# 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?

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How does $\eta(x)$ depend on $x$? Do you mean it's just $x$ after normalization? –  Raskolnikov Jun 15 '11 at 9:35
So $\eta(x)$ is just a unit vector that does depend on $x$. It tells you the direction that the hypercone is pointing in. The volume of the region will be independent of $x$ and $\eta(x)$ but I had to put it in so to define the region. –  alext87 Jun 15 '11 at 9:49

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 $r^n$.
I don't understand. What do you mean by "framework"? And where's the different notation? The only notation I used was $\theta$, which the OP introduced. Also I'm not sure that my post requires more sophistication, but that might depend on the perspective. –  joriki Jun 16 '11 at 9:26
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.