Volume of an n-ball - Aetherwisp

An n-ball is the $n$ -dimensional analog of a ball. In one dimension, it is a segment. In two, it is a disk. In three it is a proper ball. The volume of an $n$ -ball can be calculated generically for any dimension $n$ . One way to do so is to recognize that the volume must be proportional to the radius $R$ of the ball to the $n$ -th power (for dimensional constraints):

V_{n}=c_{n}R^{n}

where $c_{n}$ is some scaling constant. The surface of such a ball will be

S_{n}=\frac{ \partial V_{n} }{ \partial R } =nc_{n}R^{n-1}

which is the surface of the bounding $(n-1)$ -dimensional hypersphere. We can manually calculate these values for two and three dimensions as

\begin{align} \text{(2D)}\quad&V_{2}=\pi R^{2}&S_{2}=2\pi R \\ \text{(3D)}\quad&V_{3}=\frac{4\pi}{3}R^{3}&S_{3}=4\pi R^{2} \end{align}

The trick is to see the following integral:

\int_{-\infty}^{\infty} e^{-(x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2})} \ dx_{1}\ldots dx_{n}=\pi^{n/2}=\int_{0}^{\infty}e^{-R^{2}}S_{n}(R)\ dR=nc_{n}\int_{0}^{\infty}e^{-R^{2}}R^{n-1}\ dR

The $\pi^{n/2}$ comes by just solving the integral normally. The equality with the second integral comes from doing a coordinate shift to Polar coordinates, introducing the variable $R^{2}=x_{1}^{2}+\ldots+x_{n}^{2}$ and finding that the coordinate change function is exactly $S_{n}(R)$ . To determine $c_{n}$ , we just need to recognize that the last integral is a Gaussian integral and, by substituting $R^{2}=t$ and expressing it as a Gamma function, we get

V_{n}(R)=\frac{\pi^{n/2}}{\Gamma\left( \frac{n}{2}+1 \right)}R^{n}