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Computing a double integral with given vertices

  1. Dec 6, 2009 #1
    1. The problem statement, all variables and given/known data[/b]

    Use the transformation that takes the unit square to a triangle to compute the integral

    [tex]\int\int_{B}2x+3y dA[/tex]

    Where B is a triangular region with vertices (0,0), (5,2), and (3,4).


    3. The attempt at a solution

    What I did was I drew the region on an xy plane, I split the triangle up into two triangles and found my limits of integration by drawing the lines made by connecting the vertices. Because I split the triangle up into two, I needed to add two seperate double integrals.

    This is what I got.

    [tex]\int_{x=0}^{x=3}\int_{y=5x/2}^{y=4x/2}(2x+3y) dydx + \int_{x=3}^{x=5}\int_{y=5x/2}^{y=-x+7}(2x+3y) dydx[/tex]

    With the first integral I got

    [tex]\dfrac{-651}{8}[/tex]

    for the second integral i got

    [tex]\dfrac{-4739}{12}[/tex]

    I figured that if I add both of them together I would get the volume underneath that region (the whole thing), but the number I got, [tex]\frac{11431}{24}[/tex] seems to large.

    I also don't think I'm doing it the method wanted.
    Can anyone please direct me where to go from here, i'm somewhat lost.

    Thank you
     
  2. jcsd
  3. Dec 6, 2009 #2

    LCKurtz

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    I didn't check your work because, as you suspect, that isn't the method you have been asked to use. Your question refers to "the" transformation that takes the unit square to a triangle. I'm guessing your text has given you that example. Look it up and share the equations of that transformation with us and then we can talk.:smile:
     
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