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Volume of an ellipsoid using double integrals

  1. Jul 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Using double integrals, calculate the volume of the solid bound by the ellipsoid:

    x²/a² + y²/b² + z²/c² = 1

    2. Relevant data

    must be done using double integrals

    3. The attempt at a solution

    i simply can't find a way to solve this by double integrals, i did with triple integral, but my teacher won't accept it, this is the last question and i can't find a solution for it, a step-by-step would be awsome.

    Thanks.
     
    Last edited: Jul 7, 2013
  2. jcsd
  3. Jul 8, 2013 #2

    Simon Bridge

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    If it were a solid of rotation you could do it in a single-integral right?
    Basically you have to use reasoning in the place of one of the integrals ...

    Find the volume between $$z=c^2\left ( 1-\frac{x^2}{a^2}-\frac{y^2}{b^2} \right )$$ and the x-y plane.
     
  4. Jul 8, 2013 #3

    mfb

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    The first integration in the triple integral is trivial, and afterwards you get a double-integral - you can use this double-integral (ignoring the first step) to solve the problem.
     
  5. Jul 8, 2013 #4

    HallsofIvy

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    Essentially the same idea: projecting the ellipsoid to the xy-plane (z= 0) gives the ellipse [itex]x^2/a^2+ y^2/b^2= 1[/itex]. The two heights at each (x, y) point are [itex]z= \pm c\sqrt{1- x^2/a^2- y^2/b^2}[/itex]. The difference, [itex]2c\sqrt{1- x^2/a^2- y^2/b^2}[/itex], is the length of thin rectangle above that point and is the function to be integrated. For each x, The ellipse goes from [itex]y= -b\sqrt{1- x^2/a^2}[/itex] to [itex]y= b\sqrt{1- x^2/a^2}[/itex]. And, over all, x goes from -a to a. The volume of the ellipse is given by
    [tex]\int_{-a}^a\int_{-b\sqrt{1- x^2/a^2}}^{b\sqrt{1- x^2/a^2}} 2c\sqrt{1- x^2/a^2- y^2/b^2} dydx[/tex].

    As mfb says, this is the same as if you started with the triple integral
    [tex]\int_{-a}^a\int_{-b\sqrt{1- x^2/a^2}}^{b\sqrt{1- x^2/a^2}}\int_{-c\sqrt{1- x^2/a^2- y^2/b^2}}^{c\sqrt{1- x^2/a^2- y^2/b^2}} dzdydx[/tex]
    and integrated once.
     
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