# Jacobian transformation problem

## 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:
[Broken]

Last edited by a moderator:

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

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
Gold Member
Note that for $x<0$ you have
$$\int \mathrm{d}x \frac{1}{x}=\ln(-x)+\text{const}.$$

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, $\displaystyle \int \frac{dx}{x} = \log|x| + C$.

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
Homework Helper
Gold Member
2020 Award
$4 < u < 8$ , $-15 < v < -5$ OR $5 < v < 15$
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.

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?

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.