Recursive formula of volume of sphere

Let $\mathcal{V}_n(1)$ be the volume of the sphere centered at $0$ and radius $1$ in $\mathbb{R}^n$ . Prove that for $n \geq 3$ it holds that

$\mathcal{V}_n(1) = \frac{2\pi}{n} \mathcal{V}_{n-2}(1)$

Solution

The volume of the sphere in $\mathbb{R}^n$ is given by:

$\mathcal{V}_n (1) = \idotsint \limits_{x_1^2+x_2^2+\cdots+x_n^2 \leq 1} 1 \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right )$

Parametrize the sphere by

$\begin{matrix} x_1 &= & r \cos \theta_1 \\ x_2 & = & r \sin \theta_1 \cos \theta_2 \\ x_3 & = & r \sin \theta_1 \sin \theta_1 \cos \theta_3 \\ \vdots \\ x_{n-1} & = & r \sin \theta_1 \sin \theta_2 \cdots \sin \theta_{n-2} \cos \theta_{n-1} \\ x_n & = & r \sin \theta_1 \sin \theta_2 \cdots \sin \theta_{n-1} \end{matrix}$

taking

$\begin{matrix} 0 \leq r \leq 1\\ 0 \leq \theta_i \leq \pi & & \text{forall} \;\; i=1, 2, \dots , n-2 \\ 0 \leq \theta_{n-1} < 2 \pi \end{matrix}$

It then follows from the Change of Variables formula that the rectangular volume element $\mathrm{d} \mathcal{V} = \mathrm{d}x_1 \; \mathrm{d}x_2 \cdots \mathrm{d}x_n$ can be written in spherical coordinates as

$\begin{align*} \mathrm{d}\mathcal{V} &= \left | \det \left ( \frac{\partial x_1}{\partial \left ( r , \theta_j \right )} \right ) \right | \mathrm{d} r \; \mathrm{d} \left ( \theta_1, \theta_2, \dots, \theta_{n-1} \right ) \\ &= r^{n-1} \sin^{n-2} \theta_1 \sin^{n-3} \theta_2 \cdots \sin \theta_{n-2} \mathrm{d} r \; \mathrm{d} \left ( \theta_1, \theta_2, \dots, \theta_{n-1} \right ) \end{align*}$

Thus,

$\begin{align*} \mathcal{V}_n(1) &= \idotsint \limits_{x_1^2+x_2^2+\cdots+x_n^2 \leq 1} 1 \, \mathrm{d} \left ( x_1, x_2, \dots, x_n \right ) \\ &\!\!\!\!=\int_{0}^{2\pi} \int_{0}^{\pi} \cdots \int_{0}^{1} r^{n-1} \sin^{n-2} \theta_1 \cdots \sin \theta_{n-2} \mathrm{d} r \; \mathrm{d} \left ( \theta_1, \dots, \theta_{n-1} \right ) \\ &= \int_{0}^{2\pi} \, \mathrm{d} \theta_{n-1} \int_{0}^{1} r^{n-1} \, \mathrm{d}r \int_{0}^{\pi} \sin^{n-2} \theta \, \mathrm{d} \theta \cdots \int_{0}^{\pi} \sin \theta \, \mathrm{d} \theta \\ &= \frac{2\pi}{n} \int_{0}^{\pi} \sin^{n-2} \theta \, \mathrm{d} \theta \cdots \int_{0}^{\pi} \sin \theta \, \mathrm{d} \theta \end{align*}$

Hence,

(1) $\begin{equation*} \int_{0}^{\pi} \sin^{n-2} \theta \, \mathrm{d} \theta \cdots \int_{0}^{\pi} \sin \theta \, \mathrm{d} \theta =\frac{n}{2\pi} \mathcal{V}_{n}(1) \end{equation*}$

In particular since $n-4 = (n-2)-2$ we get that:

(2) $\begin{equation*} \int_{0}^{\pi} \sin^{n-4} \mathrm{d}\theta \cdots \int_{0}^{\pi} \sin \theta \, \mathrm{d} \theta = \frac{n-2}{2\pi} \mathcal{V}_{n-2} (1) \end{equation*}$

Using $(2)$ as well as Wallis’ integral we are able to prove the result. Let us assume that $n$ is even, then:

$\begin{align*} \mathcal{V}_n(1) &=\frac{2\pi}{n} \int_{0}^{\pi} \sin^{n-2} \theta \, \mathrm{d} \theta \int_{0}^{\pi} \sin^{n-3} \theta \, \mathrm{d} \theta \int_{0}^{\pi} \sin^{n-4} \theta \, \mathrm{d} \theta \cdots \\ &\quad \quad \quad \quad \quad \cdots \int_{0}^{\pi} \sin \theta \, \mathrm{d} \theta \\ &=\frac{2\pi}{n}\int_{0}^{\pi} \sin^{n-2} \theta \, \mathrm{d} \theta \int_{0}^{\pi} \sin^{n-3} \theta \, \mathrm{d} \theta \cdot \frac{n-2}{2\pi} \, \mathcal{V}_{n-2} (1) \\ &= \frac{2\pi}{n} \cdot \left ( \frac{n-3}{n-2} \cdots \frac{1}{2} \cdot \pi \right ) \cdot \left ( \frac{n-4}{n-3} \cdots \frac{2}{3} \cdot 2 \right ) \cdot \frac{n-2}{2\pi} \, \mathcal{V}_{n-2} (1)\\ &= \frac{2\pi}{n} \cdot \mathcal{V}_{n-2} (1) \end{align*}$