Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 2 / 13D / b / ii / Solution

For the lower semicircleFor a general potential, the Beltrami first integral isSubstitution of the semicircle therefore requiresThus, up to an arbitrary additive constant and a multiplicative strength,The additive constant is absorbed by the length multiplier. The semicircle has length $\pi a$, as required.