Past exam of the mathematical methods II course of the University of Oxford / 2016 resit / 9 / b / Solution

In the positive quadrant, the inverse transformation is single-valued:

x = v / u, y = uv .

The two rays become

1 \leq u \leq 2

and the two hyperbolas become

1 \leq v \leq 3

, so the curved original region becomes a rectangle. These are the middle two panels of the figure in part 2i. The forward Jacobian determinant is

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

Since

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

, the required double integral is

\int_{1}^{2} \int_{1}^{3} \frac{v ^{2}}{2 u ^{2}} d v d u = \frac{1}{2} (\frac{1}{2}) (\frac{26}{3}) = \frac{13}{6} .

The signs of

x, y

and

u > 0

justify the chosen square-root branch and the simplified absolute Jacobian determinant.

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