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Double Integral

  1. Feb 27, 2010 #1
    [tex]\int\int(yx^2-2xy^2)dydx[/tex]

    limits for the first are 0 [tex]\longrightarrow[/tex] 3

    limits for the second are -2 [tex]\longrightarrow[/tex] 0

    solve! help
     
  2. jcsd
  3. Feb 27, 2010 #2

    tiny-tim

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    Hi der.physika! :smile:

    (type "\int_0^3\int_{-2}^0" :wink:)

    Just split it into two integrals, ∫∫ yx2 dydx and ∫∫ 2xy2 dydx …

    what do you get? :smile:
     
  4. Feb 27, 2010 #3
    Okay so I took your advice and split the integral and I got 6, is that the correct answer?
     
  5. Feb 27, 2010 #4

    tiny-tim

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    erm :redface: … i can't check your answer without seeing your calculations, can I? :wink:
     
  6. Feb 27, 2010 #5
    [tex]\int\int(yx^2dydx)-2\int\int(xy^2dydx)[/tex]

    [tex]\int[\frac{1}{2}y^2x^2]=\int(-2x^2dx)=[\frac{-2}{3}x^3]=\frac{-54}{3}[/tex]

    [tex]-2\int\int(xy^2dydx)=-2\int[\frac{1}{3}y^3x]=-2(-12)=24=\frac{72}{3}[/tex]

    [tex]=\frac{72}{3}+\frac{-54}{3}=\frac{18}{3}=6[/tex]

    is this okay?
     
    Last edited: Feb 27, 2010
  7. Feb 27, 2010 #6

    tiny-tim

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    Sorry, I'm totally confused. :confused:

    Where are your half-way integrals, ie after just one integration?

    (and btw, which integral is going from 0 to 3, ∫ dx or ∫ dy?)
     
  8. Feb 27, 2010 #7
    How do you put in limits on the integral? I don't know how to put the code into LaTex
     
  9. Feb 27, 2010 #8

    tiny-tim

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    I showed you above … type "\int_0^3" and "\int_{-2}^0"

    (you need {} if the limit has more than one character, eg -2)
     
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