# Cannot finish calculating a double integral with change of coordinates

1. Oct 29, 2012

### Hernaner28

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

$$\displaystyle f\left( x,y \right)=\frac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}$$

on the region:
$$\displaystyle D=\left\{ \left( x,y \right)\in {{\mathbb{R}}^{2}}:0\le x\le 1,{{x}^{2}}\le y\le 2-{{x}^{2}} \right\}$$

TIP: Use change of coordinates:

$$\displaystyle x=\sqrt{v-u}$$
$$\displaystyle y=v+u$$

2. Relevant equations

3. The attempt at a solution

Alright, what I did was to sketch the two regions. After that I calculated the jacobian of the change of coordinates function:

$$\displaystyle g\left( u,v \right)=\left( \sqrt{v-u},v+u \right)$$

$$\displaystyle \left| Jg \right|=\frac{1}{\sqrt{v-u}}$$

So the new integral becomes

$$\displaystyle \int\limits_{0}^{1}{\int\limits_{u}^{1}{\frac{v-u}{v-u+{{\left( u+v \right)}^{2}}}\cdot \frac{1}{\sqrt{v-u}}dv}du}=\int\limits_{0}^{1}{\int\limits_{u}^{1}{\frac{\sqrt{v-u}}{v-u+{{\left( u+v \right)}^{2}}}dv}du}$$

And here I'm stuck. I don't know how to continue. I don't like the integrand. Any help?

Thanks!

2. Oct 30, 2012

### SammyS

Staff Emeritus
Yes, that integrand is a mess.

Your work all looks to be correct.

Are you sure your function is correct? $\displaystyle \ f\left( x,y \right)=\frac{{{x}^{2}}}{{{x}^{2}}+{{y}}}$ would work out much more nicely.

3. Oct 30, 2012

### Hernaner28

Yes, I'm sure. It's $$\displaystyle f\left( x,y \right)=\frac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}$$ and they suggest that change of coordinates.

4. Oct 30, 2012

### jackmell

Suppose now you were given just that integral over that triangle you're integrating over. Can you make a change of variable to make it easier to integrate? What about now just:

$$w=v-u$$
$$z=u+v$$

No guarantees though ok. You gotta' just try it to get use to doing that if you're going to get good in math. :)

Last edited: Oct 30, 2012