# Integral homework problem is tough

1. Mar 21, 2012

### Ted123

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

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

3. The attempt at a solution

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

If I made the substitution $u^2=x^2+y^2$ how do I go from here?

2. Mar 21, 2012

### dirk_mec1

Re: Integral

Use polar coordinates.

3. Mar 21, 2012

### Ted123

Re: Integral

So letting $$x=r\cos \theta$$$$y=r\sin\theta$$ what would be the limits of the $r$ integral and the $\theta$ integral if $x\in [0,1]$ and $y\in [0,1]$?

4. Mar 21, 2012

### tiny-tim

Come on, Ted!

θ obviously goes from 0 to π/2

and, for a fixed value of θ, r goes from 0 to … ?

5. Mar 21, 2012

### Ted123

Re: Integral

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

but what do I do about $\ln(0)$?

6. Mar 21, 2012

### tiny-tim

No!! It's a square!

How can r go from 0 to 1 to make a square?

7. Mar 21, 2012

### Ted123

Re: Integral

If $0 \leq r \leq \sqrt{2}$ then I'm still faced with a $[\ln(r)]^{\sqrt{2}}_0$

8. Mar 21, 2012

### tiny-tim

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 )

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