\iiint_\Omega\mathbf{A}dV \\= \iint_{\partial \Omega}\left[\mathbf{r}\times(\hat{n}\times\mathbf{A})\right]dS + 2\iiint_\Omega\mathbf{A}dV $$ Unfortunately I'm not sure if there's a sneaky way to write $\mathbf{A}$ as the divergence of some tensor $\mathbf{T}$ that depends on $\mathbf{A}$ in a simple way. But it certainly makes the volume integral simpler.