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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 $0$.

helios321
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  • It's hard for me to imagine what you are talking about, then. – Martin Argerami Apr 13 '20 at 06:37
  • @MartinArgerami e.g. https://math.stackexchange.com/questions/183007/improper-use-of-stokes-and-divergence-theorem-find-the-problem?rq=1 seems to be applied here. – helios321 Apr 13 '20 at 07:02
  • You explicitly say in your question that you are not talking about that. – Martin Argerami Apr 13 '20 at 07:27
  • See here: https://math.stackexchange.com/questions/3606419/using-stokes-theorem-to-show-the-integral-of-curl-f-is-zero-over-closed-s/3606579#3606579 – Christian Blatter Apr 13 '20 at 08:38
  • @MartinArgerami What I mean to say is that Stoke's theorem only makes sense for open surfaces with piecewise smooth boundary, and that to make a conclusion about a closed surface we could apply the theorem to reach a corollary. – helios321 Apr 13 '20 at 09:05
  • @ChristianBlatter My real issue is whether Stokes theorem can be applied in the following manner: https://imgur.com/sMBnwnx where $S$ is the surface of the unit sphere. – helios321 Apr 13 '20 at 09:12

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