Jacobian Transformation - new domain of integration

[IniquiTrance]

Homework Statement

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\}

Homework Equations

The Attempt at a Solution

I've made the transformation:

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.

 

[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}<br /> \tau e_1 &amp; (0 &lt; \tau \leq 1) \\<br /> e_1 + (e_2 - e_1)(\tau - 1) &amp; (1 &lt; \tau \leq 2) \\<br /> e_2 + (-e_2)(\tau - 2) &amp; (2 &lt; \tau \leq 3)<br /> \end{cases}

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

The boundary of the domain of integration goes as c(\tau) \to c&#039;(\tau) = \underline f[c(\tau)]. The result of applying the Jacobian operator to c above is

c&#039;(\tau) = \begin{cases}<br /> (e_1 + e_2)\tau &amp; (0 &lt; \tau \leq 1) \\<br /> (e_1+e_2) - 2 e_2 (\tau - 1) &amp; (1 &lt; \tau \leq 2) \\<br /> (e_1 - e_2) + (e_2 - e_1)(\tau - 2) &amp; (2 &lt; \tau \leq 3)<br /> \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.

 

[algebrat]

omg, TLDR Muphrid!

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

Homework Equations

The Attempt at a Solution

I've made the transformation:

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.

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.

 

[vela]

Homework Statement

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\}

Homework Equations

The Attempt at a Solution

I've made the transformation:

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.

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.