If $U$ is uniformly distributed on $S^{2}$, then its first component is uniformly distributed on $(-1,1)$. Assume $\mathbf{U} = (U_{1}, U_{2}, U_{3})' \sim unif(S^{2})$. How would I show that $U_{1} \sim unif(-1,1)$?
I don't know if I'm confusing myself, because I can't see this as being true for higher dimensions.
 A: Relating to the comment of Nate, there's also the "well known" formula for the area of a surface of revolution, namely if $f:[a,b]\rightarrow \mathbb R$ is smooth and positive, we can rotate its graph $\Gamma_f = \{(x,y)\ |\ x \in [a,b],\ y = f(x)\}$ around the $x$-axis and obtain a surface $\Sigma$. The area of that surface is
$$
\mathrm{Area}(\Sigma) = 2\pi \int_a^bf(x)\sqrt{1+f'(x)^2}dx.
$$
It's likely that your familiar with that from calculus or classical differential geometry. Alternatively, you can consult Wikipedia http://en.wikipedia.org/wiki/Surface_of_revolution

In particular, the function $g:[-1,1]\rightarrow\mathbb R,\ x \mapsto \sqrt{1-x^2}$ produces the sphere $S^2$ as a surface of revolution. We have to be careful about the end points $-1$ and $1$ of the interval $[-1,1]$, but since we're in a probabilistic setting and consider uniform distribution on $S^2$, we can ignore these points by invoking a measure zero argument.

Now, let's denote uniform probability measure on $S^2$ by $P$, i.e. for
$A \subseteq S^2$ Lebesgue measurable, we have
$$
P(A) = \mathrm{Area}(A) / \mathrm{Area}(S^2) = \mathrm{Area}(A)/4\pi
$$ (you can get $\mathrm{Area}(S^2)$ from the formula above :-). Moreover, let's denote Lebesgue measure on $\mathbb R$ by $\lambda$ and uniform probability measure on $[-1,1]$ by $Q$, i.e. for $B \subseteq [-1,1]$ Lebesgue measurable, we have 
$$
Q(B) = \lambda(B) / \lambda([-1,1]) = \lambda(B) / 2.
$$ Finally, let
$$
T:S^2\rightarrow [-1,1],\ (x,y,z) \mapsto x
$$ be the projection onto the first coordinate. We want to show that the image of $P$ under $T$ is $Q$, i.e.
$\forall B \subseteq [-1,1]$ Lebesgue measurable: $P(T^{-1}(B)) = Q(B)$. By construction of Lebesgue measure, this is equvalent to 
$$
\forall [a,b] \subseteq [-1,1]:\ P(T^{-1}([a,b])) = Q([a,b]).$$
Now, $T^{-1}([a,b])$ is the slice of $S^2$ "around" the interval $[a,b]$ on the $x$-axis. We can calculate its area with the formula from above, namely
$$
\mathrm{Area}(T^{-1}([a,b])) = 2\pi\int_a^bg(x)\sqrt{1+g'(x)}dx.
$$
Here, the integrand is
$$g(x)\sqrt{1+g'(x)^2} = \sqrt{1-x^2}\sqrt{1+\frac{x^2}{1-x^2}} = \sqrt{1-x^2}\sqrt{\frac1{1-x^2}} = 1.
$$
From this we get
$$
P(T^{-1}([a,b])) = \frac{\mathrm{Area}(T^{-1}([a,b]))}{4\pi} = \frac{2\pi(b-a)}{4\pi} = \frac{b-a}2 = Q([a,b]),
$$
which is exactly what we want.
