Show identity about divergence theorem I'm trying to show the following integral equality, but I really can't come up with a proof. The context here is the one of an introductive book to continuum mechanics, so everything is smooth and there are no differetiability problems.

Let $R$ be a control volume in the space, with $v$ a velocity field and $r(x)=x-0$ the position vector. Then $$\int_{\partial R}r \times (v\cdot n)vdA = \int_R r \times \operatorname{div}(v \otimes v)dV$$

I'm trying to work with components, so I take the divergence of the l.h.s $r \times (v \cdot n) v$, which has i-th component $\varepsilon_{ijk} x_j v_ln_lv_k$ and hence I take the $i$-th derivative, since I want to compute its divergence:
$$\varepsilon_{ijk}  \frac{\partial}{\partial x_i}  \bigl(x_j v_ln_lv_k \bigr) $$ but after this I really get lost in the computations. How should I move from here? Is there a clever way to move?
Any help is highly appreciated
 A: In order to use the divergence theorem correctly, you need to
know the vector or tensor that is dotted with the directed area element.
Since $n$ does not appear on the right-hand side, I assume it
is the unit vector in the direction of the area element.
In components, using your notation, with unit vectors written
as $\hat e_i$, and $\partial_\alpha \equiv \frac{\partial}{\partial x_\alpha}$, the left-side integral is then
\begin{equation}
\int_{\partial R} \hat e_i \epsilon_{ijk} x_j v_k v_\ell dA_\ell
\end{equation}
where $dA_\ell$ is $n_\ell dA$ the $\ell$th component of the directed area element.
Applying the divergence theorem
\begin{equation}
\int_{\partial R} \hat e_i \epsilon_{ijk} x_j v_k v_\ell dA_\ell
= \int_R \partial_\ell\left [\epsilon_{ijk} x_j v_k v_\ell \hat e_i\right]
dV \,.
\end{equation}
The derivative of $x_j$ is $\partial_\ell x_j = \delta_{\ell j}$,
and that term gives $\epsilon_{ijk}v_k v_j\hat e_i$ which is zero. The
remaining term becomes
\begin{equation}
\int_{\partial R} \hat e_i \epsilon_{ijk} x_k v_k v_\ell dA_\ell
= \int_R x_j \hat e_i \epsilon_{ijk}
\partial_\ell\left [v_k v_\ell \right]
dV \,.
\end{equation}
Writing $\partial_\ell [v_k v_\ell]$ as the $k$th component of
$\vec \nabla\cdot (v \otimes v)$ gives the result.
