 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/12/e/solution

The handle and the two end pieces have disjoint interiors; their common boundary circles contribute no volume. Hence the total mass is

M = M_{h} + 2 M_{e}

. With

δ = α - γ

, the results above give

M = 2 π ρ a^{3} [γ (β^{2} - δ^{2}) + \frac{2 β ^{3}}{3} + β^{2} δ - \frac{δ ^{3}}{3}] .

Equivalently, since

γ + δ = α

M = 2 π ρ a^{3} [α β^{2} + \frac{2 β ^{3}}{3} - γ (α - γ)^{2} - \frac{( α - γ ) ^{3}}{3}] .

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