Surface Areas and Volumes of Hyperspheres

I noticed that the formulae for the sizes of the surface (or rather, boundary) and volume (interior) of circles, spheres and hyperspheres seem to have discontinuities (or ‘jumps’) in the index of \pi.

However, this is only a result of simplifying a uniform formula for particular cases.

The indices I prefer for the boundary function are not standard, and thus I use an asterisk (\ast) to highlight the difference. See the note on notation below for the details.

Reduced Formulae

Here’s a table of the first few values for higher-dimensional spheres of radius r.

Dimension
n
Boundary
S^{\ast}_n(r)
Interior
V_n(r)
1 2 2 r
2 2 \pi r \pi r^2
3 4 \pi r^2 \dfrac{4}{3} \pi r^3
4 2 \pi^2 r^3 \dfrac{1}{2} \pi^2 r^4
5 \dfrac{8}{3} \pi^2 r^4 \dfrac{8}{15} \pi^2 r^5
6 \pi^3 r^5 \dfrac{1}{6} \pi^3 r^6
7 \dfrac{16}{15} \pi^3 r^6 \dfrac{16}{105} \pi^3 r^7

And for those of you who, like me, are ‘tau-ists’:

Dimension
n
Boundary
S^{\ast}_n(r)
Interior
V_n(r)
1 2 2 r
2 \tau r \dfrac{1}{2} \tau r^2
3 2 \tau r^2 \dfrac{2}{3} \tau r^3
4 \dfrac{1}{2} \tau^2 r^3 \dfrac{1}{8} \tau^2 r^4
5 \dfrac{2}{3} \tau^2 r^4 \dfrac{2}{15} \tau^2 r^5
6 \dfrac{1}{4} \tau^3 r^5 \dfrac{1}{48} \tau^3 r^6
7 \dfrac{2}{15} \tau^3 r^6 \dfrac{2}{105} \tau^3 r^7

General Formulae

But there are some regular formulae:

S^{\ast}_n(r) = \dfrac{2 \cdot \pi^{\frac{n}{2}}}{\left( \frac{n-2}{2} \right)!} r^{n-1}

V_n(r) = \dfrac{\pi^{\frac{n}{2}}}{\left( \frac{n}{2} \right)!} r^n

And for tau-ists:

S^{\ast}_n(r) = \dfrac{ \tau^{ \frac{n}{2} } }{{2^{ \frac{n - 2}{2} }} {\cdot {\left( \frac{ n -2 }{2} \right) }!} } r^{n-1}

V_n(r) = \dfrac{\tau^{\frac{n}{2}}}{2^{\frac{n}{2}} \cdot \left( \frac{n}{2} \right)!} r^n

where, in particular, the factorial of the half-integers may be calculated from

\left( - \dfrac{1}{2} \right)! = \sqrt{\pi} = \sqrt{\dfrac{\tau}{2}}

The relationships between these formulae are:

S^{\ast}_n(r) = (2 \pi r) V_{n-2}(r)

V_n(r) = \dfrac{r}{n} S^{\ast}_{n}(r)

These formulae ‘split’ for even and odd values thus

S^{\ast}_{2k}(r) = \dfrac{2 \cdot \pi^{k}}{(k-1)!} r^{2k-1}

S^{\ast}_{2k+1}(r) = \dfrac{2^{k+1} \pi^k}{(2k-1)!!} r^{2k}

V_{2k}(r) = \dfrac{\pi^{k}}{k!} r^{2k}

V_{2k+1}(r) = \dfrac{2^{k+1} \pi^{k}}{\left( 2k+1 \right)!!} r^{2k+1}

and

S^{\ast}_{2k}(r) = \dfrac{\tau^{k}}{2^{k-1} (k-1)!} r^{2k-1}

S^{\ast}_{2k+1}(r) = \dfrac{2 \cdot \tau^k}{(2k-1)!!} r^{2k}

V_{2k}(r) = \dfrac{\tau^{k}}{2^{k} k!} r^{2k}

V_{2k+1}(r) = \dfrac{2 \cdot \tau^{k}}{\left( 2k+1 \right)!!} r^{2k+1}

For reference, here is a table of a few half-integral values of the factorial function. (Incidentally, this function seems to have no simple closed form in terms of fundamental functions for values other than the half-integers.)

