Show identity about divergence theorem

Question

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

Answer 1

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.