 Squared-radius difference substitution

In the open first quadrant this is one-to-one, with $x=\sqrt{(u+v)/2}$, $y=\sqrt{(u-v)/2}$ and Jacobian determinant $-8xy$. Its image satisfies $u>v$ and $u+v>0$. In particular $1\le v\le u\le U$ is a triangular image for a circular-hyperbolic region. A factor $xy$ in the original integrand cancels the inverse-Jacobian factor $1/(8xy)$; moreover $x^4-y^4=uv$ and $4x^2{y}^2=u^2-v^2$. Boundary degeneracy on $y=0$ is confined to a measure-zero set and must not be confused with an interior failure of invertibility.