<s>[INST] Random point uniform on a sphere If $X=(x,y,z)$ is a random point uniform on the unit sphere in $\mathbb{R}^3$,  Are the  coordinates $x$, $y$, $z$ uniform in interval $(-1,1)$?
 [/INST] Let $Z_1$,$Z_2$, $Z_3$ be independent identically distributed standard normal random variables. Then the random vector
$$
    \vec{X} = \left( \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+Z_3^2}}, \frac{Z_2}{\sqrt{Z_1^2+Z_2^2+Z_3^2}}, \frac{Z_3}{\sqrt{Z_1^2+Z_2^2+Z_3^2}} \right)
$$
is uniformly distributed on a unit sphere (e.g. sphere point picking on MathWorld). 
The distribution of components of $\vec{X}$ is clearly symmetric, so for $-1<x<0$:
$$\begin{eqnarray}
   \mathbb{P}\left( \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+Z_3^2}} \leqslant x \right) &=& 
    \mathbb{P}\left( Z_1^2 \geqslant x^2  (Z_1^2+Z_2^2+Z_3^2) | Z_1 < 0\right) \\ &=& 
    \frac{1}{2}\mathbb{P}\left( \frac{2 Z_1^2}{Z_2^2+Z_3^2} \geqslant \frac{2 x^2}{1-x^2} \right)
\end{eqnarray}
$$
The ratio $\frac{2 Z_1^2}{Z_2^2+Z_3^2}$ follows $F_{1,2}$-distribution (wiki), hence:
$$
  \mathbb{P}\left( \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+Z_3^2}} \leqslant x \right) = \frac{1}{2}\left(1 - \sqrt{\frac{\frac{2x^2}{1-x^2}}{2+ \frac{2x^2}{1-x^2}}}\right) = \frac{1}{2} \left(1- \sqrt{x^2}\right) = \frac{1+x}{2}
$$
which the cumulative distribution function of the uniform distribution indeed.

Added:
As suggested by @did, consider the case of $\mathbb{R}^d$. In that case the cumulative distribution function of the component, for $-1<x<0$ reads:
$$
  \mathbb{P}\left( \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+\cdots+Z_d^2}} \leqslant x\right) = 
   \frac{1}{2}\mathbb{P}\left(\frac{(d-1) Z_1^2}{Z_2^2+\cdots+Z_d^2} \geqslant \frac{(d-1)x^2}{1-x^2} \right)
$$
The ratio $\frac{(d-1) Z_1^2}{Z_2^2+\cdots+Z_d^2}$ follows $F$-distribution with $1$ and $d-1$ degrees of freedom, thus:
$$
 \mathbb{P}\left( \frac{Z_1}{\sqrt{Z_1^2+Z_2^2+\cdots+Z_d^2}} \leqslant x\right) = \frac{1}{2} \frac{\operatorname{B}\left(1-x^2 ;\frac{d-1}{2}, \frac{1}{2}\right)}{\operatorname{B}\left(\frac{d-1}{2}, \frac{1}{2}\right)}
$$
where $\operatorname{B}\left(z ;a, b\right)$ is the incomplete Beta function. Differentiating we get the probability density:
$$
    f(x) = \frac{1}{\operatorname{B}\left(\frac{d-1}{2}, \frac{1}{2}\right)} \left(1-x^2\right)^{\tfrac{d-3}{2}} [-1<x<1]
$$
This clearly shows that the distribution is not uniform for $d\not=3$.
</s>