Jacobian transformation problem

[oferon123]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

After integration i get 0.5*ln(v) 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:

 

[Millennial]

They make use of the oddness of all four functions provided to calculate only one of those areas and then double the result.

 

[oferon123]

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?

 

[vanhees71]

Note that for x&lt;0 you have
\int \mathrm{d}x \frac{1}{x}=\ln(-x)+\text{const}.

 

[Millennial]

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 f(-x) = -f(x). You can check that without drawing anything.

On your mistake, you seem to be missing a very simple fact: \displaystyle \int \frac{dx}{x} \neq \log(x) + C, but instead, \displaystyle \int \frac{dx}{x} = \log|x| + C.

 

[oferon123]

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!

 

[haruspex]

4 &lt; u &lt; 8 , -15 &lt; v &lt; -5 OR 5 &lt; v &lt; 15

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.

 

[oferon123]

I tried to play with u=xy , v=\frac{y}{x} 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?

 

[Millennial]

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 xy = c 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.