Author Topic: volume of the unit ball (Read 2203 times) Tweet Share

humph · « **on:** February 01, 2008, 12:55:14 pm »

figured ahmad might be interested in this, at the very least. lovely proof i found of finding the volume of a ball in $\mathbb{R}^n$ . you need to know the definition and properties of the gamma function beforehand though.

$\text{An open ball $B_{r}\left(x\right)$ of radius $r>0$ centred at $x \in \mathbb{R}^n$ is the subset of $\mathbb{R}^n$ defined by} \[B_{r}\left(x\right) = \left\{y \in \mathbb{R}^n : \left|y - x\right|<r\right\} \] We're interested in the $n$-dimensional volume $V_{n}$ of a ball in $\mathbb{R}^n$. We use the symbol $\omega_{n}$ to denote the $n$-dimensional volume $V_{n}$ of the unit open ball in $\mathbb{R}^n$ centred at $0$; that is, \[\omega_{n}=V_{n}\left(B_{1}\left(0\right)\right) \] For any open ball of radius $r>0$ centred at $x \in \mathbb{R}^n$, we have \[V_{n}\left(B_{r}\left(x\right)\right)=\omega_{n}r^{n} \] This can be proved by elementary properties of volume. Now we note that $\int_{\mathbb{R}} {e^{-x^{2}} dx} = \sqrt{\pi}$ (there are many ways to prove this). Then it follows that for $x = \left(x_{1}, \ldots , x_{n}\right) \in \mathbb{R}^n$, \[\int_{\mathbb{R}^{n}} {e^{-\left|x\right|^{2}} dx} = \int_{\mathbb{R}} \cdots \int_{\mathbb{R}} {e^{-\left(x^{2}_{1} + \ldots + x^{2}_{n} \right)} dx_{1} \cdots dx_{n}} \] \[= \int_{\mathbb{R}} {e^{-x^{2}_{1}} dx_{1} } \cdots \int_{\mathbb{R}} {e^{-x^{2}_{n}} dx_{n} } = \sqrt{\pi} \cdots \sqrt{\pi} = \pi^{n/2} \] Now let $x \in \mathbb{R}^n$ and $y \in \left(0, \infty \right)$, and let $D$ be the region $D=\left\{\left(x,y\right) : \left|x\right| < \sqrt{y}\right\} \subset \mathbb{R}^{n+1}$. Then \[\int_{D} {e^{-y} dy dx} = \int_{\mathbb{R}^{n}} \int^{\infty}_{\left|x\right|^{2}} {e^{-y} dy dx} = \int_{\mathbb{R}^{n}} {e^{-\left|x\right|^{2}} dx} = \pi^{n/2} \] by the above. On the other hand, \[\begin{split}\int_{D} {e^{-y} dx dy} & = \int^{\infty}_{0} \int_{\left|x\right| < \sqrt{y}} {e^{-y} dx dy} \\ & = \int^{\infty}_{0} \int_{B_{\sqrt{y}}\left(0\right)} {e^{-y} dx dy} \\ & = \int^{\infty}_{0} {V_{n} \left(B_{\sqrt{y}}\left(0\right)\right) e^{-y} dy} \\ & = \int^{\infty}_{0} {\omega_{n} \sqrt{y}^{n} e^{-y} dy} \\ & = \omega_{n} \int^{\infty}_{0} {y^{n/2} e^{-y} dy} \\ & = \omega_{n} \Gamma \left(\frac{n}{2} + 1\right) \\ & = \omega_{n} \frac{n}{2} \Gamma \left(\frac{n}{2}\right) \end{split}\] Then, as $\int_{D} {e^{-y} dx dy} = \int_{D} {e^{-y} dy dx}$, it follows that \[\omega_{n} \frac{n}{2} \Gamma \left(\frac{n}{2}\right) = \pi^{n/2} \] and hence that \[\omega_{n}=\frac{2\pi^{n/2}}{n\Gamma\left(\frac{n}{2}\right)} \]$

enwiabe · « **Reply #1 on:** February 01, 2008, 04:24:11 pm »

I get a red 'x' and then a whole bunch of LaTeX junk after it.

Daniel15 · « **Reply #2 on:** February 01, 2008, 04:35:19 pm »

Please split it up into multiple LaTeX tags... It's passing that whole expression (everything in the tex tag) in the image URL, and the browser is probably choking on it (works in Opera, but probably messed up in IE)

Neobeo · « **Reply #3 on:** February 01, 2008, 04:43:16 pm »

I'm using Opera, but all I see a small box saying "Latex failed, probably due to an error in your expression."

In any case, I've seen this before in some book I no longer remember. It is especially interesting to note that the "volume" can actually drop to less than 1 when you get to 13th dimension and higher.

Ahmad · « **Reply #4 on:** February 01, 2008, 04:51:25 pm »

I thought a little bit about the hypervolume of a hypersphere last year. I've heard about the solution which uses the Gaussian Integral, but haven't seen the solution before (until your post), which is interesting. Shows the power of reversing the order of integration.

Another method which I'd post but I'm off to eat, is to find a recursion by snapping off two integrations using polar coordinates.

Daniel15 · « **Reply #5 on:** February 04, 2008, 01:11:17 pm »

Quote

I'm using Opera, but all I see a small box saying "Latex failed, probably due to an error in your expression."

Here, I saved the image as a static image:

humph · « **Reply #6 on:** February 04, 2008, 02:49:24 pm »

Quote from: Ahmad on February 01, 2008, 04:51:25 pm

I thought a little bit about the hypervolume of a hypersphere last year. I've heard about the solution which uses the Gaussian Integral, but haven't seen the solution before (until your post), which is interesting. Shows the power of reversing the order of integration.

Another method which I'd post but I'm off to eat, is to find a recursion by snapping off two integrations using polar coordinates.

that's the standard technique that's usually shown. another way is to use the recursion between the surface areas of spheres. i just really liked this technique that i found though - used it in the paper that i'm writing up for my summer course.

sorry about the dodgy image stuffs, btw. i'm using firefox and it works just fine, but i didn't stop to think whether it might mess up in other browsers. i just copied & pasted all the text from the latex file that i'd written up.

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