It is well-known that the volume of the unit ball in n-space is $\pi^{n/2}/\Gamma(n/2+1)$. Do you know of a proof which explains this formula? Any proof which does not treat the cases $n$ even and $n$ odd separately (like using an explicit expression for $\Gamma(n/2+1)$ in terms of factorials) should be fine.

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    $\begingroup$ Let's reserve soft-question tag for stuff without formulas :) $\endgroup$ – Ilya Nikokoshev Nov 19 '09 at 20:26
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    $\begingroup$ More memorably, the formula is $\pi^{n/2} \! \left/ \Pi(n/2) \right.$, using Euler's original notation $\Pi(s)$ for what we now call $\Gamma(s+1)$ (so $\Pi(n) = n!$). I learned this from P.X.Gallagher though for all I know it might go back to Euler Himself. $\endgroup$ – Noam D. Elkies Oct 30 '15 at 1:00

It is easier to take the derivative, and consider the volume of the $(n-1)$-sphere (i.e., the "surface area" of the boundary of the ball).

Start with the integral $\int_{\mathbb{R}^n} e^{-x_1^2 - \cdots - x_n^2} dx_1 \cdots dx_n$. Fubini's theorem lets you decompose this into a product of $1$-dimensional integrals, and you get $\pi^{n/2}$. Since the integrand is spherically symmetric, you can change to the integral $\int_0^\infty \mathrm{vol}(S^{n-1}(r)) \, e^{-r^2} dr$, where $S^{n-1}(r)$ is the unit $(n-1)$-sphere of radius $r$. The volume of this sphere is $r^{n-1}$ times the volume of the unit sphere, so solving for that, you get $\frac{\pi^{n/2}}{\int_0^\infty r^{n-1} e^{-r^2} dr}$. A change of coordinates ($u = r^2$) in the denominator yields the integral defining $\Gamma(n/2)$.


I like to write it as $\omega_n = \frac{\pi^\frac{n}{2}}{\frac{n}{2}!}$ (I've seen $\omega_n$ used both for the measure of the unit ball and for that of its boundary, but eh, I had to attach some name to it for below). I guess using the factorial notation for non-integers isn't too popular, though.

Alternatively, induction. It's true for $n=1$ (since $\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{2}$) and $n=2$.

So then:

$\omega_{n+2} = \int_{x_1^2 + \dots + x_{n+2}^2 \leq 1}dx = \int_{x_{n+1}^2+x_{n+2}^2 \leq 1}\int_{x_1^2 + \dots + x_n^2 \leq 1 - (x_{n+1}^2+x_{n+2}^2)}d(x_1,\dots,x_n)d(x_1,x_2).$

Polar coordinates in the plane give us

$\omega_{n+2} = \int_0^{2\pi}\int_0^1\sqrt{1-r^2}^n\omega_n r dr d\phi = 2\pi\omega_n \int_0^1(1-r^2)^{\frac{n}{2}}rdr = \pi\omega_n \int_0^1(1-r^2)^{\frac{n}{2}}2rdr.$

Substitute $s=1-r^2$ and get

$\omega_{n+2} = \pi\omega_n\int_0^1s^\frac{n}{2}ds = \omega_n \frac{2\pi}{n+2} = \frac{\pi^\frac{n}{2}}{\frac{n}{2}!}\frac{2\pi}{n+2} = \frac{\pi^\frac{n+2}{2}}{\frac{n+2}{2}!}.$

  • $\begingroup$ That's meant to be $d(x_1,\dots,x_n)d(x_{n+1},x_{n+2})$, of course. $\endgroup$ – Jakob Katz Nov 19 '09 at 21:07
  • $\begingroup$ I really like this one. I've never seen it proved this way, I've only ever seen the $e^{-x^2}$ way and some even more incredible way my linear algebra teacher did using inner product spaces. $\endgroup$ – user78249 Nov 2 '16 at 20:36
  • $\begingroup$ Years ago P.X.Gallagher noted the even more memorable form $$ \pi^{n/2} \bigl/ \Pi(n/2) \bigr. $$ where $\Pi(x)$ is Euler's original notation for what we now call $\Gamma(x+1)$. $\endgroup$ – Noam D. Elkies Nov 2 '16 at 21:01

Here is an argument similar to that of S. Carnahan, but more direct (without sphere surface area.)

Denote the volume of a unit ball in $\mathbb{R}^n$ by $c_n$, then the volume of a ball of radius $r$ equals $c_nr^n$.

We use the (Lebesgue-Stieltjes?) formula $\int fd\mu=\int_0^\infty \mu\{x:f(x)>t\}dt$ for a positive function $f$ on a measure space $(X,\mu)$.

Let $(X,\mu)$ be $\mathbb{R}^n$ with Lebesgue measure and $f(x)=e^{-\sum x_i^2}$. Then $$\pi^{n/2}=\left(\int_{-\infty}^\infty e^{-x^2}dx\right)^n=\int fd\mu=\int_0^1 c_n(-\log t)^{n/2}dt=\\=c_n\int_0^\infty s^{n/2}e^{-s}ds=c_n\Gamma(n/2+1),$$ where we substitute $t=e^{-s}$.


The direct way to do this is to integrate $1$ using the spherical volume element $$ d^nV = r^{n-1} \sin^{n-2}(\phi_1) \, \sin^{n-3}(\phi_2) \, \cdots \sin(\phi_{n-2}) \, dr \, d\phi_1 \, \cdots \, d\phi_{n-1}. $$ Here, for $\phi_k$ goes from $0$ to $\pi$ for $k = 1, \dots, n-2$, and $\phi_{n-1}$ goes from $0$ to $2 \, \pi$.

