Tuesday, 13 March 2012

Volume of the $n$-ball

The following is a simple and beautiful proof, shown to me to my great delight while I was in high school, that the $n$-ball of radius $r$ has $n$-volume
\[\frac{\pi^{n/2} r^n}{\Gamma(\frac{n}{2}+1)}\]
Although I expect most people will know it, I believe that everybody should see it. I don't know the history of it, and would be interested in learning.

Suppose that the $n$-ball $B^n$ of radius $1$ has $n$-volume $V_n(1)$. Then, by considering linear transformations, the $n$-ball $rB^n$ of radius $r$ has $n$-volume $V_n(r) = V_n(1) r^n$. Moreover, differentiating this with respect to $r$ should produce the surface $(n-1)$-volume of the $(n-1)$-sphere $S^{n-1} = \partial B^n$: thus we expect $\text{vol}^{(n-1)}(S^{n-1}) = V_n(1) n r^{n-1}$.

Now consider the integral
\[I = \int_{\mathbf{R}^n} e^{-\pi|x|^2}\,dx.\]
We will compute $I$ in two different ways. On the one hand,
\[I = \int_{-\infty}^{+\infty}\cdots\int_{-\infty}^{+\infty} e^{-\pi x_1^2 -\cdots - \pi x_n^2}\,dx_1\cdots dx_n = \left(\int_{-\infty}^{+\infty} e^{-\pi x^2}\,dx\right)^n = 1\]
On the other hand, the form $I$ suggests introducing a radial coordinate $r=|x|$. Computing this way,
\[I = \int_0^\infty e^{-\pi r^2} (V_n(1) n r^{n-1})\,dr = \frac{n V_n(1)}{\pi^{n/2}} \int_0^\infty e^{-t^2} t^{n-1}\, dt\qquad\quad\]
\[\qquad\quad= \frac{n V_n(1)}{2 \pi^{n/2}} \int_0^\infty e^{-s} s^{n/2-1}\,ds = \frac{n V_n(1)}{2 \pi^{n/2}} \Gamma(n/2) = \frac{V_n(1)\Gamma(n/2+1)}{\pi^{n/2}}.\]

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