Content of n dimensional ball.

The n dimensional unit ball is that set of n-tuples <x1, … xn> such that Σj xj2 ≤ 1. We will call the content of that ball Vn. The volume of the general n dimensional ball with radius r is, of course, rnVn.

We work backwards. The derivative of (1−x2)n is −2nx(1−x2)n−1.
The integral of x(1−x2)n is thus −(1−x2)n+1/(2(n+1)). This works for real values of n other than −1.

If we plot coordinates x1 and x2 on a plane we get the unit disk and all the points in the ball thus correspond to some point on the disk. The other coordinates of the point in the ball that go to a point on the disk at radius r are constrained thus: x32 + x42 + … + xn2 ≤ 1 − r2. That set of points are in an n−2 dimensional sphere with volume Vn−2(1 − r2)(n−2)/2. We divide the disc into concentric annular rings with area 2πrdr with r running from 0 to 1. For the content of the n dimensional ball we have the integral
Vn =
∫[r=0, 1] Vn−2(1 − r2)(n−2)/2 2πrdr =
2πVn−2∫[r=0, 1]r(1 − r2)(n−2)/2dr =
2πVn−2[1/(2((n−2)/2+1))] =
2πVn−2/n.

If you believe that the unit 1D ball is the interval from −1 to 1, with content 2, then the 3D ball has 4/3 π which is what Pappus got. We thus begin V1 = 2 to begin our inductive definition. More generally we get V5 = 2πV3/5 = 2π(2πV1/3)/5.
V2n+1 = (2π)n/(n(n−2)… 1).

You may believe that a 0D ball has content 1, or if not that a 2D ball (disk) has content π. Then we get
V6 = (2π)V4/6 = (2π)2V2/(6∙4) = (2π)3V0/(6∙4∙2) = π3/(3∙2∙1).

The odd and even dimensions unify to Vn = πn/2/(n!) if we were to agree that n! = n((n−1)!) for fractional values of n and further that (−½)! = √π. This may seem too cute except that
n! = ∫[x=0, ∞]xne−xdx for positive integers and ∫[x=0, ∞]x−½e−xdx= √π.
Further n! = Γ(n+1) and Γ(½) = √π officially.

Finally we have for the n dimensional ball of radius r the content rnVn = rnπn/2/(n/2)!