Volume of a ball in R^3

Statement

Let $R > 0$, and let $B = \{(x, y, z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 \le R^2\}$. Then $$\operatorname{vol}(B) = \frac{4}{3} \pi R^3.$$

Remarks

In spherical coordinates, $\operatorname{vol}(B) = \int_0^{2\pi} \int_0^\pi \int_0^R \rho^2 \sin \varphi \, d\rho \, d\varphi \, d\theta = 2\pi \cdot 2 \cdot R^3 / 3$. By slices, the cross-section at height $z$ is a disc of area $\pi (R^2 - z^2)$, and $\int_{-R}^{R} \pi (R^2 - z^2) \, dz = \tfrac{4}{3} \pi R^3$. This is Cavalieri's principle. This is the case $n = 3$ of the volume of a ball in R^n, since $\Gamma(5/2) = \tfrac{3}{4} \sqrt{\pi}$. Archimedes found this volume, by comparison with a cone and a cylinder.

Explanations

L1 human

The ball of radius $R$ fills two thirds of the cylinder of radius $R$ and height $2R$ around it, since that cylinder has volume $2 \pi R^3$. Any of the standard methods of multiple integration gives the same value.