 Solution

For an oriented piecewise smooth surface $S$ with boundary $\partial S$, and a vector field $F$ continuously differentiable on a neighborhood of $S$, Stokes theorem statesHere $n$ is the chosen unit normal vector, $dS$ is the unsigned surface area element, and $dr$ is the tangent displacement along the boundary. The right-hand rule pairs the boundary orientation with $n$. If there are several boundary components, the line integral includes all of them with their induced orientations.

A simply connected surface is connected and every closed loop on it can be continuously contracted to a point while remaining on the surface. A disc is an example. An annulus is not simply connected: a loop winding once around its hole cannot contract within it.

Stokes' theorem is valid for both types of surface, under the smoothness and orientation hypotheses above. Simple connectivity is not a requirement of the theorem. For a concrete annular example, take $F=(-y/2,x/2,0)$ and $a\le r\le R$ in the horizontal plane. Its curl is $(0,0,1)$, so the left side is $\pi (R^2-a^2)$. The right side is the counterclockwise outer-circle integral $\pi R^2$ minus the counterclockwise inner-circle integral $\pi a^2$, giving exactly the same answer. Omitting the inner boundary would be an incorrect application of the theorem.