Deriving the volume of an n-ball

In geometry, the volume of a sphere is a special case of the n-dimensional volume of a ball in n-dimensional Euclidean space.

Derivation of the volume of an n-ball

With a proof by induction, we calculate the volume of an n-ball of radius R to be

$V^{(n)}[R] = \frac{\pi^{\frac{n}{2}} R^n}{\Gamma(\frac{n}{2} + 1)}$

where Γ is the gamma function. Because the proof for the volume of an n-ball depends upon the volume of an (n − 2)-ball there are two base cases, zero dimensions and one dimension.

Base cases

In zero dimensions a 0-ball is defined to be just one point, with a 0-volume of 1. In one dimension, the 1-ball of radius R is defined to be the interval [−R, R] and has 1-volume (length) of 2R.

$\begin{align} V^{(0)}[R] &= 1\\ V^{(1)}[R] &= 2R \end{align}$

These are consistent with the desired formula because $\scriptstyle\Gamma(1) \;=\; 1$ and $\scriptstyle\Gamma\left(\frac{3}{2}\right) \;=\; \frac{\sqrt{\pi}}{2}$.

General case

For the inductive argument, we will prove the formula for $\scriptstyle V^{(n)}[R]$ by assuming that we have already proved that

$V^{(k)}[R] = \frac{\pi^{\frac{k}{2}} R^k}{\Gamma(\frac{k}{2} + 1)}$

for k < n dimensions. We prove the formula using integration in polar coordinates:

$\begin{align} V^{(n)}[R] & = \int_0^R \int_0^{2 \pi} V^{(n-2)}\left[\sqrt{R^2-r^2}\right] \, r \, d\theta \, dr \\ & = 2 \pi \int_0^R V^{(n-2)}\left[\sqrt{R^2-r^2}\right] \, r \, dr \\ & = 2 \pi \int_0^R \frac{\pi^{\frac{n-2}{2}}}{\Gamma(\frac{n}{2})} \, \left({R^2-r^2}\right)^{\frac{n-2}{2}} \, r \, dr \\ & = \frac{\pi^{\frac{n}{2}}}{\frac{1}{2}\Gamma(\frac{n}{2})} \int_0^R \, \left({R^2-r^2}\right)^{\frac{n-2}{2}} \, r \, dr \\ & = \frac{\pi^{\frac{n}{2}}}{ \frac{n}{2}\Gamma(\frac{n}{2})} \left[- \left(R^2-r^2\right)^{\frac{n}{2}} \right]_{r=0}^{r=R} \\ & = \frac{\pi^{\frac{n}{2}} R^n}{\Gamma(\frac{n}{2} + 1)} \end{align}$

QED.

In particular, the volume formula for an n-ball can be reconstructed from the base cases and the recursion

$V^{(n)}[R] = \left(\frac{2 \pi R^2}{n}\right) V^{(n-2)}[R]$

Alternative derivation of the volume of an n-ball

General formula (recursive form)

Denote by V⁽ⁿ⁾[r] the volume of the n-ball of radius r. Then

$V^{(1)}[r] = 2r \,$

because this is just a line segment twice as long as the radius; i.e., $\scriptstyle\{x \,\in\, \mathbb R:\; |x| \,\le\, r\}$.

For n ≥ 1, we have:^[1]

$V^{(n+1)}[r] = \int_{-r}^r V^{(n)}\left[\sqrt{r^2-x^2}\,\right]\,dx$

Volume is proportional to n^th power of radius

We shall first show by induction that the volume of an n-ball is proportional to the n^th power of its radius. We have already noted that this is true in one dimension. Suppose it is true for n dimensions; i.e.,:

V⁽ⁿ⁾[r] = rⁿV⁽ⁿ⁾[1]

Then:

$V^{(n+1)}[r] = \int_{-r}^r V^{(n)}\left[\sqrt{r^2-x^2}\,\right] dx$

Now, here the fly is definitely in the ointment. Let's see what we can do.

$\begin{align} V^{(n+1)}[r] &= r \int_{-1}^1 V^{(n)}\left[\sqrt{r^2-(rx)^2}\,\right]\, dx\\ V^{(n+1)}[r] &= r \int_{-1}^1 V^{(n)}\left[r\sqrt{(1-x^2)}\,\right] dx\\ V^{(n+1)}[r] &= r \int_{-1}^1 r^n V^{(n)}\left[\sqrt{(1-x^2)}\,\right] dx = r^{n+1}V^{(n+1)}[1] \end{align}$

Now we have established that for all n ≥ 1, the volume of an n-ball is proportional to the n^th power of its radius; that is, if we denote the volume of the unit n-ball by $\scriptstyle V^{(n)}[1]$, we have:

$\begin{align} V^{(n)}[r] &= r^n V^{(n)}[1]\\ V^{(n+1)}[1] &= \int_{-1}^1 \left(\sqrt{1-x^2}\,\right)^n V^{(n)}[1]\, dx\\ V^{(n+1)}[1] &= V^{(n)}[1]\int_{-1}^1 \left(\sqrt{1-x^2}\,\right)^n\, dx \end{align}$

