Let \(C\) be a circle in the \(xy\)-plane, oriented counterclockwise, and not passing through the origin. Prove that \[\oint_C \frac{-y}{x^2+y^2}\,dx + \frac{x}{x^2+y^2}\,dy\] equals \(0\) if the origin is outside the circle, and \(2\pi\) if the origin is inside the circle.
Solution
Let \[ P(x,y)=\frac{-y}{x^2+y^2}, \qquad Q(x,y)=\frac{x}{x^2+y^2}. \] Away from the origin, \[ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0. \]
If the origin lies outside \(C\), then \(P,Q\) are continuously differentiable on the disk bounded by \(C\). Green’s theorem therefore gives \[ \oint_C P\,dx+Q\,dy=0. \]
If the origin lies inside \(C\), choose a small positively oriented circle \(C_\varepsilon\) centered at the origin and contained inside \(C\). Apply Green’s theorem to the annular region between \(C\) and \(C_\varepsilon\). Its positively oriented boundary is \(C-C_\varepsilon\), so \[ \oint_C P\,dx+Q\,dy-\oint_{C_\varepsilon}P\,dx+Q\,dy=0. \] Parametrize \(C_\varepsilon\) by \[ x=\varepsilon\cos t,\qquad y=\varepsilon\sin t,\qquad 0\le t\le2\pi. \] Then \[ P\,dx+Q\,dy =\sin^2t\,dt+\cos^2t\,dt=dt. \] Hence \[ \oint_C P\,dx+Q\,dy =\int_0^{2\pi}dt =2\pi. \]