Vector curl form of the divergence theorem (${\int }_{\partial V}v\times ndS=-{\int }_V\nabla \times vdV$)

For a continuously differentiable vector field $v$ on a bounded sufficiently regular region $V$, apply the divergence theorem to $a\times v$ for an arbitrary constant vector $a$. The vector calculus identity $\nabla \cdot (a\times v)=-a\cdot (\nabla \times v)$ and the cyclic scalar triple product identity show that the dot products of both sides of the displayed formula with $a$ agree. Since $a$ is arbitrary, the vector integrals agree. Here $n$ is the outward unit normal vector. Reversing the order on the boundary gives the equivalent positive-sign formula $\int n\times vdS=\int \nabla \times vdV$.