 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-v/1/a/solution

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

v = x y

. Its Jacobian determinant in the first quadrant is

2 (x^{2} + y^{2}) > 0

, and the inverse is uniquely determined by

x^{2} = \frac{u ^{2} + 4 v ^{2} + u}{2}, y^{2} = \frac{u ^{2} + 4 v ^{2} - u}{2} .

Both are positive for

v > 0

, so the transformation is one-to-one on the region. It becomes the rectangle

- 2 \leq u \leq 4

1 \leq v \leq 3

. The integrand is

uv

, and

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

. Thus

\iint_{A} (x^{3} y - x y^{3}) d x d y = \int_{1}^{3} \int_{- 2}^{4} \frac{uv}{2 u ^{2} + 4 v ^{2}} d u d v = \int_{1}^{3} v (v^{2} + 4 - v^{2} + 1) d v = \frac{13 13 - 10 10 - 5 5 + 2 2}{3} .

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