# Thread: Hypersphere and integrals

1. ## Hypersphere and integrals

I tried googling about integrals for finding volume/surface area of the hypersphere or other regions in 4-space. I didn't find much about it though. Would "quadruple" integrals be used? or some other method.

I also tried searching amazon for a book I could buy for reference.

I'd like to challenge myself to some problems

thanks!

2. ## Re: Hypersphere and integrals

You would use quadruple integrals for the general case.

It's a direct extension of the 3D case but now "volume" has an additional component.

Note that finding integration boundaries for all but the simplest shapes is going to get complicated very quickly.

3. ## Re: Hypersphere and integrals

If I were do the region of the unit hypersphere is there spherical coordinates for 4-space? would it be the same three varaibles from 3-space plus another variable that is an angle measurement?

4. ## Re: Hypersphere and integrals

Originally Posted by Jonroberts74
If I were do the region of the unit hypersphere is there spherical coordinates for 4-space? would it be the same three varaibles from 3-space plus another variable that is an angle measurement?
no. It's just an additional euclidean direction that we just as simple humans can't imagine. Or at least most of us can't imagine it.

The positive and negative directions of this are sometimes called ana and kata, in the same sense of up and down or east and west.

5. ## Re: Hypersphere and integrals

I found on wolfram mathworld, the hyper-surface area of the n-sphere of unit radius it must satisfy $S_n\int_{0}^{\infty} e^{-r^2}r^{m-1}dr = \Bigg(\int_{-\infty}^{\infty}e^{-x^{2}}dx\Bigg)^n$

so for the unit hypersphere in 4-space it is $\Bigg(\int_{-\infty}^{\infty}e^{-x^{2}}dx\Bigg)^4 = \pi^2$ But this isn't correct, it should be $2\pi^2=S_n$ so what did I get wrong here??

because I read more on the wolfram site and saw just below that $S_n = \frac{2\pi^{n/2}}{\Gamma(\frac{1}{2}n)} = 2\pi^2;n=4$

and for volume, $V_n = \int_{0}^{R} S_n r^{n-1}dr=\frac{S_n R^n}{n}$

which gives $\frac{2\pi^2}{4}=\frac{\pi^2}{2}$

6. ## Re: Hypersphere and integrals

Originally Posted by Jonroberts74
I found on wolfram mathworld, the hyper-surface area of the n-sphere of unit radius it must satisfy $S_n\int_{0}^{\infty} e^{-r^2}r^{m-1}dr = \Bigg(\int_{-\infty}^{\infty}e^{-x^{2}}dx\Bigg)^n$
I think you are misreading this. It is not saying that $S_n$ is equal to that but that the equation is satisfied by it- to get $S_n$ you need to divide by that integral multiplying $S_n$: $S_n= \frac{\Bigg(\int_{-\infty}^{\infty}e^{-x^{2}}dx\Bigg)^n}{\int_{0}^{\infty} e^{-r^2}r^{n-1}dr}$.

so for the unit hypersphere in 4-space it is $\Bigg(\int_{-\infty}^{\infty}e^{-x^{2}}dx\Bigg)^4 = \pi^2$ But this isn't correct, it should be $2\pi^2=S_n$ so what did I get wrong here??

because I read more on the wolfram site and saw just below that $S_n = \frac{2\pi^{n/2}}{\Gamma(\frac{1}{2}n)} = 2\pi^2;n=4$

and for volume, $V_n = \int_{0}^{R} S_n r^{n-1}dr=\frac{S_n R^n}{n}$

which gives $\frac{2\pi^2}{4}=\frac{\pi^2}{2}$