 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-3/10c/solution

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

For a one-to-one continuously differentiable change of variables with nonzero Jacobian determinant in the interior, the change of variables formula is

\int_{R} f (x, y) d x d y = \int_{R} f (x (u, v), y (u, v)) det \frac{\partial ( x , y )}{\partial ( u , v )} d u d v .

The absolute value is necessary even if the transformation reverses orientation.

The original region is the first-quadrant area between the hyperbola and circle, above the horizontal-axis segment. More explicitly,

0 \leq y \leq \frac{1}{2}, 1 + y^{2} \leq x \leq 2 - y^{2} .

Its vertices are

(1, 0)

(2, 0)

and

(3/2, 1/ 2)

x = \frac{u + v}{2}, y = \frac{u - v}{2},

and the image is the triangle

1 \leq v \leq u \leq 2

(1, 1), (2, 1), (2, 2)

det (2 x 2 x 2 y - 2 y) = - 8 x y .

It is nonzero in the interior; the zero on

y = 0

lies on a measure-zero boundary, so the integral formula applies by taking interior limits. Since

x^{5} y - x y^{5} = x y uv

and

4 x^{2} y^{2} = u^{2} - v^{2}

, the inverse-Jacobian factor cancels

x y

I = \frac{1}{8} \int_{1}^{2} \int_{1}^{u} uv e^{u^{2} - v^{2}} d v d u .

Integrating in

v

first and then

u

I = \frac{1}{16} \int_{1}^{2} u (e^{u^{2} - 1} - 1) d u = \frac{1}{32} (e^{3} - 1) - \frac{3}{32} .

Therefore

I = \frac{e ^{3} - 4}{32} .

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