Jacobian Transformation - new domain of integration

1. Jun 6, 2012

IniquiTrance

1. The problem statement, all variables and given/known data

I need to compute the following using the Jacobian:

$\int\int_D \frac{x-y}{x+y} dxdy$

Where $D = \left\{(x,y):x\geq 0, y\geq 0, x+y \leq 1\right\}$

2. Relevant equations

3. The attempt at a solution

$s=x+y \qquad t = x-y$

My problem is finding the new domain of integration. I can see that $0\leq s \leq 1$.

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

Any help is much appreciated.

2. Jun 6, 2012

Muphrid

Let $r = x e_1 + y e_2$ and $r' = s e_1 + t e_2$ with $f(r) = r'$ and the coordinates s,t as you defined them.

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

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

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

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

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

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 $s \geq t \geq -s$.

3. Jun 6, 2012

algebrat

omg, TLDR Muphrid!

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

4. Jun 7, 2012

vela

Staff Emeritus
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.