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Jacobian Transformation - new domain of integration

  • #1
190
0

Homework Statement



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]

Homework Equations





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.
 

Answers and Replies

  • #2
834
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)
\end{cases}[/tex]

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)
\end{cases}[/tex]

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].
 
  • #3
428
1
omg, TLDR Muphrid!

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

Homework Equations





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.
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].
 
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
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1,268

Homework Statement



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]

Homework Equations





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.
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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