= Solution
Since $\mathbf a$ is constant, the <product rule> gives $\nabla\cdot(\mathbf a\psi)=\mathbf a\cdot\nabla\psi$. Applying the <divergence theorem> therefore gives
$$
\mathbf a\cdot\int_V\nabla\psi\,dV=\int_S\psi\,\mathbf a\cdot d\mathbf S=\mathbf a\cdot\int_S\psi\,d\mathbf S.
$$
This holds for every constant <vector> $\mathbf a$, so each component of the difference vanishes. Consequently the <gradient> has the boundary representation
$$
\boxed{\int_V\nabla\psi\,dV=\int_S\psi\,d\mathbf S.}
$$
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