# 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:

Related Calculus and Beyond Homework Help News on Phys.org
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
2019 Award
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
$4 < u < 8$ , $-15 < v < -5$ 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.