Integrating $ \int_0^2 \int_0^ \sqrt{1-(x-1)^2} \frac{x+y}{x^2+y^2} dy\,dx$ in polar coordinates I'm having a problem integrating $ \displaystyle\int_0^2  \int_0^ \sqrt{1-(x-1)^2} \frac{x+y}{x^2+y^2} \,dy\,dx$. I drew the graph, and it looks like half a circle on top of the $x$ axis.
I tried dividing it into two parts, with Area 1: $0\leq \theta\leq\frac{ \pi }{4} $ with $\csc\theta\leq r\leq 2\cos\theta$ and 
Area 2: $\frac{ \pi }{4}\leq \theta\leq\frac{ \pi }{2} $, but I dont know how to set r boundaries with this. Is it just $0\leq r\leq \csc\theta$? no.... this can't be right.... Could someone please help me out?

You need to convert the equation of the circle $(x-1)^2+y^2 = 1$ into polar coordinates. 
Substitute $x = r\cos \theta$ and $y = r\sin \theta$: 
$(r\cos\theta - 1)^2 + (r\sin\theta)^2 = 1$
$r^2\cos^2\theta - 2r\cos\theta + 1 + r^2\sin^2\theta = 1$
$r^2(\cos^2\theta + \sin^2\theta) - 2r\cos\theta + 1 = 1$
$r^2-2r\cos\theta + 1 = 1$
$r^2-2r\cos\theta = 0$
$r(r-2\cos\theta) = 0$
Do you now see what the equation of the circle is in polar coordinates? 
