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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 $1$?)

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$\newcommand{\Reals}{\mathbf{R}}$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 $$ \int \cos^{n}\phi\, d\phi = \frac{1}{n}\left[\cos^{n-1}\phi \sin\phi + (n-1) \int \cos^{n-2}\phi\, d\phi\right]. $$

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    Thank you! I'm having trouble implementing your answer, though. For instance for $n=2$ and $\theta=\frac{2}{3}\pi$ I get a "volume" (area) of $\frac{2}{3}\pi,$ but by hand I get $\frac{\theta}{2}=\frac{1}{3}\pi.$ What do you get? Are you perhaps defining the aperture differently than I? – 2017-01-02
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    Ah, yes: Implicitly I was taking $\theta \leq \pi/2$. If $\pi/2 < \theta < \pi$, subtract the "complementary" volume from the volume of the unit ball. For $\theta = 2\pi/3$, the complementary angle is $\pi/3$; the formula gives an area of $\pi/3$, so the correct area is $\pi - \pi/3 = 2\pi/3$. (This is "expected": If $\theta = 2\pi/3$, the wedge subtends an angle $2\theta = 4\pi/3$. :) – 2017-01-02
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    So, just to be clear, if we call the volume you calculated in your answer $v_n$, then $c_n-v_n$ would be the appropriate expression when $\pi/2<\theta < \pi$? – 2017-01-02
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    Yes (provided you also replace $\theta$ with $\pi - \theta$). – 2017-01-02
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    I think making the substitution $\theta \rightarrow \pi - \theta$ is enough (so that shouldn't be subtracted from $c_n$, I think)? .. At least that gives the correct result for the example (when plugging it into Mathematica). – 2017-01-02
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    I'm skeptical: The solid with $0 < \theta < \pi/2$ is less than half the ball, so its complement has larger volume (the sum of the volumes is $c_{n}$). Incidentally, the splitting of the integral on the left works for $0 \leq \theta \leq \pi$ (the first integral runs right to left if $\pi/2 < \theta$, so the effect is to subtract the volume of a cone from a "zone" of the unit ball), but to get the second integral on the right one needs $\arcsin(\cos\theta) = \frac{\pi}{2} - \theta$, which is true only if $\theta < \pi/2$. – 2017-01-02
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    You are probably right. My confusion arises due to the assumptions I give when evaluating in Mathematica. I write: f[n_, t_] = Assuming[Element[n, Integers] && n > 0 && Element[t, Reals] && Pi/2 < t < Pi, Integrate[Cos[x]^n, {x, t - Pi/2, Pi/2}]]; cn[n_] = Pi^(n/2)/Gamma[n/2 + 1]; g[n_, t_] = cn[n - 1]*(Sin[(Pi - t)]^(n - 1)*Cos[(Pi - t)]/n + f[n, t]) and when evaluating g[2, 2/3 Pi], it gives me the desired $\pi/3.$ Note that I didn't subtract the volume from that of the unit sphere. – 2017-01-02
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    I'm sorry, I'm still having a hard time getting the correct results: If I implement your original formula for $n=2,\theta=\pi/4,$ it should be obvious that the result should be $\pi/8,$ but I get $\pi/4$ from your formula instead. Do you have any ideas as to what might be going wrong? Thank you. – 2017-01-03
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    My apology: What I've been calling $\theta$ is _one-half_ the aperture. (From habit, I measure cone angles from the axis to a generator....) Please ask if that doesn't reconcile my formulas and your calculations. – 2017-01-03
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    That seemed to do it! That also made it clear why you assumed $\theta\leq \pi/2$ - of course the cone is convex! What I meant I was interested in was the aperture going from $\pi/2$ to $\pi$ and that is covered by your original range of $\theta$. Thanks again, now it all works like a charm (and I can get on with my bachelor's thesis!) :D – 2017-01-03