{
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  "body": "\n## Statement\n\nLet $n \\ge 1$, let $R \u003e 0$, and let $B_n(R) = \\{x \\in \\mathbb{R}^n : \\|x\\| \\le R\\}$.\nThen\n$$\\operatorname{vol}(B_n(R)) = \\frac{\\pi^{n/2}}{\\Gamma\\!\\left(\\frac{n}{2} + 1\\right)} \\, R^n,$$\nwhere $\\Gamma(s) = \\int_0^\\infty t^{s - 1} e^{-t} \\, dt$ is the gamma function.\n\n## Remarks\n\nThe [[change of variables|00N5FHP1CV]] $x = R u$ shows that $\\operatorname{vol}(B_n(R)) = R^n \\operatorname{vol}(B_n(1))$.\nSince $\\Gamma(1) = 1$, $\\Gamma(1/2) = \\sqrt{\\pi}$ 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 $n = 1, 2, 3, 4$.\nOne 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|0R0T1VWH5X]], and by integration over spheres of radius $r$.\nThe volume of the unit ball tends to $0$ as $n \\to \\infty$.\n\n## Explanations\n\n### L1 human\n\nThe volume of a ball grows like the $n$-th power of its radius, times a constant that depends on the dimension.\nThe gamma function gives this constant in one formula for all dimensions.\n",
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