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Change of Variables

  1. Mar 25, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose D is the parallelogram in the xy-plane with vertices P(-1,5), Q(1,-5), R(5,-1), S(3,9)
    [tex] \int\int ^{}_{D} (6x+12y) dA [/tex]
    HINT: Use transformation x = [tex]\frac{1}{6}[/tex](u+v) and y = [tex]\frac{1}{6}[/tex] (-5u+v).

    2. Relevant equations



    3. The attempt at a solution
    Calculating the Jacobian I get
    dx/du = 1/6 dx/dv = 1/6
    dy/du = -5/6 dy/dv = 1/6
    which gives me
    1/36 - (-5/36) = 1/6
    From there I need the new intervals of integration.
    P-Q
    [tex]\frac{5-(-5)}{-1-(1)}[/tex] = 10/-2 = -5
    y = 5x + 5 -> plugging in . . .
    -5/6u+1/6v = -5/6u - 5/6v + 5 -> u cancels out and then we get 6/6v = 5 which means v = 5
    S-R
    [tex]\frac{9-(-1)}{3-5}[/tex] = -5
    y = -5x + 9
    -5/6u+1/6v = -5/6u+5/6v + 9 -> v = 9
    R-Q
    [tex]\frac{(-1)-(-5)}{5-1}[/tex] = 1
    y = x-1 -> -5/6u + 1/6v = 1/6u +1/6v -1 -> u = 1
    S-P
    [tex]\frac{(9-(5)}{3-(-1)}[/tex] = 1
    y = x-9 -> -5/6u +1/6v = 1/6u + 1/6v - 9 -> u = 9

    Plugging in the x and y values and the Jacobian into the integral we get the following new integral.
    [tex] \int^{9}_{5}\int^{9}_{1} (u+v + (-10u+2v)) * 1/6 dudv [/tex]
    which simplifies to . . .
    [tex] \int^{9}_{5}\int^{9}_{1} (-4/3u+1/2v) dudv [/tex]
    After the first integration I get . . .
    [tex] \int^{9}_{5} (-2/3u^2+1/2v*u) dudv [/tex]
    Plugging in 9 and 1 I get
    -54-9/2v+50/3-5/2v = -37.33333 - 7v
    Integrating what I got I get
    -37.33333v - 7/2v2
    And after plugging in 9 and 5 I get -345.333
    which is wrong. So what am I doing wrong?
     
  2. jcsd
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