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Cannot finish calculating a double integral with change of coordinates

  1. Oct 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Integrate:

    [tex] \displaystyle f\left( x,y \right)=\frac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}[/tex]

    on the region:
    [tex] \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\}[/tex]

    TIP: Use change of coordinates:

    [tex] \displaystyle x=\sqrt{v-u}[/tex]
    [tex] \displaystyle y=v+u[/tex]

    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:

    [tex] \displaystyle g\left( u,v \right)=\left( \sqrt{v-u},v+u \right)[/tex]

    [tex] \displaystyle \left| Jg \right|=\frac{1}{\sqrt{v-u}}[/tex]

    So the new integral becomes

    [tex] \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}[/tex]

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

    Thanks!
     
  2. jcsd
  3. Oct 30, 2012 #2

    SammyS

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    Yes, that integrand is a mess.

    Your work all looks to be correct.

    Are you sure your function is correct? [itex] \displaystyle \ f\left( x,y \right)=\frac{{{x}^{2}}}{{{x}^{2}}+{{y}}}[/itex] would work out much more nicely.
     
  4. Oct 30, 2012 #3
    Yes, I'm sure. It's [tex] \displaystyle f\left( x,y \right)=\frac{{{x}^{2}}}{{{x}^{2}}+{{y}^{2}}}[/tex] and they suggest that change of coordinates.
     
  5. Oct 30, 2012 #4
    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:

    [tex]w=v-u[/tex]
    [tex]z=u+v[/tex]

    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
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