= Solution
{c}
Apply the <divergence theorem> to $\mathbf F=u\nabla v-v\nabla u$. The <product rule> gives
$$
\nabla\cdot\mathbf F=u\nabla^2v-v\nabla^2u,
$$
because the two <gradient> <inner products> cancel. Also $\mathbf F\cdot\mathbf n=u\partial_nv-v\partial_nu$. Thus
$$
\boxed{\iiint_R(u\nabla^2v-v\nabla^2u)\,dV
=\iint_{\partial R}(u\partial_nv-v\partial_nu)\,dS.}
$$
This is <Green's second identity>, under the usual regularity assumptions permitting these derivatives and boundary <integrals>.
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