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Jacobian transformation problem

  • Thread starter oferon123
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  • #1
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Homework Statement


Find surface inside four boundary curves:
[itex]xy = 4 , xy=8 , y=5x , y=15x[/itex]
using the transformation: [itex]u=xy , v=\frac{y}{x}[/itex]

Homework Equations


I'm getting the new bounds to be:
[itex]4 < u < 8[/itex] , [itex] -15 < v < -5 [/itex] OR [itex] 5 < v < 15[/itex]
Jacobian is [itex]\frac{1}{2v}[/itex]


The Attempt at a Solution



After integration i get [itex]0.5*ln(v)[/itex] which I can't put -15 nor -5 in it..
On the solution they gave us, they completely ignore the surface at x,y < 0 and go only for 5 < v < 15 , which I assume is only relevant for the surface at x,y > 0...
As if they calculated only the half of the area.
I understand all I need to do is double this area, but where am I going wrong?
I attach this sketch i've drawn trying to understand what's going on:
[Broken]
 
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Answers and Replies

  • #2
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They make use of the oddness of all four functions provided to calculate only one of those areas and then double the result.
 
  • #3
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Yes, I do understand that.
However I don't understand how your'e supposed to see that unless you draw everything.
More important - even if I do draw it, I don't understand why I get stuck with the method and can't integrate over -15 < v < -5
I obviously do something wrong if I can't, the question is what

Maybe it's got something to do with the transformation not being injective?
 
  • #4
vanhees71
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Note that for [itex]x<0[/itex] you have
[tex]\int \mathrm{d}x \frac{1}{x}=\ln(-x)+\text{const}.[/tex]
 
  • #5
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Honestly, it is not too hard to see that those functions are odd - I spotted it before even looking at your sketch of the functions. A function is odd if it satisfies [itex]f(-x) = -f(x)[/itex]. You can check that without drawing anything.

On your mistake, you seem to be missing a very simple fact: [itex]\displaystyle \int \frac{dx}{x} \neq \log(x) + C[/itex], but instead, [itex]\displaystyle \int \frac{dx}{x} = \log|x| + C[/itex].
 
  • #6
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Oh your'e right, thanks alot!

But now I have this new question I asked..
If this transformation is not injective, how come it works?
for example: for x=-1, y=-1 and x=1 y=1 we get the same v=1

Thanks again!
 
  • #7
haruspex
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[itex]4 < u < 8[/itex] , [itex] -15 < v < -5 [/itex] OR [itex] 5 < v < 15[/itex]
I don't understand how you get -15 < v < -5. In both regions, x and y have the same sign, so u and v are always positive.
 
  • #8
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I tried to play with [itex]u=xy , v=\frac{y}{x}[/itex] a little and that's what i got.
If I'm wrong then where is the negative region in xy is transformed to in uv region?
The way I drew it - one goes to the upper square at uv, and the other goes to lower square.. Is it not?
 
  • #9
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What haruspex is saying is that you can't get negative values for u and v in the regions you want to find the area for. Negative values would show up in the second and the fourth quadrants, but not in the first and the third; and you don't have the curves [itex]xy = c[/itex] for c constant in those quadrants. It will still give the correct answer though, because it just happens to integrate into the same thing; albeit out of luck.
 

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