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Jacobian transformation problem

  1. Jul 9, 2013 #1
    1. The problem statement, all variables and given/known data
    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]

    2. Relevant 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]

    3. 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:
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jul 9, 2013 #2
    They make use of the oddness of all four functions provided to calculate only one of those areas and then double the result.
  4. Jul 9, 2013 #3
    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?
  5. Jul 9, 2013 #4


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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]
  6. Jul 9, 2013 #5
    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].
  7. Jul 9, 2013 #6
    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!
  8. Jul 9, 2013 #7


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    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.
  9. Jul 10, 2013 #8
    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?
  10. Jul 10, 2013 #9
    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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