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Integrating to find the volume

  1. May 15, 2005 #1

    Aki

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    Help, I'm trying to find the hypervolume of a hypersphere and I'm stuck on this:

    [tex]V^4= 2(\frac{4\pi} {3}) \int_0^r (\sqrt{r^2-x^2}) ^3 dx [/tex]

    I don't know how to do the integration, and I can't expand the [tex](\sqrt{r^2-x^2}) ^3 [/tex]
    The answer should be [tex] \frac{\pi ^2}{2}r^4 [/tex]

    Please help, thanks
     
    Last edited: May 15, 2005
  2. jcsd
  3. May 15, 2005 #2
    can you please state the original problem in it's entirety?

    what are your coordinate conversions (i.e. what is [itex]x[/itex] in terms of [itex]r[/itex], [itex]\theta[/itex], etc.)?

    often times by converting to other coordinates (hyperspherical coordinates, in this case) the integral becomes simpler.
     
  4. May 15, 2005 #3

    dextercioby

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    Anyway,it's useless.The hypersphere has zero volume.

    Daniel.
     
  5. May 15, 2005 #4

    HallsofIvy

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    The hypersphere does have a "hyper-volume" (or, more generally, a "measure")!

    Aki: in order to integrate [tex]V^4= 2(\frac{4\pi} {3}) \int_0^r (\sqrt{r^2-x^2}) ^3 dx [/tex], did you consider using a trigonometric substitution, say x= r sin(θ)?
     
  6. May 15, 2005 #5

    dextercioby

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    That's the BALL,the hypersphere (the S_{3} being the smallest in dimension) is a hypersurface and has zero hypervolume.

    Daniel.
     
  7. May 15, 2005 #6

    Aki

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    [tex]x=r \cos \theta , \sqrt {r^2 - x^2} = r \sin \theta , dx = -r \sin \theta d\theta [/tex]
     
  8. May 15, 2005 #7
    The volume enclosed by a hypersphere.
     
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