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Jacobian Transformations

  • Thread starter bodensee9
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  • #1
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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!!
 

Answers and Replies

  • #2
Dick
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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.
 
  • #3
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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.
 
  • #4
Dick
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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).
 
  • #5
arildno
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Note that absolute values are non-negative, so for example, |x|<=1, that is [itex]-1\leq{x}\leq{1}[/itex]

An equivalent inequality to the one you have been given is:
[tex]|x|-1\leq{y}\leq{1}-|x|[/tex]
In the region [itex]x\geq{0}[/itex], this translates to:
[tex]x-1\leq{y}\leq{1-x}, 0\leq{x}\leq{1}[/tex]
Make a similar translation for negative x's!
 
Last edited:
  • #6
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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?
 
  • #7
arildno
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Of course you could use y rather than x as the "outer" variable.
 
  • #8
Dick
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Yes. That's exactly what you should do.
 
  • #9
arildno
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Yes. That's exactly what you should do.
Eeh?
What's wrong about using x as the outer variable instead?
 
  • #10
Dick
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Eeh?
What's wrong about using x as the outer variable instead?
Nothing! Either is just fine. I was replying to bodensee.
 
  • #11
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Thanks very much for all your help!
 

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