= Solution
All the following <distributions> are conditional on the point lying in the unit disk. Initially the <joint probability density> of $(S,T)$ is $1/4$ on the square. The disk has area $\pi$, so the conditioning event has <probability> $\pi/4$. Dividing by this probability gives the constant conditional <density> $1/\pi$ on the disk, proving its <uniform distribution> there.
The <polar coordinates> $S=r\cos\theta,T=r\sin\theta$ have <Jacobian determinant> $r$. Consequently
$$
\boxed{f_{R,\Theta}(r,\theta)=\frac r\pi,\qquad 0<r<1,\quad0\leq\theta<2\pi.}
$$
Integrating the <joint probability density> gives $f_R(r)=2r$ and $f_\Theta(\theta)=1/(2\pi)$. Since the joint density equals the product of these marginal densities on a product support, \b[$R$ and $\Theta$ are independent]. In particular $U=R^2$ has the <uniform distribution> on $(0,1)$.
The second <change of variables> preserves the angle and changes the radius to
$$
\boxed{Q=\sqrt{-2\log(R^2)}=\sqrt{-2\log U},\qquad\Psi=\Theta.}
$$
The event $R=0$ has probability zero, so arbitrary definitions at that point have no effect on the distribution. For $q\geq0$, monotonicity of the <logarithm> gives
$$
P(Q\leq q)=P(U\geq e^{-q^2/2})=1-e^{-q^2/2},
\qquad f_Q(q)=q e^{-q^2/2}\quad(q>0).
$$
Since $Q$ is a function of $R$, it remains <independent> of the uniform angle. Hence the new <joint probability density> is
$$
\boxed{f_{Q,\Psi}(q,\psi)=\frac{q}{2\pi}e^{-q^2/2},\qquad q>0,\quad0\leq\psi<2\pi.}
$$
Finally, $(X,Y)=(Q\cos\Psi,Q\sin\Psi)$ has polar <Jacobian determinant> $q$. Dividing the preceding density by $q$ gives the Cartesian <joint probability density>
$$
f_{X,Y}(x,y)=\frac1{2\pi}e^{-(x^2+y^2)/2}
=\left(\frac{e^{-x^2/2}}{\sqrt{2\pi}}\right)
\left(\frac{e^{-y^2/2}}{\sqrt{2\pi}}\right).
$$
This factors into two standard <normal distribution> densities on $\mathbb R^2$. Thus \b[$X$ and $Y$ are independent, each with mean zero and variance one]. This derivation is the radial form of the <Box-Muller transform>.
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