Let $X, Y$ be independent standard normal random variables. Show that $$U=X^2+Y^2 \hbox{ and } V=\frac{X}{\sqrt{X^2+Y^2}}$$ are independent.
The transformation is $T(x,y)=(u,v)$ where $T: (-\infty,\infty)\times (-\infty,\infty)\to [0,\infty)\times [-1,1]$.
Note that $T$ is not 1-1 because of $T(x,y)=T(x,-y)$.
But $T$ is 1-1 on $(-\infty,\infty)\times (0,\infty)$ and on $(-\infty,\infty)\times (-\infty,0)$.
We consider $$T^{-1}_1(u,v)=(\sqrt{u}v, \sqrt{u(1-v^2)})$$ and $$T^{-1}_2(u,v)=(\sqrt{u}v, -\sqrt{u(1-v^2)})$$.
The joint density of $X, Y$ is given by $f(x,y)=\frac{1}{2\pi}e^{-(x^2+y^2)/2}$.
By the transformation formula, the joint density $g(u,v)$ of $U, V$ is given by $$g(u,v)=f(\sqrt{u}v, \sqrt{u(1-v^2)})|J_1|+f(\sqrt{u}v, -\sqrt{u(1-v^2)})|J_2|$$ where $J_i$ is the Jacobian of $T^{-1}_i$ $i=1, 2$. Note that for $i=1, 2$
$$|J_i|=\frac{1}{2\sqrt{1-v^2}}.$$
Therefore, $g(u,v)=\frac{1}{2}e^{-u/2}\frac{1}{\pi\sqrt{1-v^2}}$ with $0\leq u<+\infty, -1<v<1$, and hence $U$ and $V$ are independent.