Introduction

This post is an extension of my Medium article: The Math Behind “The Curse of Dimensionality”. If you have any questions or feedback, feel free to leave a comment below or reach out on LinkedIn.

Here, I derive the formula for the volume of the n-ball using integral calculus and Wallis integrals.

An n-ball is a generalization of the ordinary “ball” to arbitrary dimensions. It is defined as the set of points in the n-dimensional Euclidean space that are at a fixed distance from a central point (we will condider 0, the origin). The n-ball has important applications, particulary in machine learning and data analysis.

The equation of an n-ball is given by:

\[x_1^2 + x_2^2 + \ldots + x_{N}^2 \leq r^2\]

where \(r\) is the radius of the ball.

Derivation of the volume formula

One of the most important properties of the n-ball is its volume. The volume of an n-ball of radius \(r\) is given by:

\[V_N(r) = \frac{\pi^{(N+1)/2}}{\Gamma((N+1)/2)} r^{N+1}\]

where \(\Gamma\) is the gamma function. In the rest of this post, we will derive this formula.

We define the unit (radius equal to 1) n-ball as the following space: \(\mathcal{B}_n = \{x_1, \dots, x_m; \sum\limits_{i=1}^n x_i^2 \leq 1 \}\)

We note \(V_n = V_n(1)\) the volume of this ball. And more generally \(V_n(R)\) the volume of the n-ball with radius \(R\). From now on, we will use \(V_n\) to refer to the volume of the unit n-ball:

\[V_n = \int_{x \in \mathcal{B}_n} dx_1 dx_2\dots dx_n\]

First, note that the volume of the n-ball of radius \(R\) is \(R^n V_n\): simply use the change of variable \(y_i \leftarrow x_i / R\) in the expression of \(V_n(R)\).

Now, let’s simplify \(V_n\):

\[\begin{align*} V_n &= \int_{x_1^2 + \dots + x_n^2 \leq 1} dx_1 dx_2\dots dx_n \\ &= \int_{x_1^2 \leq 1} \left( \int_{x_2^2 + \dots + x_n^2 \leq 1 - x_1^2} dx_2\dots dx_n \right) dx_1 \\ &= \int_{x_1^2 \leq 1} V_{n-1}\left(\sqrt{1-x_1^2}\right) dx_1 \end{align*}\]

We can now replace the expression of the volume of the \(n-1\) ball of radius \(\sqrt{1-x_1^2}\) with the previous relation:

\[\begin{align*} V_n &= \int_{x_1^2 \leq 1} V_{n-1}\left(\sqrt{1-x_1^2} \right) dx_1 \\ &= V_{n-1}\int_{x_1^2 \leq 1} \left(\sqrt{1-x_1^2}\right)^{n-1} dx_1 \\ &= V_{n-1}\int_{-1}^{1} \left(\sqrt{1-x^2}\right)^{n-1} dx \end{align*}\]

We use the change of variable \(x = \cos(\theta)\) (so \(dx = -\sin(\theta) d\theta\)) to simplify the integral:

\[\begin{align*} V_n &= V_{n-1}\int_{-1}^{1} \left(\sqrt{1-x^2}\right)^{n-1} dx \\ &= -V_{n-1}\int_{\pi}^{0} \sin^{n-1}(\theta) \sin(\theta) d\theta \\ &= V_{n-1}\int_{0}^{\pi} \sin^{n}(\theta) d\theta \\ &= 2 V_{n-1}\int_{0}^{\pi/2} \sin^{n}(\theta) d\theta \\ &= I_n V_{n-1} \end{align*}\]

We note \(I_n = 2\int_{0}^{\pi/2} \sin^{n}(\theta) d\theta\).

Note that \(\int_{0}^{\pi/2} \sin^{n}(\theta) d\theta\) is a famous integral in mathematics: it is called the Wallis integral and often denoted by \(W_n\). It can be derived using integration by parts (I will prove this formula in the appendix). Depending on the parity of \(n\), the integral can be expressed as:

\(W_{2p} = \frac{\pi}{2} \frac{(2p)!}{2^{2p}(p!)^2}\) and \(W_{2p+1} = \frac{2^{2p} (p!)^2}{(2p+1)!}\)

Using our recursive relation, we know that: \(V_n = I_n I_{n-1} \dots I_2 V_1\) and \(V_1 = V_1(1) = 2\) (the length of the segment \([-1, 1]\)).

We are going to make use of a very useful property:

\[I_{2p}I_{2p+1} = 4W_{2p}W_{2p+1} = \frac{\pi}{2} \frac{(2p)!}{2^{2p}(p!)^2} \frac{2^{2p} (p!)^2}{(2p+1)!} = \frac{2\pi}{2p+1} = \frac{\pi}{p+1/2}\]

We will also need the Gamma function. All you need to know is these 2 expressions:

\[\Gamma(n) = (n-1)!\]

and

\[\Gamma(n+1/2) = (n-1/2)\times \dots \times 1/2 \times \pi ^{1/2}\]

Now, we can easily compute the volume \(V_n\) by grouping successive terms, depending again on the parity of \(n\).

