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Math Help - Hypersphere and integrals

  1. #1
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    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!
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  2. #2
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    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.
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    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?
    Last edited by Jonroberts74; August 21st 2014 at 03:44 PM.
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    Re: Hypersphere and integrals

    Quote Originally Posted by Jonroberts74 View Post
    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.
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    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}
    Last edited by Jonroberts74; August 21st 2014 at 11:13 PM.
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    Re: Hypersphere and integrals

    Quote Originally Posted by Jonroberts74 View Post
    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}
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