Solving Double Integrals with Polar Coordinates: What's the Correct Approach?

[Ted123]

Homework Statement

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

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?

 

[dirk_mec1]

Use polar coordinates.

 

[Ted123]

Use polar coordinates.

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 $y\in [0,1]$?

 

[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 … ?

 

[Ted123]

Come on, Ted!

θ obviously goes from 0 to π/2

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

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 $\ln(0)$?

 

[tiny-tim]

Well if $0 \leq \theta \leq \frac{\pi}{2}$ and $0 \leq r \leq 1$

No! It's a square!

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

 

[Ted123]

No! It's a square!

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

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

 

[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 )