Double integral from rectangular region to polar $$\int_{-1}^{1} \int_{0}^{\sqrt{3+2y-y^2}} \cos(x^2 + (y-1)^2)) dy dx $$ 
How do I go about solving this? I tried substituting $x$ and $y$ for $r\cos(\theta)$ and $r\sin(\theta)$ and $x^2 + y^2 = r^2$. I can't seem to figure it out. So far I believe it's a rectangular region.
$$I=\int_{-1}^{1} \int_{0}^{\sqrt{3+2y-y^2}} \cos(x^2 + (y-1)^2)) dx dy $$ 

The change of variables in polar coordinates with center $(0,1)$ : 
$\quad \begin{cases}
x=r\cos(\theta)\\
y=1+r\sin(\theta) 
\end{cases}$ .  

$$I=\int_{-\pi/2}^0\int_{0}^{2} \cos(r^2) rdr d\theta $$
$$I=\int_{0}^2\cos(r^2) rdr \int_{-\pi/2}^0 d\theta $$
$$I=\left(\frac12\sin(2^2)\right)\left(\frac{\pi}{2}\right)$$
$$I=\frac{\pi}{4}\sin(4)$$
$$\int_0^{-\frac{3\pi}{2}}\int_0^2\cos\left(r^2\left(\cos^2\theta\right)\ +\left(r\left(\sin\theta\right)-1\right)^2\right)rdrd\theta$$

This will do it!