CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-24

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The volume of a radius- $r$ closed $n$ -ball is $π^{n / 2} r^{n} / Γ (n / 2 + 1)$

Statement

For $n \geq 1$ and $r \geq 0$ , $V_{n} (r) = π^{n / 2} r^{n} / Γ (n / 2 + 1)$ , where $n$ is an integer.

Facts & Assumptions

Given: Positive integer $n$ and radius $r \geq 0$ .

[F1]

One has $V_{1} (r) = 2 r$ for $r \geq 0$ , and for $n \geq 2$ and $r \geq 0$ , $V_{n} (r) = V_{n - 1} (1) \int_{- r}^{r} (r^{2} - t^{2})^{(n - 1) / 2} d t$ (Closed Euclidean balls are Jordan measurable and their volumes satisfy the slicing recursion).

[F2]

For every $n \geq 1$ , $V_{n} (1) = π^{n / 2} / Γ (n / 2 + 1)$ (The closed form for the volume of the unit $n$ -ball).

Proof

technique · direct

1.1

If $r = 0$ , the ball is a singleton of content zero, and the right side is zero because $n \geq 1$ .

givencases

1.2

Suppose $r > 0$ . For $n = 1$ , $V_{1} (r) = 2 r = r V_{1} (1)$ . For $n \geq 2$ , substitute $t = r u$ in [F1]; the power and differential contribute $r^{n - 1}$ and $r$ , so comparison with [F1] at radius $1$ gives $V_{n} (r) = r^{n} V_{n} (1)$ .

F1algebra

2.1

Insert [F2] into step 1.2 and combine it with the zero-radius case of step 1.1. This gives the displayed formula for every allowed $n, r$ .

step 1.1step 1.2F2cases-exhaustive∎

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Sources

University of Toronto MAT237Y1, The Gamma Function and the Beta Function, §2.4(a,e) (standard reference, not scraped)

The volume of a radius-rr closed nn-ball is πn/2rn/Γ(n/2+1)πn/2rn/Γ(n/2+1)