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Math Help - Volume between 2 paraboloids

  1. #1
    MHF Contributor arbolis's Avatar
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    Volume between 2 paraboloids

    Today was my test and I've absolutely no idea about if I'll pass it or not. (On the cut, or I don't know how to say)
    I had to calculate the volume between the paraboloids z=x^2+y^2 and z=\frac{4}{3}-\frac{x^2}{3}-\frac{y^2}{3}, if I remember well.
    I've thought about it and I think that using Cartesian coordinates will be easier than spherical and cylindrical ones.
    However I tried to calculate the volume via cylindrical coordinates: (r,\theta,z). I tried to set up the triple integral : z goes from r^2 to \frac{4}{3}-\frac{r^2}{3}.
    \theta from 0 to 2 \pi.
    And r goes from 0 to .... I wasn't sure. But I strongly believe I already made an error.

    I have 2 questions : How can I calculate the volume they ask using cylindrical coordinates?
    With what method would you solve the problem?
    Thank you very much!

    Oh, and before I forget, I know they cross each other with a circular ( x^2+y^2=1) section.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by arbolis View Post
    Today was my test and I've absolutely no idea about if I'll pass it or not. (On the cut, or I don't know how to say)
    I had to calculate the volume between the paraboloids z=x^2+y^2 and z=\frac{4}{3}-\frac{x^2}{3}-\frac{y^2}{3}, if I remember well.
    I've thought about it and I think that using Cartesian coordinates will be easier than spherical and cylindrical ones.
    However I tried to calculate the volume via cylindrical coordinates: (r,\theta,z). I tried to set up the triple integral : z goes from r^2 to \frac{4}{3}-\frac{r^2}{3}.
    \theta from 0 to 2 \pi.
    And r goes from 0 to .... I wasn't sure. But I strongly believe I already made an error.

    I have 2 questions : How can I calculate the volume they ask using cylindrical coordinates?
    With what method would you solve the problem?
    Thank you very much!

    Oh, and before I forget, I know they cross each other with a circular ( x^2+y^2=1) section.

    in Cartesian coordinate

    \int_{-1}^{1}\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int^{\frac{4-x^2-y^2}{3}}_{x^2+y^2} dzdydx

    in polar

    \int_{0}^{2\pi}\int_{0}^{1}\int_{r^2}^{\frac{4-r^2}{3}}r dzdrd\theta

    Edit : I forget the jacobian
    is there anything is not clear ??
    Last edited by Amer; June 27th 2009 at 11:09 AM.
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  3. #3
    MHF Contributor Amer's Avatar
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    to determine the boundaries I prefer to sketch the graph like this

    Volume between 2 paraboloids-cool.jpg


    and I put in it the important points then I see z change from which curve to which curve in our graph we see that z change from x^2+y^2 ,to...,\frac{4-x^2-y^2}{3}
    since the graph of the first curve is lower than the second one after that we do a dropping to the x-y plane since we first determine the boundaries of z if we drop of the whole solid that we have we will get a circle with radius 1 this is the intersect between two paraboloids like this

    Volume between 2 paraboloids-coool.jpg


    from the picture we see that y change from -\sqrt{1-x^2} to \sqrt{1-x^2}

    then x change from -1 to 1

    in polar it is the same for z and for r and theta I think it is clear

    I am sorry for my bad graph
    Last edited by Amer; June 27th 2009 at 11:09 AM.
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  4. #4
    MHF Contributor arbolis's Avatar
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    Oh... I wasn't far at all! I think I even tried it, but I made an arithmetic mistake and I gave up on this exercise.
    Thanks Amer, everything is clear.
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