# Mass in a plane region

1. Dec 24, 2011

### namu

1. The problem statement, all variables and given/known data

Find the mass of the plane region R in the first quadrant of the (x,y)-plane bounded by the hyperbolas

$xy=1 \,\,\,\,\,\,\,\,\,\, xy=2\,\,\,\,\,\,\,\,\,\, x^2-y^2=3\,\,\,\,\,\,\,\,\,\, x^2-y^2=5$

Assume the density at the point (x,y) is $\rho=x^2+y^2$

2. Relevant equations

$$m=\int \int_R \rho(x,y)dxdy$$

3. The attempt at a solution

I am stuck at finding a suitable change of variables to transform this into a "nice" region so I don't have to perform 3 seperate integrals. Even if I took the long way (3 integrals) the point of intersection is not easy to find analytically. What is a clever change of variables that I can use?

I have tried the following:

$u=xy \,\,\,\,\,\,\,\,\,\, v=x^2-y^2$

then I can't find a nice expression for $\rho(u,v)$

I also tried

$x=u/v \,\,\,\,\,\,\,\,\,\, y=v$

but then solving for v is ugly.

I even tried

$u=x^2 \,\,\,\,\,\,\,\,\,\, v=y^2$

which gave another ugly region.

Last edited: Dec 24, 2011
2. Dec 24, 2011

### vela

Staff Emeritus
Your first set of transformations is the one you want. Now consider $4 u^2+v^2$.

3. Dec 24, 2011

### namu

Im still lost. I want $\rho=x^2+y^2$.
$4u^2=x^2+y^2$
$v^2=x^4-2x^2y^2+y^4$

I'm still getting nowhere.

4. Dec 25, 2011

### vela

Staff Emeritus
How did you get 4u2=x2+y2 from u=xy?

5. Dec 25, 2011

### SammyS

Staff Emeritus
Isn't $4u^2=4(xy)^2=4x^2y^2\,?$

Now, add that to $x^4-2x^2y^2+y^4\,?$

Factor that !

6. Dec 25, 2011

### namu

Sorry, I made a typo since I was copy pasting

$4u^2=4x^2 y^2$

7. Dec 25, 2011

### namu

Ok, round 2, here it goes.

$4u^2=4x^2 y^2$
$v^2=x^4-2x^2y^2+y^4$
$4u^2+v^2=x^4+2x^2 y^2+y^4=(x^2+y^2)^2=\rho^2$

Thank you!