= Squared-radius difference substitution
{title2=$u=x^2+y^2,\quad v=x^2-y^2$}
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 $1\leq v\leq u\leq U$ 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 $4x^2y^2=u^2-v^2$. Boundary degeneracy on $y=0$ is confined to a measure-zero set and must not be confused with an interior failure of invertibility.
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