Uniformly distributed points on spherical surface Let $x=(x_1,\ldots,x_n)$ be uniformly distributed on the $(n-1)$-dimensional spherical surface $S^{n-1}(n^\frac{1}{2})$ of radius $n^\frac{1}{2}$. I'm trying to show that as $n\to\infty$, $x_1$ converges in distribution to a standard normal random variable.
As noted in the introduction of this article, 
for $-n^\frac{1}{2}<a<b<n^\frac{1}{2},$
the uniform measure of the spherical zone $a\leq x_1<b$ is
$$
\frac{\int_a^b (n-x^2)^{n/2-1}dx}{\int_{-n^\frac{1}{2}}^{n^\frac{1}{2}}(n-x^2)^{n/2-1}dx}.
$$
I know this is probably very straightforward, but I'm stuck right now---could someone please explain the above formula for the uniform measure?
EDIT
This result can alternatively be obtained as follows: We introduce auxiliary iid standard normal random variables $y_1,\ldots,y_n$ and
$$
z_j=\frac{y_j}{(y_1^2+\cdots+y_n^2)^{1/2}n^{-1/2}}.
$$
Then we show that $z=(z_1,\cdots,z_n)\in S^{n-1}(n^\frac{1}{2})$, that $x_1$ and $z_1$ have the same distribution, and that $z_1$ converges in distribution to $y_1.$
 A: The region, $R_{x_1, a,b}:=\left\{ x\in S^{n-1}(n^\frac{1}{2}) : a\leq x_1<b \right\}$, defines a collection of "$n-2$"-dimensional surfaces. 


*

*For $n=3$, the sphere is a $2$-dimensional surface and the region is a collection of $1$-dimensional circles that lie in the sphere, each uniquely identified by $x_1$.

*For $n=2$, the sphere is a $1$-dimensional sphere (i.e. circle) and the region is a collection of pairs of $0$-dimensional points, each uniquely identified by $x_1$.


The stated measure, $$\frac{\int_a^b (\sqrt{n-x^2})^{n-2}\mathrm dx}{\int_{-n^\frac{1}{2}}^{n^\frac{1}{2}}(\sqrt{n-x^2})^{n-2}\mathrm dx}\,,$$ gives the uniform distribution over $S^{n-1}(\sqrt{n})$ when applied to the aforementioned collection of "$n-2$"-dimensional surfaces. It is simply a ratio of "$n-1$"-volumes (note that the volume of $S^{n-1}(r)$ is proportional to $r^{n-1}$). So, in the case of $n=2$, it is a ratio of lengths. 
And, in the $2$-dimensional case, it describes a uniform distribution of pairs of points (i.e. a pair for each $x_1 \in [a,b)$) on the circle, and not the distribution of the projection of these points. That is, for the uniform measure $P$ over $S^1$,
$$
P\left(x\in R_{x_1, a,b}\right) = \frac{\int_a^b \mathrm dx}{\int_{-n^\frac{1}{2}}^{n^\frac{1}{2}}\mathrm dx} = \frac{b-a}{2\sqrt{n}}\,.
$$    
