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Jacobian Transformation.

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data

    Use the given transformation to evaluate the given integral.


    [tex]\int[/tex][tex]\int(x-3y)DA[/tex]
    R.

    where R is the triangular region with vertices (0,0), (2,1) and (1,2) ; x = 2u + v , y = u + 2v



    Trial :

    Using the points given I came up with these equations for the triangle lines :

    y = 2x;
    y = 1/2x;
    y = 3 - x;

    Now just substituting x = 2u and y = u+2v I get this :

    u = 0
    v = 0
    v = 3(1-u)

    plotting this , I see that 0<= u <= 1 and 0<= v <= 3(1-u)

    And from the original integral the integrand is x - 3y. Again substituting
    for x and y with the given transformation coordinate for u and v I get :

    (x-3y)DA = -(u+5v) dv*du

    Now the integral becomes :

    [tex]\int^{1}_{0} du[/tex][tex]\int^{3-3u}_{0} -u - 5v dvdu[/tex]

    In which I get the answer -4/5 which is not correct. Any help?
     
  2. jcsd
  3. Nov 29, 2009 #2

    lanedance

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    Homework Helper

    ok so the variable transformation sounds alright, but where is the Jacobian...? (as in the title)

    as do you the variable change, you are effectivley rotating & squeezes the axes, [tex] dA = dx dy \neq dudv [/tex]. So to preserve the original area you need to include the jacobian
     
    Last edited: Nov 29, 2009
  4. Nov 29, 2009 #3
    Since :

    x = 2u + v , y = u + 2v

    The jacobian is :

    |2 1|
    |1 2|
    = 2*2 - 1*1 = 3 ;

    The the answer is -4/5* 3 = -12/5 ?

    I still did something wrong I guess.
     
  5. Nov 29, 2009 #4
    the limit for v is 0<y<1-u because v=1-u not 3(1-u) and you also need to include the jacobian in your equation, but you can just multiply the final answer you get by (3) instead of doing all your integrals after multiplying with the jacobian. And after doing this question I got -3 as the final answer.
     
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