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Triple Integral: Having trouble finding my y bounds

  1. Nov 2, 2008 #1
    1. The problem statement, all variables and given/known data

    [tex]I=\int\int\int_E x^2e^ydV[/tex] where E is bounded by the parabolic cylinder

    z=1-y^2 and the planes z=0 x=1 and x=-1

    I know that the graph is a parabola that opens downwards and that has symmetry wrt the x-axis. It also stretches along the x axis toward + and - infinity.

    Can I get a hint here? :)
     
  2. jcsd
  3. Nov 2, 2008 #2

    Avodyne

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    Science Advisor

    Did you mean
    [tex]I=\int\int\int_E x^2e^y\,dV[/tex] ?
     
  4. Nov 3, 2008 #3

    HallsofIvy

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    It's pretty close to being trivial. The parabola z= 1- y2 has z= 0 at y= -1 and y= 1 so projecting down on to xy=plane, we get the square -1< x< 1, -1< y< 1. The z- integral is taken from 0 to 1- y2, the y integral from -1 to 1, and the x integral from -1 to 1.
     
  5. Nov 4, 2008 #4
    I don't see any difference. So yes. :smile:
     
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