# 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!

## 1 Answer

Yes, that is right. The boundary of the annulus between the two concentric circles is the union of the two circles, and the natural orientation is such that the outer circle is positively oriented, and the inner circle negatively, so

$$\int_{r < \lvert (x,y)\rvert < R} \operatorname{curl} F \,dS = \int_{\lvert (x,y)\rvert = R} F\cdot d\vec{s} - \int_{\lvert (x,y)\rvert = r} F\cdot d\vec{s}.$$

• Would I be right in thinking if both the circles were positively orientated, it would be the addition of the two integrals? – Mike Miller Nov 9 '13 at 23:07
• Yes, but that wouldn't be compatible with an orientation of the surface. Roughly speaking, if you draw small circles with arrows on the surface, such that all these circles "turn" in the same direction (clockwise or counterclockwise), when the circles approach the boundary curves, you get an induced orientation of the boundary curve. Here, the orientation induced from an orientation of the annulus is clockwise on one of the two boundary circles, and counterclockwise on the other. (Which is which depends on whether you look from above or below, and which orientation you choose.) – Daniel Fischer Nov 9 '13 at 23:14
• Thanks. I see what you're saying. I ask as the question I'm looking at asks to find the circulation of the two circles (positive orientation) and relate it to the flux. In this case the inner boundary has zero circulation anyway - so that would suggest orientation wouldn't matter anyway (I think). – Mike Miller Nov 9 '13 at 23:26
• 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
• That's what I was thinking. Thanks for the help. :-) – Mike Miller Nov 9 '13 at 23:34