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N-dimensional sphere

  1. Aug 12, 2005 #1
    This was done while I was in a stat-mech class a long, long time ago.......didn't understand it very well at the time. It was an aside kind of thing the prof did, but I'm still curious about this:

    How do you find the surface area of an n-dimensional sphere?


    Thanks! :approve:
     
  2. jcsd
  3. Aug 12, 2005 #2

    DaveC426913

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    Just guessing,
    The surface is the derivative of the volume.


    In 3D, volume is: [tex]4 \pi r^3[/tex] and surface is: [tex]\frac{4}{3} \pi r^2[/tex]

    So,
    A 4D sphere's surface area will be : [tex]4 \pi r^3[/tex] while its volume will be [tex]16 \pi r^4[/tex]

    I think.
     
  4. Aug 12, 2005 #3
  5. Aug 12, 2005 #4
    I was kind of hoping someone could explain the derivation to me. I'm not sure if the link to mathworld is the solution or not. Even if it is I don't understand that any better without some kind of explanation.

    I'm 99.999% sure the other response is not going in the right direction, but I appreciate the responses! :biggrin:
     
  6. Aug 12, 2005 #5

    mathwonk

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    do it by inductive reasoning. i.e. to get the surface are you just need to differentiate the volume wrt radius. to get the volume you integrate the n-1 volume of a slice.

    try to recall how to get the volume ofa 3 sphere by integrating area of circular slices, then apply that one dimension up.
     
  7. Aug 12, 2005 #6

    saltydog

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    I think the following web site is a good one:

    Area and Volume formulas for n-dim spheres and balls

    I'll go over the integrals too! :smile:

    Edit: Ok I did:

    Hey guys, I found the volume formulas for n-balls interesting: simply integrate a cross-section of the object over appropriate bounds. For the ball, it was:

    [tex] V_2=2\int_0^{\pi/2} (line)dheight[/tex]

    [tex]V_3=2\int_0^{\pi/2} (circle)dheight [/tex]

    [tex]V_4=2\int_0^{\pi/2} (sphere)dheight [/tex]

    and so on if you check the reference I gave above. I wonder if this is the case in general. That is, if I calculate the volume of my coffee cup then would the volume of a 4-D coffee cup be:

    [tex]V_4=\int_a^b \text{(3-D coffee cup)} dp[/tex]

    for appropriate values of a, b, and p.

    (Hope I don't make it confussing for you Fizixx). :confused:
     
    Last edited: Aug 12, 2005
  8. Aug 12, 2005 #7

    saltydog

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    Got some questions:

    Can we say in general to calculate the volume of ANY object in n-dimensional space, integrate the n-1 dimensional object over appropriate bounds? What then is the volume of the n-D torus? Same dif? I would say so. Maybe we need to calculate it.

    Also, I found it interesting in the MathWorld reference that the volume of the sphere reaches a maximum value at around 5-D and then drops getting closer and closer to zero as the dimension is increased. Is that the case with all n-D volumes? Does this mean that my coffee cup holds smaller and smaller amounts of n-D coffee?

    I can see it now: Salty's n-D Coffee Cafe' . . . where the coffee never runs out. :smile:
     
  9. Aug 14, 2005 #8
    saltydog...and others.....

    Great in concept, but it does not answer my question, unfortunately. I think I asked my question on the wrong board.

    Providing an aswer that tritely says to just integrate your way up the dimension-ladder despite isn't sufficient, but I do apreciate the responses a great deal, and thanks to everyone for sending a response, but I'm closing this thread and going elsewhere for further inquiry. :smile:

    Good day all, and hapy thoughts to everyone! o:)
     
  10. Aug 14, 2005 #9
    Did you read my link? ( http://mathworld.wolfram.com/Ball.html )


    It gives an explicit formula for the hypercontent ('volume') of the n-ball, in terms of the gamma function:

    [tex]V_n(r) = \frac{\pi ^{n/2} r^n}{\Gamma \left( 1 + \frac{1}{2}n \right)}[/tex]

    From which the 'area' of a hypersphere's surface naturally follows, in the usual way:

    [tex]S_n(r) = \frac{ dV_n }{ dr } = \frac{n \pi ^{n/2} r^{n-1}}{\Gamma \left( 1 + \frac{1}{2}n \right)}= \frac{2 \pi ^{n/2} r^{n-1}}{\Gamma \left( \frac{1}{2}n \right)}[/tex]

    If you don't like this, there's a table at the bottom of the page which works out the first ten dimensions; and a more complete listing is Sloane's A072478 and A072479 .

    edit: More of the same is here (also from Mathworld).
     
    Last edited by a moderator: Aug 14, 2005
  11. Aug 14, 2005 #10
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