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 \thetay=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 )