# 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 $\displaystyle \frac{\pi^{n/2} r^n}{\Gamma(\frac{n}{2}+1)}$

Although I expect most readers 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 $\displaystyle I = \int_{\mathbf{R}^n} e^{-\pi|x|^2}\,dx.$

We will compute ${I}$ in two different ways. On the one hand, $\displaystyle I = \int_{-\infty}^{+\infty}\cdots\int_{-\infty}^{+\infty} e^{-\pi x_1^2 -\cdots - \pi x_n^2}\,dx_1\cdots dx_n = \left(\int_{-\infty}^{+\infty} e^{-\pi x^2}\,dx\right)^n = 1.$

On the other hand, the form of ${I}$ suggests introducing a radial coordinate ${r=|x|}$. Computing this way, $\displaystyle I = \int_0^\infty e^{-\pi r^2} (V_n(1) n r^{n-1})\,dr = \frac{n V_n(1)}{\pi^{n/2}} \int_0^\infty e^{-t^2} t^{n-1}\, dt\qquad\quad$ $\displaystyle \qquad\quad= \frac{n V_n(1)}{2 \pi^{n/2}} \int_0^\infty e^{-s} s^{n/2-1}\,ds = \frac{n V_n(1)}{2 \pi^{n/2}} \Gamma(n/2) = \frac{V_n(1)\Gamma(n/2+1)}{\pi^{n/2}}.$