Double integral over a region needing polar coordinates.

[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 that's incorrect.
Any help is greatly appreciated, thanks.

 

[tiny-tim]

Hi Tropicalism! Welcome to PF!

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 that's incorrect.

"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

 

[Tropicalism]

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

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

 

[tiny-tim]

yup!

 

[Tropicalism]

yup!

Thanks! Thats all I needed.