Volume of a cone in an $n$-dimensional ball Assume that $B$ is an $n$-dimensional ball of radius $R$ centered at the origin, i.e.,
$B=\{x\in\mathbb{R}^n : \|x\|\leq R\}$.
Fix a point $x_0$ in $B$ and $\delta \in (0,\pi)$, and let $C$ be the following set
$C=\{x\in B : $ The angle between the vectors $ \vec{x_0} $ and $ \vec{x} $ is less than or euqal to $ \delta.\}$
My question is, how can we compute the $n$-dimensional measure of the set $C$ in terms of $R$ and $\delta$. Is it simply going to be $K\delta R^n$ for some constant $K$? Or is it more complicated than that? The answer for the case $n=2$ is clearly $ \delta R^2 $.
 A: For $\delta \in (0,\pi)$, let $$ K_\delta := \bigg \{ x\in \mathbb R^n \text{ s.t. } x_1 > \frac 1 {\tan (\delta /2 )}\vert x' \vert \bigg \}  $$ where $x'=(x_2,\dots,x_n) \in \mathbb R^{n-1}$ and $x=(x_1,x')$. The set $K_\delta$ is an infinite cone centred around the vector $e_1$. Then $C = K_\delta \cap B_R$.
By rescaling, $$\vert K_\delta \cap B_R \vert =\vert K_\delta \cap B_1 \vert R^n,   $$ so it is enough to calculate $\vert K_\delta \cap B_1 \vert$. Here I am using the standard notation that, given a (Lebesgue measurable) set $A \subset \mathbb R^n$, $\vert A \vert$ is the Lebesgue measure of $A$.
Next, we can write $K_\delta \cap B_1$ as the disjoint union of $E_\delta$ and $F_\delta$ where \begin{align*} E_\delta &:= (K_\delta \cap B_1) \cap \{ 0<x_1<\cos(\delta/2) \} &\text{(the conical part)}\\F_\delta &:= (K_\delta \cap B_1) \cap \{ \cos(\delta/2)<x_1<1 \} & \text{(the spherical cap)} \end{align*}

For $\vert E_\delta \vert$: For each $\rho \in (0,\cos(\delta/2))$, we have that $$ \{x_1=\rho \} \cap E_\delta = \{\rho\} \times B^{n-1}_{\rho \tan(\delta/2)}$$ where $B^{n-1}_r$ is the $(n-1)$-dimensional ball centred at $0\in \mathbb R^{n-1}$ with radius $r>0$. Hence, if $\lambda^{n-1}$ is the $(n-1)$-Lebesgue measure (if you are unfamiliar with measure theory, this is just the $(n-1)$ volume) then \begin{align*}
\vert E_\delta \vert &= \int_0^{cos(\delta/2)} \lambda^{n-1} (B^{n-1}_{\rho \tan(\delta/2)}) \, d\rho \\
&= \omega_{n-1} \tan^{n-1}(\delta/2) \int_0^{cos(\delta/2)} \rho^{n-1} \, d\rho \\
&= \frac{\omega_{n-1}}n \sin^{n-1}(\delta/2) \cos (\delta/2).
\end{align*} where $\omega_{n-1} := \lambda^{n-1}(B^{n-1}_1)$ the volume of the $(n-1)$-dimensional unit ball.

For $\vert F_\delta\vert$: For $\rho \in (\cos(\delta/2),1)$, $$ \{x_1=\rho \} \cap F_\delta = \{\rho\} \times B^{n-1}_{\sqrt{1-\rho^2}}.$$ Hence, \begin{align*}
\vert F_\delta \vert &= \int_{\cos(\delta/2)}^1 \lambda^{n-1} (B^{n-1}_{\sqrt{1-\rho^2}}) \, d \rho \\
&= \omega_{n-1}\int_{\cos(\delta/2)}^1 \big (1-\rho^2\big)^{\frac{n-1}2} \, d \rho.
\end{align*} Letting $\tau = \rho^2$ gives \begin{align*}
\vert F_\delta \vert &= \frac{\omega_{n-1}}2\int_{\cos^2 (\delta/2)}^1\tau^{-\frac12} \big (1-\tau\big)^{\frac{n-1}2} \, d \tau = \frac{\omega_{n-1}}2 \bigg [ \mathrm B \bigg ( \frac12,\frac{n+1}2 \bigg )-\mathrm B \bigg ( \cos^2 (\delta/2) ;  \frac12,\frac{n+1}2 \bigg )\bigg ] 
\end{align*} where $ \mathrm B(a,b)$ and $ \mathrm B(x;a,b)$ are the Beta function and the incomplete Beta function respectively.

