Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 3 / 10C / Solution

{title2=$\omega=(-y\,dx+x\,dy)/(x^2+y^2)$}

On the <punctured plane>, this <one-form> is closed and locally the differential of the polar angle. Its <line integral> around a closed curve is $2\pi$ times the <winding number>. A once-winding circle therefore has nonzero integral, proving that no global single-valued angle potential exists and the form is not an <exact differential form>.

For a counterclockwise <circle> of radius $R>0$, substitute $x=R\cos t$, $y=R\sin t$. The <pullback of a differential form> is
$$
\omega=\frac{R^2(\sin^2t+\cos^2t)}{R^2}\,dt=dt,\qquad \oint\omega=\boxed{2\pi}.
$$
The radius cancels. In coordinates, $P=-y/(x^2+y^2)$ and $Q=x/(x^2+y^2)$ satisfy $\partial_xQ=\partial_yP$ away from the origin, so $d\omega=0$ there. The nonzero integral is a global obstruction to being an <exact differential form>, not a nonzero local <exterior derivative>.

\Image[/differential-form-circulation.png]
{title=Smooth local circulation scales with enclosed area, whereas a closed angular form detects winding around an excluded point}

Compare this with the smooth <one-form> $\eta=(-y\,dx+x\,dy)/2$ on the whole plane. Here $d\eta=dx\wedge dy$, and its integral around a radius-$R$ <circle> is $\pi R^2$, the enclosed area. The <Generalized Stokes theorem> applies to the full disc for $\eta$. For $\omega$, that disc crosses the excluded origin, where the <one-form> is undefined. On an annulus the theorem applies, but its induced inner boundary is clockwise: the two integrals $2\pi$ and $-2\pi$ cancel. This explains exactly why closedness and nonzero circulation can coexist.

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