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Stokes' Theorem states the following: \begin{equation*} \oint_c \textbf{F}\centerdot d\textbf{r}= \int\int_S (\nabla \times\textbf{F})\centerdot nd \textbf{S}\end{equation*} for a given C that is the boundary of a surface S.

Can $S$ be a closed surface, where c is the boundary, given that n= the unit normal vector correctly oriented?

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Thank You

DifferentialEquations
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Yes, $S$ can be a closed surface. In that case $$\iint\limits_S (\nabla \times {\mathbf F}) \cdot {\mathbf n} \, dS = 0$$ because we consider the boundary of $S$ to be empty.

Mark Fantini
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  • I think what he means is $S$ being a topologically closed set of $\mathbb R^3$, but not a manifold without boundary. – Troy Woo Aug 20 '14 at 12:12
  • @MarkFantini , can't there be an C that is the boundary of a closed S? – DifferentialEquations Aug 20 '14 at 18:34
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    @DifferentialEquations: Boundries cannot be defined for a closed surface, but let's think about it and act like there is one...So we have an open surface and its boundry, let's say oriented counter clockwise. Now to close it we have to choose an arbitrary surface with the same boundry oriented clockwise, because that's the only way we can close it. Sum the boundries ccw-cw=0 of the same boundry...stokes theorem – dylan7 Aug 20 '14 at 21:01
  • Okay, thanks for the great explanation! @dylan7 – DifferentialEquations Aug 21 '14 at 03:07
  • Mark, search for duplicates of questions that were likely asked and answered previously: This question seems not to meet the standards for the site. Instead of answering it, why not look for a good duplicate target, or help the user by posting comments suggesting improvements? Please also read the meta announcement regarding quality standards. – amWhy May 12 '21 at 21:55
  • @amWhy I answered this question back in 2014. I agree that both question and answer were less than desired quality though. – Mark Fantini May 13 '21 at 22:45