 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-3/10b/c/solution

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

r = 2

, the line

y = 3 x

θ = π /3

, and

y = 1

r sin θ = 1

. The inequalities select

\frac{π}{6} < θ < \frac{π}{3}, csc θ < r < 2 .

z = \frac{x y}{x ^{2} + y ^{2}} = sin θ cos θ .

Thus the required volume is

V = \int_{π /6}^{π /3} \int_{c s c θ}^{2} sin θ cos θ r d r d θ = \int_{π /6}^{π /3} (2 sin θ cos θ - \frac{1}{2} cot θ) d θ = \frac{1}{2} - \frac{1}{4} lo g 3 .

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