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Math Help - Double integral of a function on a region delimited by quadratic functions

  1. #1
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    Question Double integral of a function on a region delimited by quadratic functions

    Hi,

    I'm looking for the steps to solve this kind of questions.

    Calculate the integral \int \int_D \! (xy)^2 \, \mathrm{d}A where D is the region delimited by the following functions and relations: y=4-x^2, y=x^2-4, x=1+y^2, x=-1-y^2

    When I try to solve it, I never find a way to a numeric answer. I always get a big function which I'm sure it's not.

    Thank you so much for giving me a hint,

    Bazinga
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  2. #2
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    Double integral of a function on a region delimited by quadratic functions

    Hi. Because the solution had a picture, I wrote it on piece of paper! Sorry for my bad hand writing.

    Attached Thumbnails Attached Thumbnails Double integral of a function on a region delimited by quadratic functions-z.jpg  
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  3. #3
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    Talking Thanks for answering!

    Double integral of a function on a region delimited by quadratic functions-greg.png


    You answer was really helpful, thanks a lot! It helped me visualize the problem. I think I found a way to solve it but I'm not quite sure. I try to follow your solution but I don't know how you build the x^4 equation.

    Anyway, I came out with a solution. I'm not so sure about it but I'm posting it if anybody can tell me it's a good or bad reasoning.

    Thanks,

    Bazinga
    Last edited by bazingasmile; April 3rd 2011 at 11:16 AM.
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  4. #4
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    Double integral of a function on a region delimited by quadratic functions

    Well about building x^4 equation. It's simple! If we want to find the coordinates of "point" we should put these two equations equal to each other.
    y=4-x^2  and x=1+y^2  then we get x^4-8x^2-x+17=0  . I can solve this equation with maple, but I would like to solve it in a classic way! The answers are x=1.767469062 and x=2.263652586. But we only need the first one. I used x=1.77 for simplicity. Then y=0.87.

    My answer is "33.83" which is equal to yours, but I solved it on my own at first and then checked it with maple!

    To solve the integrals we should use this formula: \int udv= uv - \int vdu  .
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