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Double Integral

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int_{D}\int y^2[/tex]
    where D = {(x,y) | -1 [tex]\leq[/tex] y [tex]\leq1[/tex], -y-2[tex]\leq x\leq y[/tex]

    The integral I set up is below :

    [tex]\int^{1}_{-1} \int^{y}_{-y-2} y^2 dx dy [/tex]

    From that I get the answer 0, but the book says its 4/3.

    I get 0 because It reduces to this integral :

    [tex]\int^{-1}_{1} 2y^3 + 2y[/tex]

    Any idea where I could be wrong?
  2. jcsd
  3. Nov 7, 2009 #2
    In your last step you reversed your bounds, it should be (-1,1) not (1,-1) as you wrote. Also the final step should be [tex]\int[/tex]2y3 +2y2 dy with the bounds (-1 to 1).
    Last edited: Nov 7, 2009
  4. Nov 7, 2009 #3


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    Hi tnutty! :smile:
    erm :redface:

    2y3 + 2y2 ? :wink:
  5. Nov 7, 2009 #4
    I mean to write 2y^3 + 2y^2. But its was the bounds. Thanks guys.
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