Showing $\iiint_R \mathbf{r} \times \nabla\phi \,dV = \iint_{\partial R} \mathbf{r} \times \phi \mathbf{n} \,dS$ using divergence theorem 
Question:
  Let $\phi$ be a smooth scalar field defined on a region $R \subseteq \mathbb{R}^{3}$ with a smooth boundary $\partial R$. Then show that the following identity holds:
$$\iiint_R \mathbf{r} \times \nabla\phi  \,dV = \iint_{\partial R} \mathbf{r} \times \phi \mathbf{n} \,dS$$

In my course up until this point the divergence theorem has been covered, and I can see the similarities between the integrals above and the divergence theorem identity but am unsure how to go from $\nabla . F$ in the divergence theorem to $\mathbf{r} \times \nabla\phi$ above.
How should one begin to attempt showing an identity like this given they have use of the divergence theorem?
 A: While the solution above is elegant and correct, I believe the following approach better demonstrates how the identity was obtained in the first place. 
Consider setting $\mathbf{F} = \phi\mathbf{c} \times \mathbf{r}$ in the Divergence Theorem as you quoted above, where $\mathbf{c}$ is an arbitrary yet constant vector field. By the divergence product rule for a scalar field and a vector field:  
$ \nabla \cdot (\phi\mathbf{c} \times \mathbf{r}) = \nabla \phi \cdot (\mathbf{c} \times \mathbf{r}) + \phi \nabla \cdot (\mathbf{c} \times \mathbf{r}) = \mathbf{c} \cdot (\mathbf{r} \times \nabla \phi)$, by the cyclic properties of the triple vector product and the fact that $ \nabla \cdot (\mathbf{c} \times \mathbf{r}) = \mathbf{0}$ (check this directly). 
Now note from the Divergence Theorem for the vector field $\mathbf{F}$: 
$ \iiint_R \mathbf{c} \cdot (\mathbf{r} \times \nabla \phi)  \,dV = \iint_{\partial R} (\phi\mathbf{c} \times \mathbf{r}) \cdot \mathbf{n} \,dS $
$ \implies \mathbf{c} \cdot \iiint_R  (\mathbf{r} \times \nabla \phi)  \,dV = \mathbf{c} \cdot \iint_{\partial R} \mathbf{r} \times \phi \mathbf{n}\,dS $ 
where the right hand side of the last expression follows again from the cyclic properties of the triple product. Recalling $\mathbf{c}$ is arbitrary, we're done.
A: In Cartesian coordinates we have $\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k},$  and, without showing every term, 
$$\iiint_R\mathbf{r} \times \nabla \phi \, dV  \\ = \left(\iiint_R y \frac{ \partial\phi}{\partial z} \, dV -  \iiint_R z \frac{ \partial\phi}{\partial y} \, dV \right)\mathbf{i} + \cdot  \cdot \cdot \\ = \left(\iiint_R \frac{ \partial }{\partial z}(y \phi) \, dV -  \iiint_R \frac{ \partial}{\partial y}(z \phi) \, dV \right)\mathbf{i} + \cdot  \cdot \cdot \\ =  \left(\iiint_R \nabla \cdot (y \phi \mathbf{k}) \, dV -  \iiint_R \nabla \cdot (z \phi \mathbf{j}) \, dV \right)\mathbf{i} + \cdot  \cdot \cdot \\ = \left(\iint_{\partial R} (y \phi \mathbf{k}) \cdot \mathbf{n} \, dS -  \iint_{\partial R} (z \phi \mathbf{j}) \cdot \mathbf{n} \, dS \right)\mathbf{i} + \cdot  \cdot \cdot  \\ = \left(\iint_{\partial R} y \phi n_z \, dS -  \iint_{\partial R} z \phi n_y \, dS \right)\mathbf{i} + \cdot  \cdot \cdot \\ =   \iint_{\partial R} \mathbf{r} \times \phi \mathbf{n}  \, dS.  $$
The divergence theorem familiar to you was applied in going from the third to fourth step.
Addendum 
A proof with much more compact notation uses the component form of the divergence theorem
$$\int_R \partial_i f \, dV = \int_{\partial R} f \,n_i \, dS,$$
and the expression for curl and cross-product using the Levi-Cevita symbol
$$(\mathbf{r} \times \nabla \phi)_i = \varepsilon_{ijk} \,r_j \,\partial_k \phi, \\ (\mathbf{r} \times \phi \mathbf{n})_i = \varepsilon_{ijk}\,r_j\phi \,n_k.$$
Thus,
$$\begin{align} \int_R (\mathbf{r} \times \nabla \phi)_i \, dV &= \int_{\partial R} \varepsilon_{ijk} \,r_j \,\partial_k \phi \, dS \\  &= \varepsilon_{ijk}\int_{\partial R}  \ \,\partial_k (r_j\phi) \, dS \\ &= \varepsilon_{ijk}\int_{\partial R}   \, r_j\,\phi \,n_k \, dS \\ &= \int_{\partial R}  \varepsilon_{ijk} \, r_j\,\phi \,n_k \, dS  \\ &= \int_{\partial R}  (\mathbf{r} \times \phi \mathbf{n})_i \, dS \end{align}$$