k k!
−1/2 \sqrt{\pi} = \dfrac{\sqrt{2}}{2} \sqrt{\tau} \approx 1.772
0 1
1/2 \dfrac{1}{2} \sqrt{\pi} = \dfrac{\sqrt{2}}{4} \sqrt{\tau} \approx 0.8862
1 1
3/2 \dfrac{3}{4} \sqrt{\pi} = \dfrac{3 \sqrt{2}}{8} \sqrt{\tau} \approx 1.329
2 2
1/2 \dfrac{15}{8} \sqrt{\pi} = \dfrac{15 \sqrt{2}}{16} \sqrt{\tau} \approx 3.323
3 6
1/2 \dfrac{105}{16} \sqrt{\pi} = \dfrac{105 \sqrt{2}}{32} \sqrt{\tau} \approx 11.63
4 24

And, here’s a table of a few double integrals (although we only need the odd values)

k k!!
0 1
1 1
2 2
3 3
4 8
5 15
6 48
7 7 × 5 × 3 × 1 = 105
8 8 × 6 × 4 × 2 = 384

Brief Note on the Calculus

The calculus relating these is quite interesting, but that is covered very well elsewhere. In brief, a (solid) ball is a collection of concentric spherical shells. The size of boundary of a ball is the derivation of the size of its body with respect to its radius.

Note on Mathematics Terminology

Note that the mathematical terminology for these objects is ball for a ‘solid’ body, and sphere for a ‘hollow’ shell.

Further, the dimension used relates to the intrinsic dimension, not the dimension of a minimal embedding space.

Thus a circle is a 1-sphere, because the curve of the circle is intrinsically line-like, or 1-dimensional, but a disk is a 2-ball, as a disk is intrinsically plane-like, or 2-dimensional. Likewise, a sphere (or spherical shell) is a 2-sphere, because the surface of the sphere (or, more accurately, the surface that is the sphere) is intrinsically plane-like, or surface-like, or 2-dimensional, but a ball (that is, a solid ball) is a 3-ball, as a ball is intrinsically volume-like, or 3-dimensional.

Note on Variations in Notation

I have allowed the factorial function to take fractional arguments. Some authors prefer to use the Pi function

\Pi (x) = x!

for non-integral values of x; others prefer to use the Gamma function

\Gamma (x) = \Pi (x-1) = (x-1)!

I use S^{\ast}_n(r) to denote the boundary size of an n-ball, where S comes from surface, and the asterisk (\ast) is to indicate that my indices are non-standard. Some authors prefer A, rather than S, for area. Others still prefer the indices to refer to an n-sphere, i.e. the boundary of an n+1-ball, and would thus write our functions as S_{n-1}(r) or A_{n-1}(r).

For completeness (and comparison with other references), here are the conventional formulae.

S_n(r) = S^{\ast}_{n+1}(r) = \dfrac{2 \cdot \pi^{\frac{n+1}{2}}}{\left( \frac{n-1}{2} \right) !} r^n

S_{2k}(r) = S^{\ast}_{2k+1}(r) = \dfrac{2^{k+1} \pi^k}{(2k-1)!!} r^{2k}

S_{2k+1}(r) = S^{\ast}_{2(k+1)}(r) = \dfrac{2 \cdot \pi^{k+1}}{k!} r^{2k+1}

Further Reading

There are excellent articles on Wikipedia (n-Sphere), (n-Ball Volume), on Division by Zero, and also a page by Thayer Watkins.

Regarding tau (\tau), see The Tau Manifesto from Michael Hartl.

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