 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016-resit/9/c/solution

Choose

u = x - y

and

v = x + y

. Then

0 \leq u \leq 1

1 \leq v \leq 3

, while

x = (v + u) /2

y = (v - u) /2

and the inverse Jacobian determinant has magnitude

1/2

. Hence

\iint_{R} (\frac{x - y}{x + y + 3})^{2} d x d y = \frac{1}{2} \int_{0}^{1} u^{2} d u \int_{1}^{3} (v + 3)^{- 2} d v = \frac{1}{6} (\frac{1}{4} - \frac{1}{6}) = \frac{1}{72} .

The denominator is bounded away from zero on the transformed rectangle, so no singularity or improper-integral qualification is needed.

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