Identity with integral of $f\textbf{r} \times \text{d}\textbf{S}$ Let $f$ be a smooth scalar field defined on a region $R \subseteq \mathbb{R}^3$ with smooth boundary $\partial R$. Show that $$\iint\limits_{\partial R} f\textbf{r} \times \text{d}\textbf{S} = \iiint\limits_{R} \textbf{r} \times \nabla f \ \text{d}V$$
If the left-hand side had $f\textbf{r} \cdot \text{d}\textbf{S}$, we could use Gauss's divergence theorem. But I'm not sure how to do it as it has $f\textbf{r} \times \text{d}\textbf{S}$. Any assistance would be much appreciated.
 A: Note that $f \mathbf{r} \times d\mathbf{S} = f \mathbf{r} \times \mathbf{n}\, dS$ where $\mathbf{n}$ is the outward unit normal vector and $dS$ is the differential element of surface area.  This cross product can be expressed in terms of components as
$$f \mathbf{r} \times \mathbf{n}= \varepsilon_{ijk}fr_j n_k \, \mathbf{e}_{i},$$
where $\varepsilon_{ijk}$ is the Levi-Civita symbol and the Einstein summation convention implying summation over repeated indexes is used.
We can apply the divergence theorem in component form to obtain
$$\tag{*}\int_{\partial R}f \mathbf{r} \times d\mathbf{S} = \int_{\partial R}\varepsilon_{ijk}fr_j n_k\, dS \, \mathbf{e}_{i} = \int_R\varepsilon_{ijk}\frac{\partial(f r_j)}{\partial r_k}\, dV \, \mathbf{e}_{i}\\ = \int_R\varepsilon_{ijk}r_j\frac{\partial f}{\partial r_k}\, dV \, \mathbf{e}_{i}+ \int_R\varepsilon_{ijk}f\frac{\partial r_j}{\partial r_k}\, dV \, \mathbf{e}_{i}$$
The first integral on the RHS of (*) is 
$$\int_R\varepsilon_{ijk}r_j\frac{\partial f}{\partial r_k}\, dV \, \mathbf{e}_{i} = \int_R \mathbf{r} \times \nabla f \, dV,$$
and since $\displaystyle\frac{\partial r_j}{\partial r_k} = \delta_{jk}$ is the Kronecker delta, the second integral is
$$\int_R\varepsilon_{ijk}f\frac{\partial r_j}{\partial r_k}\, dV \, \mathbf{e}_{i} = \int_R\varepsilon_{ijk}f\delta_{jk}\, dV \, \mathbf{e}_{i} = \int_R\varepsilon_{ijj}f\, dV \, \mathbf{e}_{i} = \mathbf{0},$$
since $\delta_{jk} = 0$ if $j \neq k$ and $\varepsilon_{ijk} \delta{jk} = \varepsilon_{ijj} \delta{jj} = 0 \cdot 1 = 0$ if $j = k.$
Therefore,
$$\int_{\partial R}f \mathbf{r} \times d\mathbf{S} = \int_R \mathbf{r} \times \nabla f \, dV$$
