# Stokes' Theorem and Surfaces

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.
• 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