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What am I doing wrong with this integral?

  1. Apr 14, 2009 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    [tex]\int_{-1}^3 \int_0^{3y} (x^2+y^2) dxdy[/tex]

    3. The attempt at a solution

    I've gotten [tex] \int_{-1}^3 \frac{x^3}{3}[/tex] = [tex] \int_{-1}^{3}y^3 dy[/tex] for the first part, but this can't be right?
     
  2. jcsd
  3. Apr 14, 2009 #2
    How are you getting that equals sign in there?
    Aren't you solving:
    [tex]\int_{-1}^3 \int_0^{3y} (x^2+y^2) dxdy[/tex] = ?

    Try splitting it up to help you see it more easily, maybe:
    [tex]\int_{-1}^3 \int_0^{3y}x^2dxdy+\int_{-1}^3 \int_0^{3y}y^2dxdy[/tex] = ?
     
  4. Apr 14, 2009 #3
    For the first integral I get:

    [tex] \int_{-1}^3 y^3 dy[/tex]
    Is that right so far?
     
  5. Apr 14, 2009 #4
    Well, no... because you know:

    [tex]\int_{-1}^3 \int_0^{3y}x^2dxdy=\int_{-1}^3 \frac{(3y)^3}{3}dy[/tex]
     
  6. Apr 14, 2009 #5
    Oh. Oops. :)
    Thanks
     
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