 Vector-valued Stokes identity

For an oriented smooth surface and compatible boundary orientation, apply Stokes theorem to $\varphi \mathbf{k}$ for an arbitrary constant vector $\mathbf{k}$. Since $\nabla \times (\varphi \mathbf{k})=\nabla \varphi \times \mathbf{k}$, its surface integral is $\mathbf{k}\cdot {\int }_{S}d\mathbf{S}\times \nabla \varphi$. The boundary integral is $\mathbf{k}\cdot \oint \varphi d\mathbf{x}$. Equality for every $\mathbf{k}$ proves the displayed vector identity. Reversing both orientations preserves the identity.