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Math Help - Volume Enclosed by surfaces (Triple integration)

  1. #1
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    Volume Enclosed by surfaces (Triple integration)

    Hello! I can't quite get this triple integral problem even after looking at the sticky for integration.

    I have the surfaces z = x^2 + 3(y^2) and z = 8 - x*x - y*y and I need to find the volume of the region enclosed.
    dV = dzdydx

    All help is greatly appreciated!
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  2. #2
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    Re: Volume Enclosed by surfaces (Triple integration)

    The integral you want to solve is \iiint_{x^2 + 3y^2}^{8 - x^2 - y^2}dzdxdy=\iint_D \left[8 - x^2 - y^2 - (x^2 + 3y^2)\right]dxdy

    The integration area D is given by the equation x^2 + 3y^2 = 8 - x^2 - y^2 \Leftrightarrow x^2 + 2 y^2 = 4

    EDIT: Fixed notation issues.
    Last edited by Ridley; March 18th 2012 at 09:24 PM.
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  3. #3
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    Re: Volume Enclosed by surfaces (Triple integration)

    I'm confused on why you only have one integral? How do I take this and turn it into three?
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  4. #4
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    Re: Volume Enclosed by surfaces (Triple integration)

    It's just a different notation. If it says dxdydz then you obviously have a triple integral.
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  5. #5
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    Re: Volume Enclosed by surfaces (Triple integration)

    Just to clarify, my limits are going to be for x: [-2, 2] and for y: [-sqrt(2), sqrt(2)]?
    Is that right? or should the y be sqrt(2-x^2 / 2)?
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  6. #6
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    Re: Volume Enclosed by surfaces (Triple integration)

    It's easier if you switch to polar coordinates.

     x = r\cos\theta
     \sqrt{2}y = r\sin\theta

    Don't forget to scale the new area element: dxdy = \frac{r}{\sqrt{2}}drd\theta

    If you don't want to switch to polar coordinates then you can integrate from x: \left[-\sqrt{4-2y^2}, \sqrt{4-2y^2}\right] and y: \left[-\sqrt{2}, \sqrt{2}\right]
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  7. #7
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    Re: Volume Enclosed by surfaces (Triple integration)

    Ah that makes so much more sense!
    Thanks so much!
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