Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 3 / 9A / c

On $C$, $dx=a{e}_{\varphi }d\varphi$, so $I={\int }_0^{2\pi }d\varphi =2\pi$. Locally $f=\varphi$, but no single-valued global potential exists on the punctured domain; this is why irrotationality does not force this closed integral to vanish.

The field is $\nabla f$ for $f=\rho \sin \varphi e^{-{\rho }^2-z^2}$. Hence $I=0$, either directly or by the fundamental theorem for line integrals.