 Solution

Let $S$ be a compact piecewise smooth oriented surface, $\mathbf{n}$ its chosen unit normal vector, and $\partial S$ its entire oriented boundary. Let $\mathbf{A}$ be a continuously differentiable vector field on a neighborhood of $S$. Stokes theorem statesHere $\nabla \times \mathbf{A}$ is the curl, $dS$ is the surface area element, and $d\mathbf{r}$ is the directed line integral element. The boundary follows the right-hand rule relative to $\mathbf{n}$; equivalently the surface lies to the left when walking along its boundary with one's head toward $\mathbf{n}$.

The topological requirement is orientability, together with the regularity and boundary assumptions above. Simple connectivity is unnecessary. An annulus or cylindrical side is allowed, provided every boundary component is included with its induced direction. For a concrete counterexample to a disk-only restriction, take the planar annulus $1\le x^2+y^2\le 4$ and $\mathbf{A}=(-y/2,x/2,0)$. The curl is ${\mathbf{e}}_z$, the surface integral is $3\pi$, and the outer counterclockwise and inner clockwise line integrals sum to $4\pi -\pi =3\pi$.

For $S$ in the $xy$ plane with upward normal vector and $\mathbf{A}=(P,Q,0)$, its curl has vertical component $Q_x-P_y$. Thus Stokes theorem becomes Green's theorem:The outer boundary is counterclockwise and any inner boundaries are clockwise.