use polar coordinates to evaluate the integral $\int^2_0 \int^{\sqrt{1-(1-x)^2}}_0 \frac{y}{y^2 + x^2} dydx$ use polar coordinates to evaluate the integral $\int^2_0 \int^{\sqrt{1-(1-x)^2}}_0 \frac{y}{y^2 + x^2} dydx$ 
I have no problem evaluating the integral but the limits of integraation is what I am iffy on. Would it be $ 0 < r < 2cos\theta $ and $0< \theta <\pi/2 $? or is it $ 0 <\theta < \pi $ and $0< r < 1 $ 
 A: For the corrected integral:
The upper boundary of the region of integration is $y=\sqrt{1-(1-x)^2}$, so it’s the upper half of $y^2=1-(1-x)^2$, or $(x-1)^2+y^2=1$, the circle of radius $1$ centred at $\langle 1,0\rangle$. Since $x$ runs from $0$ to $2$, the region of integration is simply the upper half of the disk of radius $1$ centred at $\langle 1,0\rangle$.
The angle $\theta$ is $0$ along the base of the region, and for $0\le\alpha<\frac{\pi}2$ the line with polar equation $\theta=\alpha$ crosses the region, so the limits of integration on $\theta$ must be $0$ and $\frac{\pi}2$. Clearly for each of these values of $\theta$ the lower limit on $r$ is $0$. The upper limit would be a bit messy, except that we know that the polar equation of the circle with radius $1$ and centre $\langle 1,0\rangle$ is ...
A: Well, first of all $\sqrt{1-(1-x^2)}=\sqrt{x^2}=x$ for $x\geq 0$. Now I encourage you to make a sketch of the area over which is being integrated (it is a triangle with vertices $(0,0),(0,2)\text{ and }(2,0)$). Now for a given value of $\theta$ (the angle with the $x$ axis) the radius $r$ can go from $0$ to $2\cos\theta$. And $\theta$ goes from $0$ to $\arctan{2/2}=\frac{\pi}{4}$.
