 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-4/11c/b/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

z

from the apex along the cone's symmetry axis. A thin disc at height

z

s (z) = \frac{R}{H} z

d M = ρ π s^{2} d z

. Its moment about a parallel diameter through its centre is

s^{2} d M /4

, while its centre is distance

z

from the required axis. Part (a)(ii) gives

d I = (z^{2} + \frac{s ^{2}}{4}) d M .

Hence

I = ρ π \int_{0}^{H} \frac{R ^{2} z ^{2}}{H ^{2}} (z^{2} + \frac{R ^{2} z ^{2}}{4 H ^{2}}) d z = ρ π (\frac{R ^{2} H ^{3}}{5} + \frac{R ^{4} H}{20}) .

Since

M = ρ π R^{2} H /3

I = M (\frac{3}{20} R^{2} + \frac{3}{5} H^{2}), α = \frac{3}{20}, β = \frac{3}{5} .

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