Solve $\iint_{D}\frac{1}{\sqrt{x^{2}+y^{2}}}dxdy$, where $D=\lbrace (x,y)\in\mathbb{R}^{2}\mid x>0, \ y<0, \ x^{2}+y^{2}≥\frac{1}{4}, \ x<1+y\rbrace$
I rapresented D and I know that polar coordinates don't work. I know that I need a substitution but I don't find the right one. I though about something like $u(x,y)=x^{2}+y^{2}, \ v(x,y)=x-y$ but it seems it's not work. I am trying to solve this for hours. Can someone give me a hint? Thanks before!
Here's your hint: try polar coordinates again.
Your region is everything in the fourth quadrant that's both red and blue. You can get your bounds for $\theta$ just based on the fact that you're in Quadrant IV. As for finding bounds on $r$: the lower bound should be easy. If you need a hint for the upper bound on $r$, here it is:
Hopefully that's enough to help you solve it!