Hypervolume of intersection of hypercylinders

Question

Let $B_n=\{(x_1, \ldots, x_n) \mid x_1^2+\cdots+x_n^2 \leq 1+ x_i^2 \ (i=1,\ldots,n)\}$.

How to compute the hypervolume of $B_n$? I solved $n=2, 3$ but it is difficult for greater dimensions.

Answer 1

Let's denote by $\mu$ the usual Lebesgue measure (our "hypervolume") on $\mathbb{R}^n$. An idea I have is to use induction, but with a preliminary simplification, connecting the present problem to computing the measure of a "piece" of the $n$-dimensional unit ball cut by a translated orthant (see $f_n(x)$ below). It leads to $ \newcommand{\dpoint}{(x_1,\ldots,x_n)\in\mathbb{R}^n} \newcommand{\ineqch}[2]{#1\leqslant #2\leqslant\ldots\leqslant x_n} \newcommand{\sphere}[2]{#1^2+\ldots+x_n^2\leqslant #2} \newcommand{\volint}[4]{\idotsint\limits_{\substack{#1\\#2}}#3 d#4\ldots dx_n}$

$$\mu(B_n)=2^n n(n-1)I_{n-2},\qquad I_n=\int_0^{r_n}\frac{x f_n(x)~dx}{(1+x^2)^{2+n/2}},\qquad r_n=\begin{cases}1/\sqrt{n},&n>0\\\infty,&n=0\end{cases}\tag{1}\label{eq1}$$ where $f_n(x)=\mu\{\dpoint\mid\sphere{x_1}{1},\ x_k\geqslant x\ (1\leqslant k\leqslant n)\}$ is computed using $$f_0(x)\equiv 1,\qquad f_{n+1}(x)=\int_{x\sqrt{n}}^{\sqrt{1-x^2}}\frac{y^{n+1}f_n(x/y)}{\sqrt{1-y^2}}~dy.\tag{2}\label{eq2}$$ The proof sketch is given below. This way, we find (you must know the first two values already) $$\mu(B_2)=4,\quad\mu(B_3)=8(2-\sqrt{2}),\quad\color{blue}{\mu(B_4)=12(\pi-2\sqrt{2}\tan^{-1}\sqrt{2})},\quad\ldots$$ While $f_3(x)$ may be computed analytically, I'm in doubt whether this can be done for $\mu(B_5)$.

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First let's deal with $f_n(x)$. For $0\leqslant x\leqslant r/\sqrt{n}$, let $$A_n(x,r)=\{\dpoint\mid\sphere{x_1}{r^2},\ x_k\geqslant x\ (1\leqslant k\leqslant n)\};$$ then clearly $\mu\big(A_n(x,r)\big)=r^n f_n(x/r)$. On the other hand, the intersection of $A_{n+1}(x,r)$ with the plane $x_{n+1}=z$ (for $x\leqslant z\leqslant\sqrt{r^2-nx^2}$, when it is not empty) is isometric to $A_n(x,\sqrt{r^2-z^2})$, which gives $$\mu\big(A_{n+1}(x,r)\big)=\int_x^{\sqrt{r^2-nx^2}}\mu\big(A_n(x,\sqrt{r^2-z^2})\big)~dz$$ and, substituting $\sqrt{r^2-z^2}=y$ and $r=1$, we obtain the recurrence $\eqref{eq2}$ above (for $n>0$; the equality $f_1(x)=1-x$ is obvious, so that the recurrence holds for $n=0$ too).

Now, the simplification. Let $S_n=\{\dpoint\mid 0\leqslant x_1\leqslant\ldots\leqslant x_n\}$. Then, by symmetry, \begin{align} \mu(B_n)&=2^n n!~\mu(B_n\cap S_n) \\&=2^n n!\volint{\ineqch{0}{x_1}}{\sphere{x_2}{1}}{}{x_1} \\&=2^n n!\volint{\ineqch{0}{x_2}}{\sphere{x_2}{1}}{x_2~}{x_2} \\&=2^n n!\int_0^{\color{red}{r_{n-1}}}x_2\volint{\ineqch{x_2}{x_3}}{\sphere{x_3}{1-x_2^2}}{}{x_3}~dx_2 \\&=2^n n(n-1)\int_0^{r_{n-1}}y\volint{x_3,\ldots,x_n\geqslant y}{\sphere{x_3}{1-y^2}}{}{x_3}~dy \\&=2^n n(n-1)\int_0^{r_{n-1}}y(1-y^2)^{\frac{n}{2}-1}f_{n-2}\left(\frac{y}{\sqrt{1-y^2}}\right)dy \end{align} as noted above, and substituting $y/\sqrt{1-y^2}=x$, we get $\eqref{eq1}$ (again excluding the $f_0$ case).