= Solution
The identity $\nabla\cdot(\boldsymbol a\times\boldsymbol v)=\boldsymbol v\cdot(\nabla\times\boldsymbol a)-\boldsymbol a\cdot(\nabla\times\boldsymbol v)$ reduces to $-\boldsymbol a\cdot(\nabla\times\boldsymbol v)$ when $\boldsymbol a$ is constant. In components, this follows by differentiating $\varepsilon_{ijk}a_jv_k$ and using the antisymmetry of the <Levi-Civita symbol>. The <divergence theorem> and the cyclic <scalar triple product> identity now give
$$
\boldsymbol a\cdot\int_S\boldsymbol v\times d\boldsymbol S
=\int_S(\boldsymbol a\times\boldsymbol v)\cdot d\boldsymbol S
=-\boldsymbol a\cdot\int_V\nabla\times\boldsymbol v\,dV.
$$
Since $\boldsymbol a$ is arbitrary,
$$
\boxed{\int_S\boldsymbol v\times d\boldsymbol S=-\int_V\nabla\times\boldsymbol v\,dV.}
$$
The order of the <cross product> determines the minus sign, as in the <vector curl form of the divergence theorem>.
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