Probability of a point sampled from a ball lying in a spherical cap/segment What is the probability that a point sampled from an n-ball lies in a spherical segment in the ball?
Alternatively, what is the probability that a point sampled from an n-ball lies in a spherical cap in the ball?
One way to tackle these questions is using sampling. Sampling uniformly from a ball is equivalent to sampling from independent normal distributions, each corresponding to one dimension. Is it possible to estimate the volume of the spherical cap/segment or give the probability estimate above?
 A: I assume the point is drawn uniformly from the sphere, in which case the probability will just be the ratio of the volume of the segment or cap to the ratio of the full sphere.
So then the question becomes what is the volume of a spherical cap with height $h$ of a sphere with radius $R$. Note that we can assume that $h\leq R$ because if $h>R$, the volume will just be the volume of the sphere minus the volume of the cap with height $h-R$. Also, we can express the volume of any spherical segment as the difference of volumes of two spherical caps.
So assume we are in dimension $n$ and we have a sphere or radius $S_n(R)$ of radius $R$ and we are given a fixed height $h$. We need to compute the volume of $C_n(R,h)$, the volume of the cap starting at the pole $(0,0,\dots,0,R)$ having height $h$.
Step 1: The volume of $S_n(R)$ is given by $V(S_n(R))=\frac{2(2\pi)^{\frac{n-1}{2}}}{n!!}R^n$, where the double factorial is defined for odd numbers as $(2k+1)!!:= (2k+1)\cdot(2k-1)\cdot(2k-3)\dots 3\cdot 1$ and for even numbers as $(2k)!!:=(2k)\cdot(2k-2)\cdot(2k-4)\dots 4 \cdot 2\cdot \sqrt{\frac{2}{\pi}}$, see https://en.wikipedia.org/wiki/Volume_of_an_n-ball.
Step 2: The volume of a cap of height $h$ can be written as $V(C_n(R,h))=\int_0^h V(S_n(R)\cap P_n(t))dt$, where $P_n(t)$ is the hyperplane in $\mathbb{R}^n$ with equation $x_n=t$.
Then
\begin{equation*}
S_n\cap P_n(t)=\{(x_1,\dots,x_n):\sum_{k=1}^n x_k^2=R^2, x_n=t\}=\{(x_1,\dots,x_{n-1},t):\sum_{k=1}^{n-1} x_k^2=R^2-t^2\},
\end{equation*} so $V(S_n\cap P_n(t))=V(S_{n-1}(\sqrt{R^2-t^2}))$. Recall that we assumed $h\leq R$.
Therefore, $V(C_n(R,h))=\frac{2(2\pi)^{\frac{n-2}{2}}}{(n-1)!!}\int_0^h (R^2-t^2)^\frac{n-1}{2}dt = \frac{2(2\pi)^{\frac{n-2}{2}}}{(n-1)!!} R^n\int_{\cos^{-1}(\frac{h}{R})}^{\frac{\pi}{2}} \sin^n(x)dx$, assuming my change of variable computations are correct. And there are known ways of iterating integration by parts to get a closed form for the antiderivative of $\sin^n$.
