# Integrating to find the volume

1. May 15, 2005

### Aki

Help, I'm trying to find the hypervolume of a hypersphere and I'm stuck on this:

$$V^4= 2(\frac{4\pi} {3}) \int_0^r (\sqrt{r^2-x^2}) ^3 dx$$

I don't know how to do the integration, and I can't expand the $$(\sqrt{r^2-x^2}) ^3$$
The answer should be $$\frac{\pi ^2}{2}r^4$$

Last edited: May 15, 2005
2. May 15, 2005

### quetzalcoatl9

can you please state the original problem in it's entirety?

what are your coordinate conversions (i.e. what is $x$ in terms of $r$, $\theta$, etc.)?

often times by converting to other coordinates (hyperspherical coordinates, in this case) the integral becomes simpler.

3. May 15, 2005

### dextercioby

Anyway,it's useless.The hypersphere has zero volume.

Daniel.

4. May 15, 2005

### HallsofIvy

Staff Emeritus
The hypersphere does have a "hyper-volume" (or, more generally, a "measure")!

Aki: in order to integrate $$V^4= 2(\frac{4\pi} {3}) \int_0^r (\sqrt{r^2-x^2}) ^3 dx$$, did you consider using a trigonometric substitution, say x= r sin(&theta;)?

5. May 15, 2005

### dextercioby

That's the BALL,the hypersphere (the S_{3} being the smallest in dimension) is a hypersurface and has zero hypervolume.

Daniel.

6. May 15, 2005

### Aki

$$x=r \cos \theta , \sqrt {r^2 - x^2} = r \sin \theta , dx = -r \sin \theta d\theta$$

7. May 15, 2005

### Icebreaker

The volume enclosed by a hypersphere.