Since $$ \int_0^\pi \sin^k \phi \, d\phi = B\left(\frac12, \frac{k+1}{2} \right) =\frac{ \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{k+1}{2}\right) } { \Gamma\left(\frac{k+2}{2} \right) }, $$ where $B(p,q)$ is the beta function, we get $$ \begin{align} \int d^nV &= \int_0^1 r^{n-1} \, dr \, \frac{ \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{n-1}{2}\right) } { \Gamma\left(\frac{n}{2} \right) } \frac{ \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{n-2}{2}\right) } { \Gamma\left(\frac{n-1}{2} \right) } \cdots \frac{ 2 \, \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{1}{2}\right) } { \Gamma\left(\frac{2}{2} \right) } \\ &= \frac{2}{n} \frac{ \left[ \Gamma\left(\frac{1}{2} \right)\right]^n } { \Gamma\left(\frac{n}{2} \right) } = \frac{ {\sqrt\pi}^n } { \Gamma\left(\frac{n}{2} + 1\right) }. \end{align} $$

Another way is to use the formula for the Fourier transform of a spherical function $$ \tilde f(k) = \int f(r) \, e^{i\mathbf k \cdot \mathbf r} \, d\mathbf r = \frac{(2\pi)^{n/2}} { k^{n/2 - 1} } \int_0^\infty f(r) J_{n/2-1}(k\,r)\, r^{n/2} \, dr, $$ where $J_{n/2-1}(k \, r)$ is the Bessel function.

In our case $f(r) = \Theta(1-r)$ is the step function. Then, by using the recurrence relation $$ \frac{d}{dx} \left[ r^{\nu} J_{\nu}(x) \right] = r^{\nu} J_{\nu-1}(x), $$ with $\nu = n/2$, we get $$ \tilde f(k) = \frac{(2\pi)^{n/2}} { k^{n/2} } J_{n/2}(k). $$ For the limit of $k\rightarrow 0$, we get the volume $$ \lim_{k \rightarrow 0} \tilde f(k) = \frac{(2\pi)^{n/2}} { k^{n/2} } \frac{(k/2)^{n/2}}{\Gamma(\frac{n}{2}+1)} = \frac{\pi^{n/2}} {\Gamma(\frac{n}{2}+1)}. $$


Adding to what Jakob Katz wrote, where the volume of the unit ball $B_{2n}$ in $\mathbb{C}^n$ is $\pi^n/n!$, the $\pi^n$ calls to mind the volume of a polydisc $B_2^n \subset (\mathbb{C}^1)^n = \mathbb{C}^n$, on which the symmetric group $S_n$ acts by permuting coordinates. The permutation action is volume-preserving, and so $\pi^n/n!$ is naturally interpreted as the volume of the orbit space $B_2^n/S_n$. Can this be somehow related to the volume of the $n$-ball?

Andreas Blass and Stephen Schanuel cooked up such an explanation:

Theorem: (Blass-Schanuel) Given $(z_1, \ldots, z_n) \in \mathbb{C}^n$, write coordinates $z_j$ in polar coordinate form $z_j = r_j e^{i \theta_j}$, and define an $S_n$-invariant map $\phi \colon B_2^n \to B_{2 n}$ by first permuting the $z_j$ so that $r_1 \geq r_2 \geq \ldots \geq r_n$ and then mapping $(z_1, \ldots, z_n)$ to $$(\sqrt{r_1^2 - r_2^2}e^{i\theta_1}, \sqrt{r_2^2 - r_3^2}e^{i(\theta_1 + \theta_2)}, \ldots, \sqrt{r_{n-1}^2-r_n^2}e^{i(\theta_1 + \theta_2 + \ldots + \theta_{n-1})}, r_n e^{i(\theta_1 + \theta_2 + \ldots + \theta_n)})$$ Then $\phi$ induces a continuous well-defined map $B_2^n/S_n \to B_{2 n}$. Furthermore, when restricted to the set $P_n$ of $(z_1, \ldots, z_n)$ for which the $r_j$ are all distinct, $\phi$ induces a smooth symplectic isomorphism mapping $P_n/S_n$ onto the set $Q_n$ of $(w_1, \ldots, w_n) \in B_{2 n}$ for which $w_j \neq 0$ for $1 \leq j \leq n-1$.

In other words, writing $z_j = x_j + i y_j$ the symplectic 2-form

$$\sum_{j=1}^n d x_j \wedge d y_j = \sum_{j=1}^n r_j d r_j \wedge d\theta_j$$

is preserved by pulling back along $\phi \colon P_n/S_n \to Q_n$. Since symplectic maps are locally volume-preserving, and since $P_n$ and $Q_n$ are almost all of $B_2^n$ and $B_{2 n}$ respectively, this gives a proof that the volume of $B_{2 n}$ is $\pi^n/n!$ (alternate to standard purely computational proofs).


  • Andreas Blass, Stephen Schanuel, On the volumes of balls (ps).
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    $\begingroup$ Thanks for publicizing my paper with Steve Schanuel. The paper also contains a slightly simpler version of the map, preserving volumes but not the symplectic structure. While Steve and I were procrastinating plans for amplifying our paper, that simpler map was independently found by Omar Hijab; see "The volume of the unit ball in $C^n$," Amer. Math. Monthly107 (2000) 259. (When Steve found this paper, he notified me with an email that began "If one postpones a task long enough, eventually it is no longer necessary.") $\endgroup$ – Andreas Blass Oct 30 '15 at 10:45

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