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Volume of n-dimensional sphere

  1. Nov 20, 2013 #1
    Hello, this may seem like a stupid question but how would one calculate the volume of an n-dimensional sphere?
    Thanks.
     
  2. jcsd
  3. Nov 20, 2013 #2

    jedishrfu

    Staff: Mentor

    I think you would extroplate the formulas for area of a circle to volume of a sphere to hypervolume of a hypersphere...

    pi*r^2
    4/3pi*r^3

    http://en.wikipedia.org/wiki/N-sphere

    Midway through the article is a cool table of the progress of n from 0 to ... and the volumes and surfaces of the hyperspheres.
     
  4. Nov 21, 2013 #3
  5. Nov 21, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Given an [itex]x_1,x_2, ..., x_n[/itex] Cartesian coordinate system, the the equation of the n-sphere of radius R, with center at the origin is [itex]x_1^2+ x_2^2+ x_3^2+ \cdot\cdot\cdot+ x_n^2= R^2[/itex].
    It is clear that, if all the other variables are 0, then [itex]x_1^2= R^2[/itex] so that [itex]x_1[/itex] ranges between -R and R to cover the entire n-sphere. In the [itex]x_1x_2[/itex] plane, all other variables 0, [itex]x_1^2+ x_2^2= R^2[/itex] so that, for fixed [itex]x_1[/itex], [itex]x_2= \pm\sqrt{R^2- x_1^2}[/itex] and so [itex]x_2[/itex] ranges between [itex]-\sqrt{R^2- x_1^2}[/itex] and [itex]\sqrt{R^2- x_1^2}[/itex] etc.

    Continuing like that, we see that the volume is given by
    [tex]\int_{-R}^R\int_{-\sqrt{R^2- x_1^2}}^{\sqrt{R^2- x_1^2}}\int_{-\sqrt{R^2- x_1^2- x_2^2}}^{\sqrt{R^2- x_1^2- x_2^2}}\cdot\cdot\cdot\int_{-\sqrt{R^2- x_1^2- x_2^2- \cdot\cdot\cdot- x_{n-1}^2}}^{\sqrt{R^2- x_1^2- x_2^2- \cdot\cdot\cdot- x_{n-1}^2}} dx_ndx_{n-1}\cdot\cdot\cdot dx_2 dx_1[/tex].

    You ought to be able to take the formulas for area of a circle (2-sphere), volume of a sphere (3-sphere) and use that integral to find the hyper-volumes of the 4-sphere, 5-sphere, etc to find a general formula.
     
  6. Nov 21, 2013 #5

    mathman

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    Science Advisor
    Gold Member

    http://en.wikipedia.org/wiki/N-sphere

    Above is a detailed discussion. Note that the development works with a generalization of spherical coordinates.
     
  7. Nov 22, 2013 #6
    Thanks all especially halls of ivy as I was just about to ask for it in integral form
     
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