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

  • $\begingroup$ As far as I remember, proving the integral theorems in a general and rigorous way is rather non-trivial. I think I have seen them fill several pages and at least one or two lemmata in math textbooks. This has to do with the fact that the integration domains can be quite arbitrary. However, cheap almost-proofs (for special integration domains like cubes etc.) can be found in electrodynamics textbooks. $\endgroup$
    – oliver
    Commented Jun 4, 2021 at 22:28
  • $\begingroup$ @oliver Uhm, but here I don't have to prove divergence thm or similar, I just need to work out derivatives in the proper way so that I can apply divergence theorem. It should be only a matter of working in components $\endgroup$ Commented Jun 4, 2021 at 22:59
  • $\begingroup$ Okay, then sorry for misunderstanding your question. $\endgroup$
    – oliver
    Commented Jun 4, 2021 at 23:07
  • $\begingroup$ No worries :-) @oliver $\endgroup$ Commented Jun 4, 2021 at 23:19
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    $\begingroup$ What is $\otimes$ here? Is that supposed to be a cross product? $\endgroup$
    – DanielSank
    Commented Jun 5, 2021 at 0:24

1 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.

  • $\begingroup$ Thanks for your answer @user200143, but I still have a problem with the left hand side: $\int_{\partial R} \hat e_i \epsilon_{ijk} r_k v_k v_\ell dA_\ell $. I think the $r_k$ should be $r_j$, be definition of cross product, so the term should be $$\int_{\partial R} \hat e_i \epsilon_{ijk} r_j v_k v_\ell dA_\ell $$ $\endgroup$ Commented Jun 5, 2021 at 7:42
  • $\begingroup$ And second, I've never used the notation with $d A_l$, how is it defined? Could you please write it in terms of $n$? I have the suspicion that it's $n_l dA$ @user200143 $\endgroup$ Commented Jun 5, 2021 at 7:46
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    $\begingroup$ I apologize for the typos and not defining $\partial_i$. I have edited the answer accordingly. $\endgroup$
    – user200143
    Commented Jun 5, 2021 at 19:05
  • $\begingroup$ No worries. $\partial_i$ was cristal clear! What I didn't know was $dA_l$ but now that's clear too :-) $\endgroup$ Commented Jun 9, 2021 at 10:31

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