Volume of a ball in R^n

Statement

Let $n \ge 1$, let $R > 0$, and let $B_n(R) = \{x \in \mathbb{R}^n : \|x\| \le R\}$. Then $$\operatorname{vol}(B_n(R)) = \frac{\pi^{n/2}}{\Gamma\!\left(\frac{n}{2} + 1\right)} \, R^n,$$ where $\Gamma(s) = \int_0^\infty t^{s - 1} e^{-t} \, dt$ is the gamma function.

Remarks

The change of variables $x = R u$ shows that $\operatorname{vol}(B_n(R)) = R^n \operatorname{vol}(B_n(1))$. Since $\Gamma(1) = 1$, $\Gamma(1/2) = \sqrt{\pi}$ and $\Gamma(s + 1) = s \, \Gamma(s)$, the formula gives $2R$, $\pi R^2$, $\tfrac{4}{3} \pi R^3$ and $\tfrac{1}{2} \pi^2 R^4$ for $n = 1, 2, 3, 4$. One proof computes $\int_{\mathbb{R}^n} e^{-\|x\|^2} \, dx = \pi^{n/2}$ in two ways: as a product of $n$ Gaussian integrals by Fubini's theorem, and by integration over spheres of radius $r$. The volume of the unit ball tends to $0$ as $n \to \infty$.

Explanations

L1 human

The volume of a ball grows like the $n$-th power of its radius, times a constant that depends on the dimension. The gamma function gives this constant in one formula for all dimensions.