Where does this formula for the volume of a n-dimensional ball come from?

Question

I recently came across the following formula for the volume of an n-dimensional unit ball: $$\frac{\pi^{n/2}}{\Gamma(n/2 + 1)}$$ Why exactly does this formula work?

Answer 1

The story so far: $$ \lim_{n \rightarrow \infty} V_n(1) = 0. $$ This has made a lot of people very angry and been widely regarded as a bad move.

Induction, but increasing the dimension by $2$ each time. You do need to know that $\Gamma(1+n) = n!$ along with the special value $$ \Gamma(\frac{1}{2}) = \sqrt \pi $$ The recursion is, in general, $$ z \Gamma(z) = \Gamma(1+z) $$

Using the wikipedia notation, you have written $V_n(1),$ so that $V_n(R) = V_n(1) R^n.$

$$ V_{n+2}(1) = \int_0^{2 \pi} \int_0^1 V_n(1) \left( \sqrt{1-r^2} \right)^n r dr d\theta = 2 \pi V_n(1) \int_0^1 \left( \sqrt{1-r^2} \right)^n r dr $$
$$ V_{n+2}(1) = \pi V_n(1) \int_0^1 \left( \sqrt{1-r^2} \right)^n (2r) dr $$

I get $$ V_{n+2}(1) = V_{n}(1) \cdot \frac{2 \pi}{n+2}. $$ Let me check that... Yes.

Here is the application by induction, given $$ V_n(1) = \frac{\pi^{n/2}}{\Gamma (\frac{n+2}{2})}, $$ we arrive at $$ V_{n+2}(1) = V_{n}(1) \cdot \frac{ \pi}{ \frac{n+2}{2}} = \frac{\pi^{n/2} \pi}{\Gamma (\frac{n+2}{2}) \frac{n+2}{2}} = \frac{\pi^{(n+2)/2}}{\Gamma ( 1 + \frac{n+2}{2}) }$$

Answer 2

I will sketch the very elegant argument given by Keith Ball in An Elementary Introduction to Modern Convex Geometry. Let we consider:

$$ I(n) = \int_{\mathbb{R}^n}\exp\left(-\sum_{k=1}^{n}x_k^2\right)\,d\mu. \tag{1}$$ By Fubini's theorem, it is just $\Gamma\left(\frac{1}{2}\right)^n = \pi^{n/2}$. Let: $$ S_R = \int_{x_1^2+\ldots+x_n^2=R^2}1\,d\mu. \tag{2}$$ By scaling, $S_R$ is just $R^{n-1} S_1$. If we apply Cavalieri's principle to $(1)$, we have: $$ I(n) = \int_{0}^{+\infty} S_R e^{-R^2}\,dR = S_1\int_{0}^{+\infty}R^{n-1}e^{-R^2}\,dR = \frac{S_1}{2}\,\Gamma\left(\frac{n}{2}\right) \tag{3}$$ hence: $$ S_1 = \frac{2\pi^{n/2}}{\Gamma\left(\frac{n}{2}\right)},\qquad S_R = \frac{2\pi^{n/2}}{\Gamma\left(\frac{n}{2}\right)}\,R^{n-1}\tag{4} $$ but by Cavalieri's principle/scaling again, if $$ V_R = \int_{x_1+\ldots+x_n\leq R^2}1\,d\mu = R^n V_1\tag{5}$$ we have: $$ \frac{d}{dR}\,V_R = S_R, \tag{6} $$ hence it follows that: $$ V_R = \frac{\pi^{n/2}}{\Gamma\left(\frac{n}{2}+1\right)}\,R^{n} \tag{7}$$ as wanted.