 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-2/5/c/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

Integrate the horizontal-disk expression over

π /2 < θ < π

. With

u = cos θ

and

sin θ d θ = - d u

V = π a^{3} \int_{- 1}^{0} (1 - u)^{2} (1 - 2 u) (1 - u^{2}) d u = π a^{3} \int_{- 1}^{0} (1 - 4 u + 4 u^{2} + 2 u^{3} - 5 u^{4} + 2 u^{5}) d u .

Evaluating the integral of this polynomial gives

V = \frac{5 π a ^{3}}{2} .

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