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Finding the volume of an ellipsoid by using the volume of a sphere

  1. Aug 22, 2012 #1
    1. The problem statement, all variables and given/known data
    So I keep coming across problems that suggest finding the Volume of an ellipsoid using the volume of a ball ie:

    Find the volume enclosed by the ellipsoid:

    (x/a)^2 + (y/b)^2 + (z/c)^2 = 1

    by using the fact that the volume of the unit ball in R^3 is 4pi/3


    2. Relevant equations



    3. The attempt at a solution

    I keep getting stuck on this one. It seems like it will be some change of variables in which abc will eventually just pop out front of the usual expression for the volume of a ball. I keep thinking that I should be able to use spherical coords and a substitution like u = x/a, v = y/b and w = z/c. But then you just get the normal unit ball integral. I feel like it might come out of the limits of the integral in terms of P. That is, the maximum length of p is a function of [itex]\phi[/itex] and [itex]\theta[/itex] in the ellipsoid case. But with the above substitution then P is 1. Or I keep thinking it might come out of dP, but that hasn't been clear either.

    I also keep thinking of plugging in for p -> p = [itex]\sqrt{u2 + v2 + w2}[/itex] = [itex]\sqrt{(psin \phi cos \theta)2/a + (psin \phi sin \theta)2/b + (pcos \phi cos \theta)2/c}[/itex] but then the p2 term would cancel out.

    Anyway, I am just going in circles. Does anybody have any suggestions?
     
  2. jcsd
  3. Aug 22, 2012 #2

    LCKurtz

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    The volume of the ellipsoid is$$
    \iiint_V 1dxdydz$$in the ##xyz## plane. Don't think in terms of spherical coordinates, think more about what the ellipsoid becomes in ##uvw## space under the transformation you are thinking about which I have highlighted. And the Jacobian has something to do with it.
     
  4. Aug 22, 2012 #3
    Is it simply that x/a = u -> dx = adu and etc?

    I think I can see it from the Jacobian in Spherical but not Cartesian. Like if u,v,w are defined like normal in spherical coordinates, then making the jacobian makes a factor of a across the first row, b on the second and c on the third which pull out in the determinant.

    But the Jacobian in cartesian still seems weird. So:

    ∫∫∫R(xyz)dV = ∫∫∫R(uvw) J dV


    But in cartesian x=ua, y=vb and z=wc only defines the "functions" in terms of one other variable (x = f(u), y=g(v) etc...) then fv(u) is 0 etc. I am pretty new to Jacobians so I am problem just have a misunderstanding with them.
     
  5. Aug 22, 2012 #4
    Ohhh! I see the Jacobian will just be a diagonal matrix, so it won't be 0. Thanks, sorry I can be a little slow.
     
  6. Aug 22, 2012 #5

    LCKurtz

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    So do you see how to finish it?
     
  7. Aug 22, 2012 #6
    I think so. Just that J then equals abc. Right?
     
  8. Aug 22, 2012 #7

    LCKurtz

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    Yes. Assuming you mean ##\left |\frac {\partial(x,y,z)}{\partial(u,v,w)}\right|##. So you get ##abc## times the volume of the unit sphere, which you already know.
     
  9. Aug 22, 2012 #8
    I'm not too familiar with the notation. So just to be sure. If I saw we have x=f(u), y=g(v), z = h(w) then is what you wrote equivalent to determinate of

    fu fv fw
    gu gv gw
    hu hv hw

    ? Sorry, I still can't figure out how to write matrices in this forum.
     
  10. Aug 22, 2012 #9

    LCKurtz

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    Yes. Right click on this to show the tex:$$
    J = \left |\begin{array}{ccc}
    a & 0 & 0\\
    0 & b & 0\\
    0 & 0 & c
    \end{array}\right| = abc$$
     
  11. Aug 22, 2012 #10

    Curious3141

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    Alternatively, you can use a Tex editor. I've found Daum equation editor (a free Chrome app) to be excellent for complicated Tex formatting.
     
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