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Volume of n-dimensional sphere equation

  1. Jul 31, 2009 #1
    I need help in understanding this problem.
    Equation of sphere in n-dimensional space is:

    [tex]x^2_1+x^2_2+...+x^2_n=R^2[/tex]

    We serch volume as [tex]V=C_nR^n[/tex]. Why? Perhaps its analogy with [tex]CR^3[/tex].

    Now we calculate this integral

    [tex]I=\int^{\infty}_{-\infty}dx_1\int^{\infty}_{-\infty}dx_2\int^{\infty}_{-\infty}dx_3...\int^{\infty}_{-\infty}dx_ne^{-a(x^2_1+x^2_2+...+x^2_n)}[/tex]
    Why we do this?

    And we get [tex](\frac{\pi}{a})^{\frac{N}{2}}[/tex]

    And then

    [tex]I=\int dV_n e^{-ar^2}[/tex]

    we get

    [tex]V_n=\frac{(\pi)^{\frac{N}{2}}}{\Gamma(\frac{N}{2}+1)}R^n[/tex]

    Can someone tell me idea of all this. Thanks
     
  2. jcsd
  3. Jul 31, 2009 #2

    Cyosis

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    Homework Helper

    Some gaps need to be filled to arrive at the answer.

    The reason we seek a volume of that form is because any volume is proportional to the dimension of the space. Intuitively think about cubic meters. A 'normal' volume would have m^3 as a unit, a 4-dimensional volume would have m^4 as unit and an n-dimensional volume would have m^n as unit. So we know that [itex]V \approx R^n[/itex]. The only thing that is missing is the proportionality constant, which we will denote [itex]C_n[/itex]. Therefore we are looking for an expression of the form [itex]V_n=C_n R^n[/itex].

    Similarly we know that the area of the n-sphere is proportional to R^(n-1), introducing another proportionality constant, [itex]S_n[/itex], we get [itex]A_n=S_n R^{n-1}[/itex]. Note that [itex]S_n[/itex] is the surface area of a unit n-sphere.
    Now think about a solid sphere as a lot of spherical shells stacked on top of each other. Adding all these shells together will give us the volume of the sphere. Therefore

    [tex]
    \begin{equation}
    V_n=\int_0^R S_n r^{n-1} dr=\frac{S_n R^n}{n}
    \end{equation}
    [/tex].

    We will need this result later on.

    Now we use a trick by equating two integrals to each other and then solving for [itex]S_n[/itex]

    The integral over all space of the Gaussian function in Cartesian coordinates is.

    [tex]
    \int_{\mathbb{R}^n} e^{-a ||\vec{x}||^2 }d\vec{x}=\left( \int_{-\infty}^\infty e^{-a x^2}dx \right)^n=\left(\frac{\pi}{a}\right)^{n/2}
    [/tex]

    Switching to spherical coordinates we should of course get the same answer.

    [tex]
    \begin{align*}
    \int_{\mathbb{R}^n} e^{-a r^2}dV_n & =\int_0^\infty S_n r^{n-1} e^{-a r^2}dr \;\;\;\;\;\;\;(\mathb{substitute \;\; ar^2=x}) \\

    & = \frac{S_n}{2 a^{\frac{n}{2}}} \int_0^\infty x^{\frac{n}{2}-1} e^{-x} dx \\

    & =\frac{S_n}{2 a^{\frac{n}{2}}} \Gamma(\frac{n}{2})

    \end{align*}
    [/tex]

    Therefore
    [tex]
    \frac{S_n}{2 a^{\frac{n}{2}}} \Gamma(\frac{n}{2})=\left(\frac{\pi}{a}\right)^{n/2}
    [/tex]

    Now solve for S_n and plug it into equation one.
     
    Last edited: Jul 31, 2009
  4. Jul 31, 2009 #3
    Thanks! I have one more question. Why you solve this integral

    [tex]\int_{\mathbb{R}^n} e^{-a ||\vec{x}||^2 }d\vec{x}[/tex]?

    Is that because is that easiest way?

    Principial you can solve some other integral? Am I right?
     
    Last edited: Jul 31, 2009
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