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Limits of integration

  1. Apr 1, 2009 #1
    Find the double integral of:

    (x+y)x dxdy

    Where R is a quadrilateral with vertices at (-4,-1), (-2,-2) (-1,1) and (-3,2)

    **I have done the diagram and i know that there will be two regions R1 and R2 but i am not sure exactly how to find the limits of int. for these two regions, any suggestions?

  2. jcsd
  3. Apr 1, 2009 #2


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    i drew this quickly, but looks like a rotated square to me, so you could think about a linear variable change to make the limits of integration independent & simple
  4. Apr 2, 2009 #3
    How about taking the gradient of the 2 sides of the square? i think that might work. i will give that a try
  5. Apr 2, 2009 #4


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    i'm not totally sure what you mean...

    I was thinking along the lines of new variables u,v, whose level curves are lines parallel to the edges of the squares
  6. Apr 2, 2009 #5


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    Neither a square nor a rectangle- it is parallelogram. Two sides are of length [itex]\sqrt{5}[/itex] and the other two of length [itex]\sqrt{10}[/itex]- and the sides are not perpendicular. However, it can be done as lanedance suggests: The lines through (-2,-2) and (-1,1) and through (-4,-1) and (-3,2) both have slope (1+2)/(-1+2)= 3. The first line is y= 3x+ 4 and the second y= 3x+ 7. If you let u= y- 3x, then the first line is u= -4 and the second u= -7. The lines through (-1, 1) and (-3, 2) and through (-2, -2) and (-4, -1) both have slope (2-1)/(-3+1)= -1/2. The first line is y= -(1/2)x+ 1/2 and the second is y= (-1/2)x- 1. If you let v= y+ (1/2) x, then the first line is v= 1/2 and the second is v= 1. Make that change of variables and don't forget to change the dxdy properly.
  7. Apr 2, 2009 #6
    So basically i am solving it for dudv by making that change of variable. sweet
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