First few steps

In the case of $\scriptstyle V^{(2)}$ we have^[2]

$V^{(2)}[1] = V^{(1)}[1]\int_{-1}^1 \sqrt{1-x^2}\,dx = 2\left.\frac{1}{2}{x\sqrt{1-x^2}+\arcsin x} \right|_{x=-1}^1 = \pi$

which is the area of the unit circle, as we expect. The next derivation, the volume of the unit sphere, is much easier:

$V^{(3)}[1] = V^{(2)}[1] \int_{-1}^1 \left(1-x^2\right)dx = \frac 4 3 \pi$

General case

Let us try to generalize this derivation for a ball of any dimension:

$\begin{align} V^{(n+1)}[1] &= V^{(n)}[1] \int_{-1}^1 \left(1-x^2\right)^{\frac{n}{2}} dx\\ {} &= V^{(n)}[1] \cdot 2\int_0^1 \left(1-x^2\right)^{\frac{n}{2}} dx \end{align}$

Here is a graph of the integrand to make it easier to visualize what is going on:

By a change of variables u = 1 − x² we have:

$\begin{align} x &= \sqrt{1-u}\\ dx &= \frac{-du}{2\sqrt{1-u}}\\ \Rightarrow V^{(n+1)}[1] &= V^{(n)}[1] \cdot 2\int_0^1 \left(1-x^2\right)^{\frac{n}{2}} \, dx\\ {} &= V^{(n)}[1] \int_0^1 u^{\frac{n}{2}}(1-u)^{-\frac{1}{2}}\, du \end{align}$

The integral on the far right is known as the beta function:

$V^{(n+1)}[1] = V^{(n)}[1] \mathrm B\left(\frac n 2 + 1, \frac 1 2 \right)$

which can be expressed in terms of the gamma function:

$V^{(n+1)}[1] = V^{(n)}[1] \frac {\Gamma\left(\frac n 2 + 1\right)\Gamma\left( \frac 1 2 \right)} {\Gamma\left(\frac n 2 + \frac 3 2\right)}$

Since $\scriptstyle\Gamma\left(\frac 1 2\right) \;=\; \sqrt \pi$, we can easily verify by induction that for all n ≥ 1:

$V^{(n)}[1] = \frac {\pi^{\frac{n}{2}}}{\Gamma\left(\frac n 2 + 1 \right)}$

General form and surface area

The "surface area" of the n-ball (i.e., the (n − 1)-dimensional volume measure of the (n − 1)-sphere) can easily be found by differentiating the volume of the n-ball with respect to the radius. So, if we denote the volume of the n-ball of radius r by

$V^{(n)}[r] = \frac {\pi^{\frac{n}{2}} r^n}{\Gamma\left(\frac n 2 + 1 \right)}$

then its "surface area" is

$S^{(n-1)}[r] = \frac \partial{\partial r} V^{(n)}[r] = \frac {\pi^{\frac{n}{2}} nr^{n-1}}{\Gamma\left(\frac n 2 + 1 \right)} = \frac {2\pi^{\frac{n}{2}}r^{n-1}}{\Gamma\left( \frac n 2 \right)}$

This is an example of a disintegration of measure in Euclidean space.^[3]

Further generalizations

The alternative method of integration can carry over to balls in other L^p spaces where p ≠ 2, which has significance for information theory and coding theory. Also, since the expressions are analytical for complex (continuous) n, they are used in dimensional regularization, a fundamental step in calculations within the standard model of elementary particles.

In fact, for the unit L^p balls, we have the recurrence relation

$V^{(n+1)}[1] = V^{(n)}[1] \int_{-1}^1 \big(1 - |x|^p \big)^{\frac{n}{p}} \, dx$

from which one may recover the formula

$V^{(n)}[r] = \frac {\left[ 2\, \Gamma\left(\frac 1 p + 1\right) r \right]^n} {\Gamma \left(\frac n p + 1 \right)}$

for the volume of a ball of radius r in $\ell^p_n$, the volume measure being, as before, that of Lebesgue in the orthonormal (cartesian) coordinates. However, when p ≠ 2, it is no longer possible to calculate the (hyper)surface area by differentiating the volume of the ball with respect to its radius, because the radius is no longer everywhere normal to the surface.

See also

• n-sphere
• Sphere packing
• Hamming bound

References

1. ^ http://www-staff.lboro.ac.uk/~coael/hypersphere.pdf
2. ^ Samuel M. Selby, editor. Standard Mathematical Tables, 18th edition. The Chemical Rubber Co., Cleveland, Ohio, 1970.
3. ^ D. Leao Jr. et al. Regular conditional probability, disintegration of probability and Radon spaces. Proyecciones. Vol. 23, No. 1, pp. 15–29, May 2004, Universidad Católica del Norte, Antofagasta, Chile PDF

External links

• http://www.brouty.fr/Maths/sphere.html (derivation in hyperspherical coordinates.)
• http://mathworld.wolfram.com/Hypersphere.html
• http://www-staff.lboro.ac.uk/~coael/hypersphere.pdf
• http://www.mathreference.com/ca-int,hsp.html