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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:

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?

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Checkout the answers here:… – 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.

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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

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