0
$\begingroup$

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

$\endgroup$
0

1 Answer 1

Reset to default
3
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ 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, 2014 at 12:12
  • $\begingroup$ @MarkFantini , can't there be an C that is the boundary of a closed S? $\endgroup$
    – DifferentialEquations
    Aug 20, 2014 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, 2014 at 21:01
  • $\begingroup$ Okay, thanks for the great explanation! @dylan7 $\endgroup$
    – DifferentialEquations
    Aug 21, 2014 at 3:07
  • $\begingroup$ 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. $\endgroup$
    – amWhy
    May 12, 2021 at 21:55

Not the answer you're looking for? Browse other questions tagged .