Does Stokes Thoerem say $\oint_{\partial R} \mathbf{F}\cdot d\mathbf{s} = \iint_R \nabla \times \mathbf{F} \cdot d\mathbf{S} $ for closed surfaces in $\mathbb{R}^3$?
My issue is that most statements of Stoke's theorem and proofs seem to imply that the surface does indeed have a non-empty boundary. Note I am not talking about applying Stoke's theorem to closed surfaces by breaking up the surface, hence showing that surface integral of a curl is always $0$.