 Cross-product form of Stokes theorem

For a continuously differentiable vector field on an oriented surface with compatibly oriented boundary, this vector identity follows by dotting both sides with an arbitrary constant vector $a$ and applying Stokes theorem to $\mathbf{F}\times a$. The curl identity $\nabla \times (a\times \mathbf{F})=a\nabla \cdot \mathbf{F}-(a\cdot \nabla )\mathbf{F}$ gives the surface integrand. In components, $[(d\mathbf{S}\times \nabla )\times \mathbf{F}{]}_i=d{S}_j{\partial }_i{F}_j-d{S}_i{\partial }_j{F}_j$, with the derivatives acting on $\mathbf{F}$. For the position vector field $\mathbf{F}=\mathbf{r}$, it equals $-2d\mathbf{S}$; hence the line integral is minus twice the oriented vector area.