= Vector curl form of the divergence theorem
{title2=$\displaystyle\int_{\partial V}v\times n\,dS=-\int_V\nabla\times v\,dV$}
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 $\int n\times v\,dS=\int\nabla\times v\,dV$.
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