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

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- Solution ii

Solution

For the lower semicircle

z (x) = - a^{2} - x^{2}, 1 + z^{'2} = \frac{a}{- z} .

For a general potential, the Beltrami first integral is

\frac{ρ Φ ( z ) + λ}{1 + z ^{'2}} = C .

Substitution of the semicircle therefore requires

- \frac{z}{a} (ρ Φ (z) + λ) = C .

Thus, up to an arbitrary additive constant and a multiplicative strength,

Φ (z) = Φ_{0} + \frac{A}{z} .

The additive constant is absorbed by the length multiplier. The semicircle has length

πa

, as required.

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