 Solution

Let $a$ be an arbitrary constant vector. By the cyclic identity for the scalar triple product,Apply Stokes theorem to $\mathbf{F}\times a$. Since $a$ is constant, the curl product identity givesConsequentlyFor precision, the operator in the last line has componentsThe partial derivatives act on $\mathbf{F}$, with $\mathbf{n}$ multiplying their values; they do not differentiate $\mathbf{n}$. Since the dot products agree for every constant $a$, the cross-product form of Stokes theorem follows:Here $\wedge$ in the paper denotes the cross product, and the printed surface integral has its area element $dS$ understood.