Evaluate the following integral
$$\iint_C x\,dx\,dy$$
on the region $C=\left\{ (x,y) : x \in \left[ 0, \sqrt{4 - y^2} \right] \wedge y \in [1,2]\right\}$.
I know $C$ is bounded by the line $y = 1$ and the upper portion of the circle $x^2 + y^2 = 4$, but I don't know how to convert to cylindrical coordinates.
First, find the intersection of the two curves. In polar coordinates, the circle has equation $r=2$ and the line $y=1$ has equation $$ r \sin\theta = 1 \implies r = \frac{1}{\sin\theta} = \csc\theta $$ So the intersection occurs when $$ 2 = \csc\theta \implies \sin\theta = \frac{1}{2} $$ $\theta = \frac{\pi}{6}$ or $\theta = \frac{2\pi}{3}$. Only $\theta = \frac{\pi}{6}$ is in the first quadrant. The other limit for $\theta$ is when $x=0$, or $\theta = \frac{\pi}{2}$.
For each $\theta$ in $\left[\frac{\pi}{6},\frac{\pi}{2}\right]$, the ray at angle $\theta$ enters the region when $r=\csc\theta$ and exits when $r=2$. Therefore $$ \iint_C x\,dA = \int_{\pi/6}^{\pi/2}\int_{\csc\theta}^2 (r\cos\theta)(r\,dr\,d\theta) = \int_{\pi/6}^{\pi/2}\int_{\csc\theta}^2 r^2 \cos\theta\,dr\,d\theta $$