Cannot finish calculating a double integral with change of coordinates

[Hernaner28]

Homework Statement

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

Homework Equations

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!

 

[SammyS]

Homework Statement

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}\,,\quad\quad y=v+u

Homework Equations

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!

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.

 

[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.

 

[jackmell]

So the new integral becomes

\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!

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. :)