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Volume of Ellipsoid

  1. Feb 15, 2006 #1
    I've been asked to derive an expression for the volume of an ellipsoid. I know what the expression is, I just don't know how to get there from the information given. All that is given is that it is defined by

    [tex]\frac{x^2}{a^2} +\frac{y^2}{b^2} +\frac{z^2}{c^2} \leq 1[/tex],

    a standard inequality/definition for an ellisoid. Not quit sure how to go about this.
     
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  3. Feb 15, 2006 #2

    HallsofIvy

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    Have you tried anything? Like first drawing a picture. Certainly it is straight forward to set up the integral. The ellipsoid's widest girth comes in the xy-plane where it is an ellipse:
    [tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}= 1[/tex]
    That is is covered by taking x from -a to a and, for each x, y from [itex]-b\sqrt{1-\frac{x^2}{a^2}[/itex] to [itex]b\sqrt{1-\frac{x^2}{a^2}}[/itex]. For each (x,y), z ranges from [itex]-c\sqrt{1-\frac{x^2}{a^2}- \frac{y^2}{b^2}}[/itex]. The volume is
    [tex]\int_{x=-a}^a\int_{y=-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}}\int_{z=-c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}^{c\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}}dzdydx[/tex]
    or
    [tex]2c\int_{x=-a}^a\int_{y=-b\sqrt{1-\frac{x^2}{a^2}}}^{b\sqrt{1-\frac{x^2}{a^2}}}\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}dydx[/tex]
     
  4. Feb 15, 2006 #3

    arildno

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    Use spherical coordinates, with a twist.
     
  5. Feb 15, 2006 #4
    I agree with the previous poster, use spherical coordinates. You have (x²/a²) + (y²/b²) + (z²/c²) = 1. Spherical coordinates woudldn't work too well with that equation the way it is. It would be much better if your ellipse was transformed into a sphere with an equation like u^2 + v^2 + w^2 = 1. This suggests a change of variables. Consider u^2 = (x²/a²) and likewise for the other components for the three obvious substitutions.
     
  6. Jul 17, 2006 #5
    if anyone could further evaluate this problem, i would greatly appreciate it. I followed the parameterization suggestion and set u=x/a, v=y/b, and w=z/c and switched the bounds on the integral. However, after integrating once I run into the problem of

    2C * sqrt(1-u^2-v^2)

    When I integrate that, I am unable to do it by hand and must do it on a computer system, which gives me a very long answer with two arctangents. I am pretty sure that will not work out to give me 4/3Piabc.

    Thanks in advance.
     
  7. Jul 17, 2006 #6

    HallsofIvy

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    That's not difficult to integrate. It is essentially [itex]\sqrt{a^2- u^2}[/itex] for which the standard substitution is the trig substitution [itex]u= a sin(\theta)[/itex]. Here that would be [itex]u= \sqrt{1- v^2}sin(\theta)[/itex]. Integrating that, with respect to u, will give something involving [itex]\sqrt{1- v^2}[/itex] which you do with another sin substitution.
     
  8. Jul 17, 2006 #7
    I'm sorry but I don't understand. The formula listed in my previous post...

    2C * sqrt(1-u^2-v^2)

    is the resulting formula after having already integrated with respect to u.
    You said "integrating that, with respect to u," but I need to integrate that with respect to v.

    Also, if I introduce a sin(theta) into the equation, I don't see a point in the future at which that constant will disappear, and thus my answer would have a trig function in it, although the correct answer should be
    4/3 * Pi * abc. Correct me if I'm wrong, I'm just not following your suggested method.
     
  9. Jul 17, 2006 #8

    Office_Shredder

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    Ok, so just do 2C*sqrt[ (1-u^2)-v^2] and integrate with respect to v. By making the trig substitution, you're actually integrating with respect to sin(theta) instead, so it goes away then
     
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