# Selecting the correct bounds for polar integrals

1. Nov 3, 2011

### blank_slate

Hi!

Here's a question I am working on:

Double integral of arctan(y/x).

where R: 1≤x2+y2≤4, 0≤y≤x.

I have the bounds for r as 1 to 2, but for θ I don't know if I should use ∏/4 to ∏/2 or 0 to ∏/2. How do I know which one?

The integration is easy, but I need help with the bounds.

Thanks.

2. Nov 4, 2011

### Zahar

My point of view :
[ 0 < y < x ] means that y and x are more then zero, [ y < x ] means that all points are below/under the line [ y = x ], line [ y = x ] in the polar coordinate system has an equation :: [ angle = pi / 4 ]. Then the bounds in polar system from 0 to pi/4.
arctg (y / x) = ( r*sin( ang ) / [ r*cos( ang ) ] ) = acrtg( tg (ang) ) = ang
Integral ( angle * R * ( d R ) * ( d angle ) ) = 3*pi*pi / 64