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Double integral parallelogram

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Let Ω ⊂ R^2 be the parallelogram with vertices at (1,0), (3,-1), (4,0) and (2,1). Evaluate ∫∫_Ω e^x dxdy.

    Hint: It may be helpful to transform the integral by a suitable (affine) linear change of variables.


    2. Relevant equations



    3. The attempt at a solution
    Ok here is what I have done:
    From the sketch of the parallelogram, I have found the limits to be x=y+1 to (4-2y) and y=-1 to 1. With this, I am able to determine the solution of the integral which is 3/2*e^2 -1 -1/2*e^6.

    Could anyone please verify this for me? Also, if my solution turns out to be right, how would I approach the hint to find the solution of this double integral? Thanks.
     
    Last edited: Feb 21, 2010
  2. jcsd
  3. Feb 21, 2010 #2

    LCKurtz

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    The sides of your parallogram are not parallel to the coordinate axes and you would have to break up the integrals into two parts; you can't go from y = -1 to 1 in one swoop. The sides of your parallelogram are of the form x - y = constants and x + 2y = constants. So if you want to transform your problem into one with constant limits you might want to consider the transformation:

    u = x - y
    v = x + 2y
     
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