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Integral homework problem is tough

  1. Mar 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Find [tex]\int \left( \int \frac{x^2-y^2}{(x^2+y^2)^2} \;dx \right) dy.[/tex]

    3. The attempt at a solution

    How would I go about finding [itex]\int \frac{x^2-y^2}{(x^2+y^2)^2} \;dx[/itex]?

    If I made the substitution [itex]u^2=x^2+y^2[/itex] how do I go from here?
     
  2. jcsd
  3. Mar 21, 2012 #2
    Re: Integral

    Use polar coordinates.
     
  4. Mar 21, 2012 #3
    Re: Integral

    So letting [tex]x=r\cos \theta[/tex][tex]y=r\sin\theta[/tex] what would be the limits of the [itex]r[/itex] integral and the [itex]\theta[/itex] integral if [itex]x\in [0,1][/itex] and [itex]y\in [0,1][/itex]?
     
  5. Mar 21, 2012 #4

    tiny-tim

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    Come on, Ted! :rolleyes:

    θ obviously goes from 0 to π/2

    and, for a fixed value of θ, r goes from 0 to … ? :smile:
     
  6. Mar 21, 2012 #5
    Re: Integral

    Well if [itex]0 \leq \theta \leq \frac{\pi}{2}[/itex] and [itex]0 \leq r \leq 1[/itex] then [tex]\int^1_0 \int^1_0 \frac{x^2-y^2}{(x^2+y^2)^2}dxdy = \int^{\frac{\pi}{2}}_0 \int^1_0 \frac{\cos(2\theta)}{r} drd\theta[/tex]

    but what do I do about [itex]\ln(0)[/itex]?
     
  7. Mar 21, 2012 #6

    tiny-tim

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    No!! It's a square!

    How can r go from 0 to 1 to make a square? :redface:
     
  8. Mar 21, 2012 #7
    Re: Integral

    If [itex]0 \leq r \leq \sqrt{2}[/itex] then I'm still faced with a [itex][\ln(r)]^{\sqrt{2}}_0[/itex]
     
  9. Mar 21, 2012 #8

    tiny-tim

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    Ted, to find the limit you have to fix θ, and see where r goes between for that value of θ.

    So your r limit will be a function of θ.

    (btw, in this particular case, you'll need separate functions for θ ≤ π/4 and θ > π/4 :wink:)
     
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