Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 12 / d

Solution

(r, θ, ϕ)

centred on the right-hand sphere, with polar axis pointing outwards along

s

r^{2} sin θ

, and the retained region is

r \leq R

with

r cos θ \geq - d

For

0 \leq r \leq d

, every polar angle

0 \leq θ \leq π

is allowed. For

d < r \leq R

, the cut requires

0 \leq θ \leq arccos (- d / r)

. Splitting the volume integral at

r = d

therefore gives

M_{e} = = = = ρ \int_{0}^{2 π} [\int_{0}^{d} \int_{0}^{π} r^{2} sin θ d θ d r + \int_{d}^{R} \int_{0}^{a r c c o s (- d / r)} r^{2} sin θ d θ d r] d ϕ 2 π ρ [2 \int_{0}^{d} r^{2} d r + \int_{d}^{R} r^{2} (1 + \frac{d}{r}) d r] 2 π ρ [\frac{2 d ^{3}}{3} + \frac{R ^{3} - d ^{3}}{3} + \frac{d}{2} (R^{2} - d^{2})] π ρ (\frac{2 R ^{3}}{3} + R^{2} d - \frac{d ^{3}}{3}) .

Substitution of

R = β a

and

d = (α - γ) a

reproduces the mass in part (c). The radial split avoids integrating over a polar angle range independent of

r

, which would incorrectly include part of the removed spherical cap.

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