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Jacobian Transformation - new domain of integration

  1. Jun 6, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to compute the following using the Jacobian:

    [itex]\int\int_D \frac{x-y}{x+y} dxdy[/itex]

    Where [itex] D = \left\{(x,y):x\geq 0, y\geq 0, x+y \leq 1\right\}[/itex]

    2. Relevant equations

    3. The attempt at a solution

    I've made the transformation:

    [itex] s=x+y \qquad t = x-y [/itex]

    My problem is finding the new domain of integration. I can see that [itex]0\leq s \leq 1[/itex].

    Yet for t, my book says it should be [itex]-s\leq t \leq s[/itex], but I cannot see why this should be.

    Any help is much appreciated.
  2. jcsd
  3. Jun 6, 2012 #2
    Let [itex]r = x e_1 + y e_2[/itex] and [itex]r' = s e_1 + t e_2[/itex] with [itex]f(r) = r'[/itex] and the coordinates s,t as you defined them.

    Now, let the boundary of the domain of integration be [itex]c(\tau)[/itex] for some scalar variable tau. There's more than one way to do this, but you can choose

    [tex]c(\tau) = \begin{cases}
    \tau e_1 & (0 < \tau \leq 1) \\
    e_1 + (e_2 - e_1)(\tau - 1) & (1 < \tau \leq 2) \\
    e_2 + (-e_2)(\tau - 2) & (2 < \tau \leq 3)

    Now, the Jacobian is a linear operator on a vector:
    [tex]\underline f(e_1) = e_1 + e_2 \\
    \underline f(e_2) = e_1 - e_2[/tex]

    The boundary of the domain of integration goes as [itex]c(\tau) \to c'(\tau) = \underline f[c(\tau)][/itex]. The result of applying the Jacobian operator to c above is

    [tex]c'(\tau) = \begin{cases}
    (e_1 + e_2)\tau & (0 < \tau \leq 1) \\
    (e_1+e_2) - 2 e_2 (\tau - 1) & (1 < \tau \leq 2) \\
    (e_1 - e_2) + (e_2 - e_1)(\tau - 2) & (2 < \tau \leq 3)

    Drawn up, this is a diagonal from the origin to (s,t) = (1,1), down to (1,-1), and then back to the origin. That's why they say [itex]s \geq t \geq -s[/itex].
  4. Jun 6, 2012 #3
    omg, TLDR Muphrid!

    First, find where the boundaries are mapped to, you have [itex]x=0[/itex], [itex]y=0[/itex] and [itex]x+y=1[/itex]. These are mapped to [itex]x=(s+t)/2=0[/itex], [itex]y=(s-t)/2=0[/itex], and [itex]x+y=s=1[/itex]. Draw these lines as the boundary in the new [itex]s,t[/itex]-space. (Two diagonals through the origin and the vertical line [itex]s=1[/itex].) If you want to do [itex]∫∫f\ dtds[/itex], then for each [itex]s[/itex] value in [itex][0,1][/itex] (the projection of the region), [itex]t[/itex] should go from the lower diagonal to the upper diagonal, that is, from [itex]t=-s[/itex] to [itex]t=s[/itex].
  5. Jun 7, 2012 #4


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    First, you should sketch the region in the xy-plane. Next, draw in what curves of constant s and constant t look like on the xy-plane. For example, s=0 corresponds to the line x+y=0; that is, the t-axis is the line y=-x. Just from the sketch, it should be pretty clear what the limits are.
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