Finding marginal distribution, unit sphere

A uniform distribution on the sphere does not have a density function in three variables, but the marginal distribution for two of the three variables does have a density. It is obtained by expressing the area element of the sphere $\sin\theta d\varphi\wedge d\theta$ in new coordinates $x$ and $y$ and then normalizing.

$$\eqalign{ \sin^2\theta&=x^2+y^2\\ 2\sin\theta\cos\theta d\theta&=2xdx+2ydy\\ \sin\theta d\theta&=\frac{xdx+ydy}{\sqrt{1-x^2-y^2}}\\ \varphi&=\arctan\left(\frac y x\right)\\ d\varphi&=\frac1{1+\frac{y^2}{x^2}}\left(\frac1xdy-\frac{y}{x^2}dx\right)\\ \sin\theta d\varphi\wedge d\theta&=\frac1{\sqrt{1-x^2-y^2}}dx\wedge dy\\ }$$

This gives the density function

$$\frac1{2\pi\sqrt{1-x^2-y^2}}$$

on the disk $x^2+y^2\leq1.$

An alternative solution (eschewing spherical coordinates) may be based on the approximation of the surface density by a bulk one as follows. Let $h>0$ (destined to go to zero). Consider the spherical shell $\Sigma_h$ of thickness $h$ about the sphere of radius $1$, i.e., the union of all the spheres of center $0$ and radius ranging between $1-h/2$ and $1+h/2$. Since this is a $3$-dimensional volume you may consider on it a uniform distribution (constant function) $f_{XYZ}^h\equiv c(h)$, with $c(h)$ constant in $x,y,z$ (but depending on $h$) chosen such that the integral of $f_{XYZ}^h$ is $1$. So $c(h)=1/\operatorname{vol}\Sigma_h$, which the following calculation reveals: $$ \operatorname{vol}\Sigma_h = \int_{1-h/2}^{1+h/2}\int_{\partial B_0(\rho)}\operatorname d S\operatorname d\rho = \int_{1-h/2}^{1+h/2}4\pi\rho^2\operatorname d\rho = \frac{4\pi}3\Big(3h+\frac{h^3}4\Big) = 4\pi h+\frac\pi3h^3 . $$ As $h\to0$ the mass of the shell stays $1$, $c(h)\approx1/(4\pi{h}\to\infty$ and the $f_{XYZ}^h$ converges (weakly, or in the sense of distributions, or whatever your mathematical religion prescribes, but intuition is what matters) to $f_{XYZ}$. Thanks to the integral being bounded to compute $f_{XY}$ it is legitimate to interchange limits (here the theory of distribution does come handy, but you can use geometric measure theory,if you prefer) to integrate $f^h_{XYZ}$ in $z$ first and then take $h\to0$. The integration in $z$ rests on some simple geometry, for all $x,y$ such that $x^2+y^2<(1-h/2)^2$, the line passing through $(x,y)$ intersect the shell twice symmetrically about the $(x,y)$-plane. Looking at the upper half (and then doubling the result) the length of the segment is given by $S^+_h(x,y)-S^-_h(x,y)$ where $S_h^\pm$ are functions whose graphs are the outer and inner boundaries of the (upper half of the spherical) shell, respectively. Explicitly $$S^\pm_h(x,y)=\sqrt{(1\pm h/2)^2-x^2-y^2}=:F(\pm h).$$ The auxiliary function $F$'s Taylor expansion of order $1$ about $0$ yields $$ S^+_h(x,y)-S^-_h(x,y) = 2F'(0)h+o(h) = \frac{h+o(h)}{\sqrt{1-x^2-y^2}} , $$ since $$F'(\xi)=\frac{1+\xi/2}{2\sqrt{(1+\xi/2)^2-x^2-y^2}}.$$ Multiplying this by $2$ (to take into account the lower half of the sphere) and the density normalizing constant $c(h)$ we obtain $$ \int_{\mathbb R} f^h_{XYZ}(x,y,z)\operatorname dz = \frac{2h+o(h)}{\sqrt{1-x^2-y^2}(4\pi h+\pi h^3/3)} = \frac{1+o(h^0)}{\sqrt{1-x^2-y^2}(2\pi+\pi h^2/6)} \to \frac1{2\pi\sqrt{1-x^2-y^2}} ,\text{ as }h\to0 . $$