 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-3/3a/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

u = \frac{x}{y}, v = x y .

Because the region lies in the positive quadrant, its four inequalities become

1 \leq u \leq α, 1 \leq v \leq α .

D

to a rectangle in the

uv

\frac{\partial ( u , v )}{\partial ( x , y )} = 1/ y y - x / y^{2} x = \frac{2 x}{y} = 2 u,

d x d y = \frac{d u d v}{2 u} .

Since

x^{2} = uv

\iint_{D} x^{2} d x d y = \int_{1}^{α} \int_{1}^{α} uv \frac{d u d v}{2 u} = \frac{1}{2} (\int_{1}^{α} d u) (\int_{1}^{α} v d v) = \frac{( α - 1 ) ( α ^{2} - 1 )}{4} .

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