= Solution
{c}
Put $r=\sqrt{x^2+y^2}$. The definitions of the <parabolic polar coordinates> give $u^2=x+r$ and $v^2=-x+r$. Subtracting and multiplying these equations yields
$$
u^2-v^2=2x,\qquad u^2v^2=r^2-x^2=y^2.
$$
Since $u,v,y\geq0$, taking the nonnegative <square root> gives
$$
\boxed{x=\frac{u^2-v^2}{2},\qquad y=uv}.
$$
The <Jacobian matrix> has columns $(u,v)$ and $(-v,u)$. Its <Jacobian determinant> is therefore
$$
\frac{\partial(x,y)}{\partial(u,v)}
=\det\begin{pmatrix}u&-v\\v&u\end{pmatrix}
=\boxed{u^2+v^2}.
$$
Here the displayed fraction denotes the <Jacobian determinant>, as in the paper. The <change of variables formula> consequently gives $dx\,dy=(u^2+v^2)\,du\,dv$. The <Jacobian determinant> is positive except at the origin, and the definitions supply the unique nonnegative inverse throughout the region $y\geq0$.
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