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  "body": "\n## Proof\n\nWe argue by induction on $n$, slicing the ball by hyperplanes.\n\n1. Here $\\operatorname{vol}$ is [[Lebesgue measure|2Z2QVWZ262]] $\\lambda^n$ on [[Euclidean space|0XPR1E85W0]] $\\R^n$. For $\\rho \\ge 0$ let $B_n(\\rho) = \\set{x \\in \\R^n : \\norm{x} \\le \\rho}$ and $V_n(\\rho) = \\lambda^n(B_n(\\rho))$. Each $B_n(\\rho)$ is closed, so it is a Borel set and Lebesgue measurable.\n2. Let $c_n = \\pi^{n/2} / \\Gamma(n/2 + 1)$, with $\\Gamma$ the [[gamma function|0V3FZMVB4K]]. We prove by induction on $n \\ge 1$ the claim $P(n)$: $V_n(\\rho) = c_n \\rho^n$ for every $\\rho \\ge 0$. This contains the statement.\n3. First, $\\Gamma(1/2) = \\sqrt{\\pi}$. For $0 \u003c \\varepsilon \u003c M$, [[substitution|0D37QFPEZE]] with $t = u^2$, $dt = 2u \\, du$ 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 $M \\to \\infty$. Since $e^{-u^2}$ is even, the [[Gaussian integral|0DFT1GP1JP]] gives $\\Gamma(1/2) = 2 \\int_0^\\infty e^{-u^2} \\, du = \\int_{-\\infty}^{\\infty} e^{-u^2} \\, du = \\sqrt{\\pi}$.\n5. By the [[functional equation|0FDCW77H44]] $\\Gamma(s + 1) = s \\, \\Gamma(s)$, we get $\\Gamma(3/2) = \\sqrt{\\pi}/2$, and $\\Gamma(1) = 0! = 1$.\n6. The case $\\rho = 0$ holds for every $n$. Indeed $B_n(0) = \\set{0}$ is the box $[0, 0]^n$ of volume $0$, so $\\lambda^n(B_n(0)) = 0 = c_n \\cdot 0^n$.\n7. Base case $n = 1$, $\\rho \u003e 0$. Here $B_1(\\rho) = [-\\rho, \\rho]$. On a bounded interval the [[Lebesgue integral|XVEWR93EMV]] of a Riemann integrable function equals its Riemann integral. So $V_1(\\rho) = \\int_{-\\rho}^{\\rho} 1 \\, dx = 2\\rho$.\n8. By step 5, $c_1 = \\sqrt{\\pi} / \\Gamma(3/2) = 2$. So $P(1)$ holds.\n9. For an integer $m \\ge 0$ let $J_m = \\int_{-1}^{1} (1 - u^2)^{m/2} \\, du$. We claim that $J_m = \\sqrt{\\pi} \\, \\Gamma(m/2 + 1) / \\Gamma(m/2 + 3/2)$.\n10. For $m = 0$: $J_0 = 2$, and the right side is $\\sqrt{\\pi} \\, \\Gamma(1) / \\Gamma(3/2) = 2$ by step 5.\n11. For $m = 1$: substitute $u = \\sin \\theta$ with $\\theta \\in [-\\pi/2, \\pi/2]$. Then $\\sqrt{1 - u^2} = \\cos \\theta$ and $J_1 = \\int_{-\\pi/2}^{\\pi/2} \\cos^2 \\theta \\, d\\theta = \\int_{-\\pi/2}^{\\pi/2} \\tfrac{1}{2}(1 + \\cos 2\\theta) \\, d\\theta = \\pi/2$.\n12. The right side for $m = 1$ is $\\sqrt{\\pi} \\, \\Gamma(3/2) / \\Gamma(2) = \\sqrt{\\pi} \\cdot (\\sqrt{\\pi}/2) / 1 = \\pi/2$. So the claim holds for $m = 1$.\n13. Let $m \\ge 2$. The function $w(u) = (1 - u^2)^{m/2}$ is continuously differentiable on $[-1, 1]$, with $w'(u) = -m u (1 - u^2)^{m/2 - 1}$, since $m/2 - 1 \\ge 0$.