Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|cite|improve this question
Let's reserve soft-question tag for stuff without formulas :) –  Ilya Nikokoshev Nov 19 '09 at 20:26

2 Answers 2

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 $\int\_{\mathbb{R}^n} e^{-x\_1^2 - ... - x\_n^2} dx\_1 \dots dx\_n$. 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 $\int\_0^\infty vol(S^{n-1}(r)) e^{-r^2} dr$, 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 = r2) in the denominator yields the integral defining $\Gamma(n/2)$.

share|cite|improve this answer

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

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

share|cite|improve this answer
That's meant to be $d(x_1,\dots,x_n)d(x_{n+1},x_{n+2})$, of course. –  Jakob Katz Nov 19 '09 at 21:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.