Stereographic Projection of Uniform Distribution on Sphere Consider a (punctured) unit sphere $S = \{ (x,y,z) \mid x^2+y^2+z^2 = 1, z\neq 1 \} $ and the plane $L = \{ (X,Y,0)  \}$ (let us use the shorthand $(X,Y)$ for $(X,Y,0)$).
We can define the stereographic projection $\Pi$ and its inverse $\Pi^{-1}$ as
$$(X,Y) = \Pi(x,y,z) =  \left(\frac{x}{1-z}, \frac{y}{1-z} \right)$$
and
$$(x,y,z) = \Pi^{-1}(X,Y) = \frac{1}{1+X^2+Y^2} ( 2X, 2Y, -1 + X^2 + Y^2).$$
Clearly $\Pi : S \to L$ is a diffeomorphism.
Now let us consider the uniform distribution on $S$: $p((x,y,z)) = \frac{1}{4\pi}$ if $(x,y,z) \in S$,  and 0 otherwise. How can I find the distribution on $L$ if we sample points from $S$ according to $p$? 
Or in other words: Let $R \sim U(S)$ be a random variable on $S$. What is the distribution of $Q = \Pi(R)$?
 A: We can introduce Spherical coordinates $\theta\in[0,\pi]$, $\phi\in[0,2\pi)$, where $\theta=0$ is the point of contact. You can show that the initial distribution doesn't depend only on $\phi$ and since the point remains on the same vertical plane, the resulting distribution doesn't depend on $r$ being expressed in polar coordinates.
Spherical cap $\theta<\theta_0$ will project to the circle $r<2\tan\frac{\theta_0}2$. Thus cumulative distribution:
$$
P(r\le R)=\frac1{4\pi}A_{\text{cap}} = \frac1{4\pi}2\pi\left(1-\cos\left(2\arctan \frac{R}2\right)\right)=\frac{R^2}{4+R^2}
$$
which is as expected is $0$, when $R=0$ and equals to $1$, when $R\to\infty$.
To get the PDF, we need to differntiate:
$$
\omega(R) = \frac{8R}{(4+R^2)^2}.
$$
Finally, if we want to express this in cartesian coordinates, we need to multiply by Jacobian:
$$
\omega(X,Y)dX\,dY=\omega(X,Y)R\,dR\,d\phi=\omega(R)\omega(\phi)dR\,d\phi\\
\omega(X,Y) = \frac1{2\pi R}\frac{8R}{(4+R^2)^2} = \frac1{\pi}\frac{4}{(4+X^2+Y^2)^2}
$$
