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Deriving Volume

  1. May 22, 2005 #1
    How would you go about deriving the volume of a prolate spheroid using calculus? Would I use the method of shells? I am not sure where to begin. Any help and tips are greatly appreciated!

    Thanks :smile:
     
    Last edited: May 22, 2005
  2. jcsd
  3. May 22, 2005 #2

    dextercioby

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    Use prolate spheroid coordinates.

    Daniel.
     
  4. May 22, 2005 #3
    I am not sure what you mean. How would I go about it using the method of shells?

    Thanks
     
  5. May 22, 2005 #4

    dextercioby

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    I don't have a clue.I've suggested another method which i'm sure would do it.

    Daniel.
     
  6. May 22, 2005 #5
    You know where I could find an explanation of that method? Do calculus books contain these topics? Also the volume of a prolate spheroid is the same as the volume of an American football which I am trying to find. If I have some dimensions, could the problem be made easier? Also is an ellipse rotated about the x-axis a prolate spheroid?

    Thanks
     
  7. May 22, 2005 #6

    dextercioby

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    Yes for the last question.Yes for the penultimate one.

    Well,then u could consider the arbitrary ellipsoid and then take 2 of the semiaxis to be equal...

    Daniel.
     
  8. May 22, 2005 #7
    So lets say the length of the football is 24 inches. Then each semi-axes are 12 inches each. So assuming we have the ellipse centered at the origin, would the intersection points be (-12,0) and (12,0). So you would use [tex] \int^{12}_{-12} \pi(f(x))^{2} dx [/tex]?

    Thanks :smile:
     
  9. May 22, 2005 #8

    dextercioby

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    Yes,it's correct.After performing that integration,u should check with

    [tex] V_{\mbox{prolate ellipsoid}}=\frac{4\pi}{3} a^{2}c [/tex]

    ,where "c"is the large semixis of the ellipse and "a" is the small semiaxis=the radius of the circle described by rotation.

    Daniel.
     
  10. May 23, 2005 #9

    Curious3141

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    Isn't proving the volume formula trivial by simple integration using rectangular coordinates ?

    Consider the ellipse [tex]x^2(\frac{a^2}{c^2}) + y^2 = a^2[/tex]. This is the equation of the curve that would give the required ellipsoid by revolution about the x-axis.

    The required enclosed volume of revolution is given by :

    [tex]V = \pi\int_{-c}^{c}{y^2}dx[/tex] which can be evaluated easily and shown to be [tex]V = \frac{4}{3}\pi a^2c[/tex]
     
  11. May 23, 2005 #10
    how would you show that if you don't mind my asking?

    Thanks
     
  12. May 23, 2005 #11

    arildno

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    courtigrad:
    The following coordinate transformation is handy in tackling ellipsoids (and thus, prolate spheroids).
    Suppose your ellipsoid is specified by the inequality:
    [tex](\frac{x}{a})^{2}+(\frac{y}{b})^{2}+(\frac{z}{c})^{2}\leq{1}[/tex]
    with x,y,z normal Cartesian coordinates.
    Introduce variables [tex]r,\theta,\phi[/tex] as follows:
    [tex]x=ar\sin\phi\cos\theta,y=br\sin\phi\sin\theta, z=cr\cos\phi, 0\leq{r}\leq1,0\leq\theta\leq{2\pi},0\leq\phi\leq\pi[/tex]

    This coordinate transformation is seen to simplify the inequality specifying the ellipsoid to [tex]r^{2}\leq{1}[/tex] (which is fulfilled with the limits placed on r)

    In order therefore to determine the volume of the ellipsoid, just use this coordinate transformation along with the change-of-variables theorem.
     
  13. May 23, 2005 #12

    Curious3141

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    Arildno's post is about the general ellipsoid. You only asked for the prolate spheroid, which means two axes are necessarily the same while the third is possibly different.

    I just realised that my work is for the oblate spheroid (in the orientation given), but it doesn't matter at all as long as you keep track of which is the major axis and which are the minor axes. The same formula will apply for the prolate spheroid as well.

    So, taking my calculation (for the oblate spheroid), can you see why the ellipse formula is the way it is ? Basically I am relating the ellipse to a circle centered at the origin with radius a, but with a stretch factor of (c/a) applied to the horizontal dimension.

    After getting that equation for the bounding ellipse, just evaluate the integral given :

    [tex]V = \pi\int_{-c}^{c}{y^2}dx = \pi\int_{-c}^{c}{a^2 - (x^2)\frac{a^2}{c^2}dx = \pi[(a^2)x - \frac{x^3}{3}(\frac{a^2}{c^2})]_{-c}^{c} = (2\pi)(\frac{2}{3})a^2c = \frac{4}{3}\pi a^2c[/tex]
     
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