In $f_X(x) \neq \frac{2 \pi \sqrt{R^2 - x^2}}{4\pi R^2}$? $X$ belongs to points uniformly distributed on the surface of a sphere. it was shown mathematically that the marginal distribution of the x-coordinate on points uniformly distributed on the surface of the sphere is also uniformly distributed.
This is just not intuitive to me from a graphical perspective. Could someone provide an explanation, preferably graphically, why this makes sense?
Let $S^2\subset{\mathbb R}^3$ be the unit sphere, and let $C:=\bigl\{(x,y,z)\bigm|x^2+y^2=1, \>-1\leq z\leq1\bigr\}$ be the cylinder touching $S^2$ along the unit circle in the $(x,y)$-plane. It is a standard fact of three-dimensional geometry that the "horizontal-radial-projection" $$\psi:\quad S^2\to C,\qquad\bigl(\sqrt{1-z^2}\cos\phi,\sqrt{1-z^2}\sin\phi, z\bigr)\mapsto(\cos\phi,\sin\phi,z)$$ is area preserving. In particular: For any $-1\leq a\leq b\leq 1$ the two planes $z=a$, $z=b$ create annular regions on $S^2$ and $C$ that have the same area.
Now on $C$ the function $Z: (x,y,z)\mapsto z$ is obviously uniformly distributed, i.e., $$f^{\rm cyl}_Z(\phi,z)={1\over4\pi}\qquad(0\leq\phi\leq2\pi,\>-1\leq z\leq1)\ .$$ Since $\psi$ is area preserving and keeps $z$ fixed it follows that the probability density of $Z$ on $S^2$ is equally given by $$f^{\rm sph}_Z(\phi,z)={1\over4\pi}\qquad(0\leq\phi\leq2\pi,\>-1\leq z\leq1)\ .$$