# Finding the volume of a 15-dimensional unit ball [using elementary multivariable calculus]?

It wouldn't surprise me if what I am asking is impossible (knowing my professor), but is there a way to find the volume of a 15-dimensional unit ball using elementary multivariable calculus?

How elementary, you might ask? We have barely gone over divergence and curl.

He offered extra credit to those able to solve this problem. I'm wondering if anyone can point me in the right direction.

So far I thought about somehow getting the volume under the following set of points:

$S = \{(x_1,x_2,x_3,...,x_{15})|\sqrt{{x_1}^2+{x_2}^2+...+{x_{15}}^2}= 1\}$

I figured the best place to ask for a hint would be here. I have been thinking about this for quite some time, and have not had much luck. Any help would be appreciated.

• – user127096 Mar 21 '14 at 1:48

The general formula and the derivation can be found here, but you do need to familiarize yourself with the gamma function $\Gamma(s) = \int_0^\infty e^{-t}t^{s - 1}dt$. It is simply a generalization of the factorial function $n!$, that is, $\Gamma(n + 1) = n!$ for $n = 0, 1, 2, 3, \ldots$
Proof: Clearly $\Gamma(n + 1) = \int_0^\infty e^{-t}t^ndt$. Integration by parts yields $$\lim_{M \to \infty} \left[ -t^n e^{-t} \right]_0^M + n\int_0^\infty e^{-t} t^{n - 1}dt.$$ By l'Hospital's Rule $\lim_{t \to \infty} t^n e^{-t} = 0$, so $\Gamma(n + 1) = n \int_0^\infty e^{-t} t^{n - 1}dt = n\Gamma(n)$. Integrating $n\Gamma(n)$ by parts $n - 1$ times gives $\Gamma(n + 1) = n \times (n - 1) \times 2 \times 1 = n!$. Q.E.D.
You also need to know the following result: $$\Gamma\left( \frac{1}{2} \right) = \sqrt\pi.$$
Proof: Using the definition of $\Gamma(s)$, $\Gamma(1/2) = \int_0^\infty e^{-t}t^{-\frac{1}{2}}dt$. The substitution $t = u^2$ leads to a Gaussian integral: $\Gamma(1/2) = 2\int_0^\infty e^{-u^2}du = \int_{-\infty}^\infty e^{-u^2}du$. Now we consider $\Gamma^2(1/2)$ instead, namely, $$\left( \int_{-\infty}^\infty e^{-u^2}du \right)^2 = \left( \int_{-\infty}^\infty e^{-x^2}dx \right) \left( \int_{-\infty}^\infty e^{-y^2}dy \right) = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2 + y^2)}dxdy.$$ Write $[\Gamma(1/2)]^2$ in polar coordinates as $\int_0^\infty \int_0^{2\pi} e^{-r^2}rdrd\theta$ and put $v = r^2$ to get $$\pi\int_0^\infty e^{-v}dv = \lim_{M \to \infty}\left[ -\frac{\pi}{e^v} \right]_0^M = \pi.$$ Thus $[\Gamma(1/2)]^2 = \pi$, but $e^{-u^2} > 0$ for all $u \in \mathbb R$, so $\Gamma(1/2) = \sqrt\pi$. Q.E.D.
By the way, the formula for the volume of the $n$-sphere can be thought of as a generalization of the result $\Gamma(1/2) = \sqrt\pi$.