Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm wondering about different ways to compute the volume of an $n$-sphere. Please see the wikipedia page for one method to compute the volume via hyperspherical coordinates:

http://en.wikipedia.org/wiki/N-sphere#Hyperspherical_volume_element

Suppose now I want to compute the volume $V(n)$ of an $n$-sphere by integrating the volumes $V(n-1)$ of a whole bunch of $(n-1)$-spheres. Assuming that this $(n-1)$-sphere is aligned with the first $n-1$ coordinate axes, I don't see a way to just integrate $(n-1)$-spheres with the variable being the last angular coordinate. I'm also more generally interested in this question when we integrate $(n-k)$-spheres with volumes $V(n-k)$ with the variables being the last $k$ angular coordinates. Any thoughts?

share|cite|improve this question
    
Checkout the answers here: math.stackexchange.com/questions/164/… – Tpofofn May 22 at 3:41

First, consider integrating for the volume of a circle along a line. $$2\int_0^r (r^2-x^2)^.5dx$$ This one's somewhat involved, as is every subsequent integral that yields an increase in pi power, because you have to use polar on the square root, but you eventually get: $$A=\pi r^2$$ Now, we want to find the volume of a sphere by adding up the volumes of circular cross sections of the sphere along the x-axis. We already know the area of a circle because we calculated it from the previous integral. The radius of the circle is sqrt(r^2-x^2) and there is symmetry, so the integral is $$ 2\pi\int_0^r r^2-x^2dx$$ $$ 2\pi r^3-2\pi r^3/3$$ $$ 4\pi r^3/3$$ You just continue. $$4\pi/3\int_0^r (r^2-x^2)^{1.5}dx$$ and so forth for higher dimension hyperspheres. The OP said something about an angle. If you want to integrate with respect to an angle, change to polar/spherical and follow the same method.

share|cite|improve this answer
    
This seems suspicious. Checking Wikipedia, it seems that we should see increasing powers of $\pi$ in the volume -- does this happen with what you've written? (Also the missing $dx$'s are physically painful :)) – pjs36 May 22 at 3:37
    
Good point, actually the extra pi's show up every two iterations, specifically when there are square roots in the integral. I originally thought the rest of the integrals were simple, but actually every second iteration requires polar because there needs to be a way to get more powers of pi. – thecat May 22 at 14:40

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.