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?

Best Regards,

Thank You


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.

  • $\begingroup$ I think what he means is $S$ being a topologically closed set of $\mathbb R^3$, but not a manifold without boundary. $\endgroup$ – Troy Woo Aug 20 '14 at 12:12
  • $\begingroup$ @MarkFantini , can't there be an C that is the boundary of a closed S? $\endgroup$ – DifferentialEquations Aug 20 '14 at 18:34
  • 1
    $\begingroup$ @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 $\endgroup$ – dylan7 Aug 20 '14 at 21:01
  • $\begingroup$ Okay, thanks for the great explanation! @dylan7 $\endgroup$ – DifferentialEquations Aug 21 '14 at 3:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.