If \(n = 2p\): \(\begin{align*} V_{2p} &= I_{2p}I_{2p-1} \dots I_2 V_1 \\ &= I_{2p} (I_{2p-1}I_{2p-2}) \dots (I_3I_2)\times 2 \\ &= \pi \times \frac{(2p)!}{2^{2p}(p!)^2} \times \frac{\pi}{p-1/2} \times \frac{\pi}{p-3/2}\times \dots \times \frac{\pi}{3/2} \times \frac{\pi}{1/2} \\ &= \frac{(2p)!}{2^{p}(p!)^2} \times \frac{\pi^{p}}{(2p-1)(2p-3)\dots (1)} \\ &= \frac{\pi^{p}}{2^p} \times \frac{(2p)!}{(p!)^2} \times \frac{(2p)(2p-2)\dots (4) (2)}{(2p!)} \\ &= \frac{\pi^{p}}{2^p} \times \frac{(2p)!}{(p!)^2} \times \frac{2^p p!}{(2p)!} \\ &= \frac{\pi^p}{p!} \\ &= \frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)} \end{align*}\)

If \(n = 2p+1\), it is a bit less complicated:

\[\begin{align*} V_{2p+1} &= I_{2p+1}I_{2p}I_{2p-1} \dots I_2 V_1 \\ &= (I_{2p+1}I_{2p})(I_{2p-1}I_{2p-2}) \dots (I_3I_2)\times 2 \\ &= \frac{\pi}{p+1/2} \times \frac{\pi}{p-1/2} \times \frac{\pi}{p-3/2}\times \dots \times \frac{\pi}{3/2} \times \frac{1}{1/2} \\ &= \frac{\pi^{p+\frac{1}{2}}}{(p+1/2)(p-1/2)\dots (1/2) \pi ^ {\frac{1}{2}}} \\ &= \frac{\pi^{p+\frac{1}{2}}}{\Gamma(p+\frac{1}{2} + 1)} \\ &= \frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)} \end{align*}\]

So we obtained a single formula for the volume of the n-ball, regardless of the parity of \(n\):

\[V_n = V_n(1) = \frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)}\]

And by extension, the volume of the n-ball of radius \(R\) is:

\[\boxed{V_n(R) = R^n V_n(1) = R^n \frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)}}\]

Note that there is a very easy way to derive the area of the surface of n-ball using this expression. We simply need to derive the volume of the n-ball of radius \(R\) with respect to \(R\) and we obtain the surface area of the n-ball of radius \(R\):

\[\frac{d}{dR} V_n(R) = n R^{n-1} V_n(1) = n R^{n-1} \frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)}\]

Paricularly, the surface area of the n-ball of radius \(1\) is:

\[S_n = n \frac{\pi^{n/2}}{\Gamma(\frac{n}{2} + 1)} = \frac{2 \pi^{n/2}}{\Gamma(\frac{n}{2})}\]



Appendix: Derivation of the Wallis integrals

The Wallis integrals are defined as:

\[W_{n} = \int_{0}^{\pi/2} \sin^{n}(\theta) d\theta\]

We will derive a recursive formula using integration by parts.

\[\begin{align*} W_n &= \int_{0}^{\pi/2} \sin^{n}(\theta) d\theta \\ &= \int_{0}^{\pi/2} \sin^{n-2}(\theta) (1- \cos^2(\theta)) d\theta \\ &= W_{n-2} - \int_{0}^{\pi/2} \sin^{n-2}(\theta) \cos^2(\theta) d\theta \\ &= W_{n-2} - \left(\left[\frac{\sin^{n-1}(\theta)}{n-1}\cos(\theta) \right]_0^{\pi/2} - \frac{1}{n-1} \int_0^{\pi/2}-\sin^{n-1}(\theta)\sin(x)d\theta\right) \\ &= W_{n-2} - \frac{1}{n-1} W_{n} \end{align*}\]

We conclude that:

\[W_n = \frac{n-1}{n} W_{n-2}\]

We can now distinguish 2 cases depending on the parity of \(n\):

  • If \(n = 2p\), we have:
\[W_{2p} = \frac{2p-1}{2p} W_{2p-2} = \frac{2p-1}{2p} \frac{2p-3}{2p-2} \dots \frac{1}{2} W_0 = \frac{2p-1}{2p} \frac{2p-3}{2p-2} \dots \frac{1}{2} \frac{\pi}{2} = \frac{\pi}{2} \frac{(2p)!}{2^{2p}(p!)^2}\]

We obtained the last equality by multiplying the denominator and the numerator by \(2p(2p-2)\dots 2\) to make \((2p)!\) appear. We then factorized each term of the denominator by 2.

  • Similarly, if \(n = 2p+1\), we have:
\[W_{2p+1} = \frac{2p}{2p+1} W_{2p-1} = \frac{2p}{2p+1} \frac{2p-2}{2p-1} \dots \frac{2}{3} W_1 = \frac{2p}{2p+1} \frac{2p-2}{2p-1} \dots \frac{2}{3} \frac{2}{1} = \frac{2^{2p} (p!)^2}{(2p+1)!}\]

We conclude that

\[\boxed{W_{2p} = \frac{\pi}{2} \frac{(2p)!}{2^{2p}(p!)^2}}\]

and

\[\boxed{W_{2p+1} = \frac{2^{2p} (p!)^2}{(2p+1)!}}\]