= Solution
<Independence> gives the joint <probability density function> $f_{X,Y}(x,y)=4e^{-(x^2+y^2)}/\pi$ on $x,y>0$. The inverse transformation is
$$
x=\sqrt{\frac{u}{1+v^2}},\qquad y=v\sqrt{\frac{u}{1+v^2}},\qquad u,v>0.
$$
Its <Jacobian matrix> has absolute determinant $1/[2(1+v^2)]$. One way to see the determinant is to use polar coordinates: $u=r^2$, $v=\tan\theta$, with $0<\theta<\pi/2$. Then $dx\,dy=r\,dr\,d\theta=du\,dv/[2(1+v^2)]$. The <change-of-variables formula for probability densities> gives
$$
\boxed{f_{U,V}(u,v)=\frac2\pi\frac{e^{-u}}{1+v^2},\qquad u,v>0,}
$$
and zero outside that support. Integrating over the complementary variable gives
$$
\boxed{f_U(u)=e^{-u}\ \ (u>0),\qquad f_V(v)=\frac2{\pi(1+v^2)}\ \ (v>0).}
$$
These are respectively a unit-rate <exponential distribution> and a unit-scale <half-Cauchy distribution>. Their product equals the <joint probability density function>, so \b[$U$ and $V$ are independent]. This <radial-ratio independence of two half-normal variables> results from the radial symmetry of the positive-quadrant <joint probability density function>.
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