In conclusion, we obtain that $$\vert K_\delta \cap B_R \vert = \frac {\omega_{n-1}}2 \bigg [ \frac2n \sin^{n-1}(\delta/2) \cos (\delta/2) +  \mathrm B \bigg ( \frac12,\frac{n+1}2 \bigg )-\mathrm B \bigg ( \cos^2 (\delta/2) ;  \frac12,\frac{n+1}2 \bigg ) \bigg ]R^n. $$

Checking $n=2,3$:
If $n=2$ then $\omega_{n-1}=\omega_1=2$ and $$B(x^2 ; 3/2,1/2) = x \sqrt{1-x^2}-\arctan \left(\sqrt{\frac{1}{x^2}-1}\right)+\frac{\pi }{2} ,$$ so \begin{align*}
\mathrm B \bigg ( \frac12,\frac32 \bigg )-\mathrm B \bigg ( \cos^2 (\delta/2) ;  \frac12,\frac32 \bigg ) &=  -  \sin(\delta/2)\cos(\delta/2)+\frac \delta2.
\end{align*} Thus, $$\vert K_\delta \cap B_R \vert = \bigg (  \sin(\delta/2) \cos (\delta/2)   -  \sin(\delta/2)\cos(\delta/2)+\frac \delta2 \bigg )R^2= \frac 1 2 \delta R^2 $$ which is the volume of sector with angle $\delta$ (not $\delta R^2$ by the way).
If $n=3$ then $\omega_{n-1}=\omega_2 = \pi$ and $$B(x^2 ; 3/2,1/2) =-\frac{2}{3}  x \left(x^2-3\right). $$ Thus, \begin{align*}
\vert K_\delta \cap B_R \vert &= \frac {\pi}2 \bigg [ \frac23 \sin^{2}(\delta/2) \cos (\delta/2) +\frac4{3} +\frac{2}{3}  \cos(\delta/2) \left(-\sin^2(\delta/2)-2\right) \bigg ]R^3 \\
&=\frac {2\pi}3 \bigg [ 1 -  \cos(\delta/2)  \bigg ]R^3 \\
&=\frac {4\pi}3  \sin^2(\delta)  R^3 
\end{align*} which is very close to what you calculated (maybe you/I made a small computation error somewhere).
A: Using spherical coordinates on the unit sphere, the number you are looking for is
$$ R^n \int_{\phi_{n-1}=0}^{2\pi} \int_{\phi_{n-2}=0}^\pi
\cdots \int_{\phi_1=0}^\delta d^{n-1}V$$
where
$$d^{n-1}V = 
\sin^{n-2}(\phi_1)\sin^{n-3}(\phi_2)\cdots \sin(\phi_{n-2})\, d\phi_1 \, d\phi_2\cdots d\phi_{n-1}$$
It is certainly not linear in $\delta$ when $n>2$.
Here I assume $x_0$ to be the "vertical" direction, i.e., $phi_1$ is the angle between the given point and $x_0$.
A: There is a relationship between the volume of a cone in an n-dimensional ball and Student's t-distribution. This occurs via a relation between the angle and the t-score:
$$\frac{t}{\sqrt{n-1}} = \frac{1}{tan(\delta)}$$
and you can express the relative volume using the cumulative distribution function for $t_{\nu}$ (with $\nu = n-1$ degrees of freedom):
$$V(\delta)/V(\pi) = 1-F_{t_\nu}\left(\frac{\nu}{tan(\delta)}\right)$$
The cumulative distribution function can be written in terms of the incomplete beta function:
$$F(t) = 1-\frac{1}{2} I(\frac{\nu}{t^2+\nu};\frac{\nu}{2},\frac{1}{2})$$
See https://stats.stackexchange.com/a/365070/164061 for (1) the intuition behind the correspondence between the cone volume and the t-distribution (2) and a derivation for $F(t)$ in terms of incomplete beta functions
