Volume of intersection of unit $n$-sphere and $n$-cone (or how to evaluate $\prod_{i=1}^{n-1} \int_{\phi_i=0}^t \sin^{n-i-1}(\phi_i)\,d\phi_{i}$) I wish to find the volume, $v_n$, of intersection between a unit $n$-sphere and a solid $n$-dimensional right circular cone whose height is larger than $1$ and whose aperture is $\theta<\pi.$
From this section of the wiki-article for an $n$-sphere, I've concluded (perhaps incorrectly) that this should be given by
\begin{align}
v_n(\theta)=\int_{\phi_1=0}^\theta\int_{\phi_2=0}^\theta\cdots \int_{\phi_{n-1}=0}^\theta\int_{r=0}^1 \mathrm{d}^nV,
\end{align}
where 
$$\mathrm{d}^nV=r^{n-1} \sin^{n-2}(\phi_1)\sin^{n-3}(\phi_2)\cdots \sin^1(\phi_{n-2})\sin^0(\phi_{n-1})\,\mathrm{d}r \,\mathrm{d}\phi_{1}\,\mathrm{d}\phi_{2} \cdots \,\mathrm{d}\phi_{n-1}.$$
We can simplify the notation a bit:
$$v_n(\theta)=\frac{1}{n}\prod_{i=1}^{n-1} \int_{\phi_i=0}^\theta \sin^{n-i-1}(\phi_i)\,\mathrm{d}\phi_{i}.$$

If this is the correct expression, how can it be evaluated? 
If it isn't the correct expression, what then is, and how can that be evaluated?
Or is there an easier way? (Perhaps involving "trimming down" the volume of an $n$-cone with height $1$?)

 A: $\newcommand{\Reals}{\mathbf{R}}$Cavalieri's theorem looks easier: Let $c_{n}$ denote the volume of the unit ball in $\Reals^{n}$, so that a ball of radius $\rho$ has $n$-dimensional volume $c_{n} \rho^{n}$.
If $x$ is a Cartesian coordinate along the cone axis with its origin at the vertex, each hyperplane orthogonal to the $x$-axis cuts your solid in an $(n-1)$-ball whose radius is
$$
\rho = \begin{cases}
  x\tan\theta & 0 \leq x \leq \cos\theta, \\
  \sqrt{1 - x^{2}} & \cos\theta < x \leq 1.
\end{cases}
$$
Consequently, the volume of your object is
\begin{multline*}
\int_{0}^{\cos\theta} c_{n-1}(x\tan\theta)^{n-1}\, dx + \int_{\cos\theta}^{1} c_{n-1} (1 - x^{2})^{(n-1)/2}\, dx \\
= c_{n-1} \left[\frac{\sin^{n-1}\theta \cos\theta}{n} + \int_{\frac{\pi}{2} - \theta}^{\frac{\pi}{2}} \cos^{n} \phi\, d\phi\right].
\end{multline*}
The second integral can, at worst, be handled using the reduction formula
$$
\int \cos^{n}\phi\, d\phi
= \frac{1}{n}\left[\cos^{n-1}\phi \sin\phi + (n-1) \int \cos^{n-2}\phi\, d\phi\right].
$$
