Jacobian Transformations

bodensee9
Can someone help me with the following?
I am supposed to evaluate
∫∫ e^(x+y)dA where the area of integration is given by the inequality |x|+|y|≤1.
So, suppose I do one of these Jacobians, and I set u = |x| and v = |y|, so wouldn’t the equation have to satisfy the inequality u+v≤1, and u≥0, v≥0? So, would wouldn’t the Jacobian be 1? But clearly I’m doing something wrong here, so any hints would be greatly appreciated!! Thanks!!

Homework Helper
The function in a change of variable should one-to-one (as your example shows - it simply omits regions of the integration domain). Differentiability is also handy. Absolute value is neither.

bodensee9
I'm not sure if I understand what you mean. Can you clarify more or point out where I'm going wrong with this? Thanks.

Homework Helper
I thought I did? You can SEE where it's going wrong, right? The formula for change of variables involving Jacobians has premises. You can't blindly use any old functions. |-1|=|1|=1. It's not one-to-one. Hence you can't blindly use it in change of variable. Jacobian isn't even defined at (0,0).

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Dearly Missed
Note that absolute values are non-negative, so for example, |x|<=1, that is $-1\leq{x}\leq{1}$

An equivalent inequality to the one you have been given is:
$$|x|-1\leq{y}\leq{1}-|x|$$
In the region $x\geq{0}$, this translates to:
$$x-1\leq{y}\leq{1-x}, 0\leq{x}\leq{1}$$
Make a similar translation for negative x's!

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bodensee9
Thanks, that is very helpful. So, can I just ask, since |x|+|y| ≤1, so this means that |x|≤1-|y|. |y| - 1 ≤ |x| ≤ 1-|y|. So this means that y – 1 ≤ x ≤ 1-y, where 0 ≤ y ≤ 1. But I don’t see why I can’t just use those parameters to integrate? So, could I do ∫∫ e^(x+y)dxdy where y – 1 ≤ x ≤ 1-y and where 0 ≤ y ≤ 1?

Homework Helper
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Of course you could use y rather than x as the "outer" variable.

Homework Helper
Yes. That's exactly what you should do.

Homework Helper
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Yes. That's exactly what you should do.

Eeh?