 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ia/paper-3/12c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

y \leq x \leq 2 y

y \geq 0

, and the reciprocal bounds exclude zero. Thus the whole region lies in the positive quadrant. Introduce ratio-product coordinates

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

The region becomes the rectangle

1 \leq u \leq 2

1 \leq v \leq 2

. The inverse map and its Jacobian determinant are

x = uv, y = v / u, \frac{\partial ( x , y )}{\partial ( u , v )} = x / (2 u) - y / (2 u) x / (2 v) y / (2 v) = \frac{x y}{2 uv} = \frac{1}{2 u} > 0.

\int_{A} \frac{x}{y} d x d y = \int_{1}^{2} \int_{1}^{2} u \frac{1}{2 u} d v d u = \frac{1}{2} .

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