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  "body": "\n## Proof\n\nWe take the case $n = 3$ of the general formula.\n\n1. Let $R \u003e 0$ and $B = \\set{(x, y, z) \\in \\R^3 : x^2 + y^2 + z^2 \\le R^2}$. This is the closed ball $B_3(R)$ of radius $R$ about $0$.\n2. By the [[volume of a ball in R^n|00YH28R457]] with $n = 3$, $\\operatorname{vol}(B) = \\pi^{3/2} R^3 / \\Gamma(5/2)$, where $\\Gamma$ is the [[gamma function|0V3FZMVB4K]].\n3. We compute $\\Gamma(1/2)$. For $0 \u003c \\varepsilon \u003c M$, the [[substitution|0D37QFPEZE]] $t = u^2$ gives $\\int_\\varepsilon^M t^{-1/2} e^{-t} \\, dt = 2 \\int_{\\sqrt{\\varepsilon}}^{\\sqrt{M}} e^{-u^2} \\, du$.\n4. Let $\\varepsilon \\to 0$ and $M \\to \\infty$. The integrand $e^{-u^2}$ is even, so the [[Gaussian integral|0DFT1GP1JP]] gives $\\Gamma(1/2) = \\int_{-\\infty}^{\\infty} e^{-u^2} \\, du = \\sqrt{\\pi}$.\n5. By the [[functional equation|0FDCW77H44]] $\\Gamma(s + 1) = s \\, \\Gamma(s)$, applied twice, $\\Gamma(5/2) = \\tfrac{3}{2} \\cdot \\tfrac{1}{2} \\cdot \\Gamma(1/2) = \\tfrac{3}{4} \\sqrt{\\pi}$.\n6. Hence $\\operatorname{vol}(B) = \\pi^{3/2} R^3 \\cdot \\frac{4}{3 \\sqrt{\\pi}} = \\frac{4}{3} \\pi R^3$.\n\n## Idea\n\nThe volume of a ball in every dimension is given by one formula with the gamma function. In dimension $3$ the gamma value $\\Gamma(5/2)$ reduces to $\\Gamma(1/2) = \\sqrt{\\pi}$, and the powers of $\\pi$ combine to give $\\tfrac{4}{3} \\pi R^3$.\n",
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