# Stokes' theorem for an annulus

From Wiki, I'm looking at this definition:

"If the surface is not closed, it has an oriented curve as boundary. Stokes' theorem states that the flux of the curl of a vector field is the line integral of the vector field over this boundary. This path integral is also called circulation, especially in fluid dynamics. Thus the curl is the circulation density."

I'm looking to take the flux of an area between two oriented concentric circles in the plane $z=0$. Would I be right in thinking I can apply Stokes' theorem in this case as it's an open surface? If that's right, as there's a 'hole' in the middle, would the flux be the circulation of the outer boundary minus the circulation of the inner boundary?

Thanks!

$$\int_{r < \lvert (x,y)\rvert < R} \operatorname{curl} F \,dS = \int_{\lvert (x,y)\rvert = R} F\cdot d\vec{s} - \int_{\lvert (x,y)\rvert = r} F\cdot d\vec{s}.$$
• Right, orientation just gives a sign. And $(-1)\cdot 0 = 0$. But in principle, the answer would be "Flux = outer circulation - inner circulation". – Daniel Fischer Nov 9 '13 at 23:32