# College Math Teaching

## June 9, 2015

### Volumes of n-balls: why?

Filed under: calculus, geometry — Tags: , , — collegemathteaching @ 11:38 am

I talked about a method of finding the hypervolume of an n-ball in a previous post. To recap: the volume of the n-ball is that “hypervolume” (I’ll be dropping “hyper”) of the region described by $\{(x_1, x_2,...x_n) | x_1^2 + x_2^2 + ...x_n^2 \leq R^2 \}$.

The formula is: $V_n = \frac{\pi^{\frac{n}{2}}}{\Gamma[\frac{n}{2} + 1]} R^n$

Here, we’ll explore this topic further, both giving a different derivation (from Greg Huber’s American Mathematical Monthly paper) and make some informal observations.

Derivation: the argument I present can be tweaked to produce a formal induction argument that, if the volume is $V_n$, it is proportional to the n’th power of the radius $R$.

Now note that if the surface area of the $n-1$ sphere is given by $W_{n-1} = w_{n-1}R^{n-1}$ we have, from the theory of differentials, $\frac{dV_n}{dR} = W_{n-1}$. Think of taking a sphere and adding just a bit $\Delta R$ to its radius; you obtain a shell of thickness $\Delta R$ all the way around the sphere which is roughly equal to the surface area times $\Delta R$ So we can rewrite this as $V_n = \int^R_0 W_{n-1} dr = \int^R_0 w_{n-1}r^{n-1} dr = w_{n-1}\int^R_0r^{n-1} dr$

To see what comes next, we first write this same quantity in two different ways: $\int \int \int ...\int_{S^{n-1}} \Pi^{n}_{i=1} dx_i = \int^R_0 w_{n-1}r^{n-1} dr = w_{n-1}\int^R_0r^{n-1} dr = \int \int \int ...\int^R_{0} r^{n-1} J(\theta_1, ..\theta_{n-1}) dr$.

The first integral is integral in rectangular coordinates within the boundary of the n-1 sphere. The rightmost integral is the same integral in generalized spherical coordinates (see Blumenson’s excellent Monthly article) where the first iterated integrals are those with angular limits with $J$ being the angular volume element. The middle integral is the volume integral. All are equal to the volume of the n-ball. The key here is that the iterated integrals evaluated over the entire n-1 sphere are equal to $w_{n-1}$.

Now integrate $e^{-r^2}$ over the region bounded by the sphere $r^2 = x_1^2 + x_2^2 + ...x_n^2$, noting that $e^{-r^2} = e^{-x_1^2}e^{-x_2^2}...e^{-x_n^2}$: $\int \int \int ...\int_{S^{n-1}} \Pi^{n}_{i=1}e^{-x_i^2} dx_i = w_{n-1}\int^R_0e^{-r^2}r^{n-1} dr = \int \int \int ...\int^R_{0} r^{n-1}e^{-r^2}J(\theta_1, ..\theta_{n-1}) dr$

Equality holds between the middle and right integral because in “angle/r” space, the r and angular coordinates are independent. Equality between the leftmost and rightmost integrals holds because this is a mere change of variables.
So we can now drop the rightmost integral. Now take a limit as $R \rightarrow \infty$: $\int \int \int ...\int_{R^{n}} \Pi^{n}_{i=1}e^{-x_i^2} dx_i = (\int^{\infty}_{-\infty} e^{-x^2} dx)^n = w_{n-1}\int^{\infty}_0e^{-r^2}r^{n-1} dr$

The left integral is just the n-th power of the Gaussian integral and is therefore $\pi^{\frac{n}{2}}$ and a substitution $r = \sqrt{u}$ turns this into $\frac{w_{n-1}}{2}\int^{\infty}_{0} u^{\frac{1}{2} -1}e^{-u}du = w_{n-1} \frac{1}{2}\Gamma[\frac{1}{2}]$ (recall $\Gamma[x] = \int^{\infty}_{0} t^{x-1}e^{-t} dt$ ).

So $w_{n-1} = \frac{2 \pi^{\frac{n}{2}}}{\Gamma[\frac{n}{2}]}$ and hence, by integration, $v_n = \frac{2 \pi^{\frac{n}{2}}}{n\Gamma[\frac{n}{2}]}= \frac{ \pi^{\frac{n}{2}}}{n\Gamma[\frac{n}{2}+1]}$

Now $v_n =V_n$ when $R = 1$. $\frac{v_n}{2^n}$ can be thought of as the percentage of the cube with vertexes $(\pm 1, \pm 1, ... \pm 1)$ that is taken up by the inscribed unit sphere.

Now we set $R = 1$ at look at a graph of hypervolume vs. n:

The first graph is the ratio of the volume taken up by the ball verses the hypervolume of the hyper cube that the ball is inscribed in. Next we see the the hypervolume peaks at n = 5 (max is between 5 and 6 )and then starts to decline to zero. Of course there has to be an inflection point somewhere; it turns out to be between n = 10 and n = 11. Now we plot the hyperarea of the n-1 sphere vs. the hypervolume of the ball that it bounds; we see that more and more of the hypervolume of the ball is concentrated near the boundary as the dimension goes up. For more: see the interesting discussion on Division by Zero.