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Substitution in double integral

  1. Jul 9, 2008 #1
    1. The problem statement, all variables and given/known data
    Use the transformation u = 3x + 2y and v = x + 4y to evaluate:

    The double integral of (3x^2 + 14xy + 8y^2) dx dy for the region R in the first quadrant bounded by the lines y = -(3/2)x + 1, y = -(3/2)x + 3, y = -(1/4)x, and y = (-1/4)x + 1.

    2. Relevant equations

    3. The attempt at a solution
    I first wrote the equations for x and y in terms of u and v and got:
    x = (2u - v)/5, y = (3v - u)/10

    Then I solved for the Jacobian and got 1/10. Next I saw that the region R has the following boundaries:
    x = 2/3 - (2/3)y
    x = 2 - (2/3)y
    y = 0

    Plugging in the equations I got earlier for x and y, I get the equivalent in terms of u and v:
    u = 2
    u = 6
    v = u/3

    So I set up my integral now like this:

    Double integral of (u * v) * (1/10) dv du, with 2 <= u <= 6, and 0 <= v <= u/3. Solving this I get the answer 16/9, while the correct answer is 64/5. Which part(s) did I do wrong here? thanks
  2. jcsd
  3. Jul 10, 2008 #2


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    Homework Helper

    I didn't check through all your working but it looks like you didn't express x and y in terms of u,v properly.
  4. Jul 10, 2008 #3
    You don't have to express x and y in terms of u and v.

    You can calculate the reciprocal of the 'inverse jacobian' (don't know if it's called that):

    [tex]\left( \frac{ \partial(u,v)}{\partial(x,y)} \right)^{-1} = \frac{ \partial(x,y)}{\partial(u,v)}[/tex]

    So if you got an equation for u, and one for v, you can calculate [itex]\frac{ \partial(u,v)}{\partial(x,y)}[/itex] and then take it's reciprocal to find the jacobian [itex] \frac{ \partial(x,y)}{\partial(u,v)}[/itex]
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