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Unit balls

  1. May 30, 2010 #1
    How do you derive the hypervolume/hypersurface of the unit ball in n dimensions? I thought it'd be trivial but oh well.
    I'm guessing switch to n-D spherical coordinates?
    Is there an argument which doesn't involve explicitly calculating too many things? :x
  2. jcsd
  3. May 31, 2010 #2


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    Howdy, well, this argument I saw in a stat mech text: Consider 2 different ways of computing the integral (<n> arguments)

    [tex] \int_{- \infty}^{+ \infty} e^{- x^2 - y^2 - z^2 - ...} dx{}dy{}dz{}... [/tex]

    First it's a product of "n" integrals of Poisson/Gauss. The result of

    [tex] \int_{- \infty}^{+ \infty} e^{- x^2 } dx[/tex] we know and just raise it to the power "n".

    Now use the (hyper) spherical coordinates for the first integral. The result should be

    [tex] \int_{0}^{+ \infty} e^{-r^2} {} \ r^{n-1} {} \ dr \times S_{n} [/tex]

    S_{n} is the area of the unit ball in "n" dimensions. The first integral you can compute and then you can get S_{n}. But S_{n} it's easily linked to the V_{n} and thus you can find V_{n}.
    Last edited: May 31, 2010
  4. May 31, 2010 #3
    Yeah, but um.
    I "know" it's gonna work out all nice, but how do you construct spherical coordinates in n dimensions, and show that the Jacobian is what it is?

    I guess dV=dr.dS is reasonable. Analysis of it seems a bit more annoying though :x
  5. May 31, 2010 #4


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    Why bother with all that mess when you already know how the measure of a shell of constant radius scales with r?

    The way I've heard it, the only tricky part is showing that the integral really can be rearranged like that without changing the value.
  6. May 31, 2010 #5
    Oops. I was kind of thinking "How do we know that's S_n"?
    But that's the definition of S_n.
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