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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/ia/paper-3/10a/solution

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Past exam of the mathematics course of the University of Cambridge / 2001 / ia / Paper 3 / 10A / Solution by

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

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

\iint_{D} F (x, y) d x d y = \iint_{E} F (x (u, v), y (u, v)) det \frac{\partial ( x , y )}{\partial ( u , v )} d u d v,

(1)

where

E

is the transformed region and the inverse map covers

D

once. The absolute value is required even if the coordinate map reverses orientation.

Here

x, y

are positive. For

u = y / x

v = x y

, the inverse is

x = v / u

and

y = uv

. On the first portion of

D

, the bounds give

v \geq 1

u \leq 4

and

v \leq u

because

x \leq 1

. Together these give

1 \leq v \leq u \leq 4

; conversely these inequalities imply

1/2 \leq x \leq 1

and the original bounds. On the second portion, they give

u \geq 1

v \leq 4

and

v \geq u

, hence

1 \leq u \leq v \leq 4

, with the converse following similarly. The two triangles combine into the rectangle $E = {1 \leq u \leq 4, 1 \leq v \leq 4}$ , sharing only the diagonal corresponding to

x = 1

The Jacobian is

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

(2)

Also

x y^{3} = u v^{2}

and

x^{2} + y^{2} = v (1 + u^{2}) / u

. The transformed integral consequently factors:

\iint_{D} \frac{4 x y ^{3}}{x ^{2} + y ^{2}} d x d y = \int_{1}^{4} \int_{1}^{4} \frac{2 uv}{1 + u ^{2}} d v d u = [\frac{v ^{2}}{2}]_{1}^{4} [lo g (1 + u^{2})]_{1}^{4} .

(3)

Therefore

\iint_{D} \frac{4 x y ^{3}}{x ^{2} + y ^{2}} d x d y = \frac{15}{2} lo g \frac{17}{2} .

(4)

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