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

For a change of variables in a double integral, first solve for $x=x(u,v)$ and $y=y(u,v)$ and transform every boundary inequality to identify the new integration region $S$. Compute the Jacobian determinant, replace the integrand by its expression in $u,v$, and useA sufficient standard hypothesis is a continuously differentiable one-to-one transformation with nonzero Jacobian determinant in the interior, with harmless measure-zero boundaries. If one coordinate chart does not cover the region injectively, split into suitable pieces rather than silently counting points multiple times. The absolute value is essential for an ordinary area integral.

In the positive quadrant, the inverse transformation is single-valued:The two rays become $1\le u\le 2$ and the two hyperbolas become $1\le v\le 3$, so the curved original region becomes a rectangle. These are the middle two panels of the figure in part 2i. The forward Jacobian determinant isSince $x^{3}y=v^2/u$, the required double integral isThe signs of $x,y$ and $u>0$ justify the chosen square-root branch and the simplified absolute Jacobian determinant.

Choose $u=x-y$ and $v=x+y$. Then $0\le u\le 1$, $1\le v\le 3$, while $x=(v+u)/2$, $y=(v-u)/2$ and the inverse Jacobian determinant has magnitude $1/2$. HenceThe denominator is bounded away from zero on the transformed rectangle, so no singularity or improper-integral qualification is needed.