Volume of a hypercone with cap What is the volume of this region: 
$
C(x,\eta(x),\theta,r) = \{{x+\lambda y\in\mathbb{R}^d: y\in\mathbb{R}^d, ||y||_2=1, y^T\eta(x)\geq \cos\theta, \lambda\in[0,r]\}}
$
where $\theta\in(0,\pi/2)$, $r>0$ and $\eta(x)$ is a unit vector. This is a rather abstract way of writing down a cone with a cap (but also in higher dimensions). 
For instance, for $d=1$ (then $\theta=0$) the region is a line of length $r$. 
For $d=2$ the region is the segment of a circle with area $\frac{\theta}{\pi}\pi r^2 = \theta r^2$. 
I'm after a general formula for $d\geq 2$. A first guess would be 
$
Vol(C(x,\eta(x),\theta,r)) = \left(\frac{\theta}{\pi}\right)^{d-1}\sigma_d r^d
$
where $\sigma_d$ is the volume of the $d$-dimensional unit ball. Can anyone show this?
 A: This could also be described as a hyperspherical sector. Its volume is the same fraction of the volume of the entire hypersphere as the hyperspherical part of its surface is of the surface of the entire hypersphere. As explained in my answer here, the surface element is $(1-\cos^2\theta)^{(d-3)/2}\mathrm d\cos\theta=(\sin^2\theta)^{(d-3)/2}\sin\theta\mathrm d\theta=\sin^{d-2}\theta\mathrm d\theta$. This needs to be integrated from $\theta$ to $1$; I don't think there's a nice closed form for the result; at least it's not the one you conjectured. 
A: Let the axis of the cone $C$ be the $x_1$-axis and assume for the moment $r=1$. Then the points $(t, x'):=(t,x_2,\ldots, x_n)\in C$ are characterized by
$$ |x'| \leq \min\{t\ \tan \theta, \sqrt{1-t^2}\}\ ,\qquad 0\leq t\leq 1\ .$$
It follows that a plane $t={\rm const.}$ intersects $C$ in an $(n-1)$-dimensional ball of radius $t\ \tan\theta$ if $t\in[0,\cos\theta]$, and of radius $\sqrt{1-t^2}$ if $t\in[\cos\theta, 1]$. It follows that the volume of $C$ is given by
$${\rm vol}(C)= \int_0^{\cos\theta}\sigma_{n-1}(t\ \tan\theta)^{n-1}\ dt + \int_{\cos\theta}^1 \sigma_{n-1}(\sqrt{1-t^2})^{n-1}\ dt=\ldots\ .$$
For arbitrary $r>0$ multiply this by $r^n$. 
