# Double integral over a region needing polar coordinates.

1. Nov 6, 2013

### Tropicalism

1. Evaluate the double integral ∫∫arctan(y/x) dA by converting to polar coordinates over the Region R= { (x,y) | 1≤x^2+y^2≤4 , 0≤y≤x }

My attempt at solving

Converting to polar using x=rcosθ and y=rsinθ I get
∫∫arctan(tan(θ))r drdθ

I understand that I have to integrate first with respect to r and then with respect to θ but I'm not sure what the region is supposed to look like over which I am integrating so I don't know what my bounds should be. I understand that 1≤x^2+y^2≤4 is basically 2 circles, one with radius 1, and one with radius 2 and my region is between them. I don't know how 0≤y≤x affects it though, I thought it would just make it the top right portion of the circle but thats incorrect.
Any help is greatly appreciated, thanks.

Last edited: Nov 6, 2013
2. Nov 6, 2013

### tiny-tim

Hi Tropicalism! Welcome to PF!
"0≤y and 0≤x" would be top right portion (the first quadrant )

"0≤y≤x" is the half of it under the diagonal x=y

3. Nov 6, 2013

### Tropicalism

Oh, ok thank you. So using polar that portion would end at θ=∏/4 correct?

4. Nov 6, 2013

yup!

5. Nov 6, 2013

### Tropicalism

Thanks! Thats all I needed.