\n14. [[Integration by parts|08Y1BBQTK9]] with the factors $w$ and $u$ gives $J_m = [u \\, w(u)]_{-1}^{1} - \\int_{-1}^1 u \\, w'(u) \\, du = m \\int_{-1}^1 u^2 (1 - u^2)^{m/2 - 1} \\, du$, because $w(\\pm 1) = 0$.\n15. Write $u^2 = 1 - (1 - u^2)$. Then step 14 reads $J_m = m (J_{m-2} - J_m)$, so $J_m = \\frac{m}{m+1} J_{m-2}$.\n16. Assume the claim of step 9 for $m - 2$. By step 5's functional equation, $\\Gamma(m/2 + 1) = \\tfrac{m}{2} \\Gamma(m/2)$ and $\\Gamma(m/2 + 3/2) = \\tfrac{m+1}{2} \\Gamma(m/2 + 1/2)$.\n17. Hence $\\sqrt{\\pi} \\, \\Gamma(m/2 + 1) / \\Gamma(m/2 + 3/2) = \\frac{m}{m+1} \\sqrt{\\pi} \\, \\Gamma(m/2) / \\Gamma(m/2 + 1/2) = \\frac{m}{m+1} J_{m-2} = J_m$. By induction on $m$, the claim of step 9 holds for all $m \\ge 0$.\n18. Inductive step. Let $n \\ge 2$ and assume $P(n - 1)$. Fix $R \u003e 0$. Write a point of $\\R^n$ as $(s, y)$ with $s \\in \\R$ and $y \\in \\R^{n-1}$, so that $\\norm{(s, y)}^2 = s^2 + \\norm{y}^2$.\n19. For $\\abs{s} \\le R$, the slice $\\set{y : (s, y) \\in B_n(R)}$ is $B_{n-1}(\\sqrt{R^2 - s^2})$. For $\\abs{s} \u003e R$ the slice is empty.\n20. Apply [[Tonelli's theorem|067SYD2J4B]] to the indicator function of $B_n(R)$ on $\\R \\times \\R^{n-1}$. With step 19 and $P(n - 1)$, $V_n(R) = \\int_{\\R} g(s) \\, ds$, where $g(s) = c_{n-1} (R^2 - s^2)^{(n-1)/2}$ for $\\abs{s} \\le R$ and $g(s) = 0$ otherwise.\n21. The function $g$ is continuous on $[-R, R]$ and zero outside. So, as in step 7, its Lebesgue integral is the Riemann integral $c_{n-1} \\int_{-R}^{R} (R^2 - s^2)^{(n-1)/2} \\, ds$.\n22. The substitution $s = R u$ turns this into $c_{n-1} R^n J_{n-1}$, since $(R^2 - R^2 u^2)^{(n-1)/2} \\cdot R = R^n (1 - u^2)^{(n-1)/2}$.\n23. By step 9 with $m = n - 1$, $c_{n-1} J_{n-1} = \\frac{\\pi^{(n-1)/2}}{\\Gamma((n+1)/2)} \\cdot \\frac{\\sqrt{\\pi} \\, \\Gamma((n+1)/2)}{\\Gamma(n/2 + 1)} = \\frac{\\pi^{n/2}}{\\Gamma(n/2 + 1)} = c_n$.\n24. So $V_n(R) = c_n R^n$ for $R \u003e 0$, and step 6 covers $R = 0$. This proves $P(n)$.\n25. By induction, $\\operatorname{vol}(B_n(R)) = \\frac{\\pi^{n/2}}{\\Gamma(n/2 + 1)} R^n$ for all $n \\ge 1$ and $R \u003e 0$.\n\n## Idea\n\nCut the $n$-ball into slices perpendicular to one axis. Each slice is an $(n-1)$-ball, so its volume is known by induction, and the volume of the $n$-ball becomes a one-variable integral. Integration by parts gives a recursion for that integral, and the recursion matches the functional equation of the gamma function.\n",
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