Answer
$\iint_R\arctan(y/x)dA=\frac{3\pi^2}{64}$
Work Step by Step
The inequality $1\leq x^2+y^2\leq 4$ describes a region between two disks center the origin with radii $1$ and $2$, so that in polar coordinates it can be written as $1\leq r\leq 2$.
Meanwhile, the inequality $0\leq y\leq x$ describes a region between the lines $y=0$ and $y=x$, so that in polar coordinates it can be rewritten as $0\leq \theta\leq \pi/4$.
Evaluate the given integral over $R$:
$\iint_R\arctan(y/x)dA=\int_1^2\int_0^{\pi/4}\arctan(r\sin \theta/r\cos \theta)r d\theta dr$
$=\int_1^2\int_0^{\pi/4}\arctan (\tan\theta)rd\theta dr$
$=\int_1^2\int_0^{\pi/4}\theta rd\theta dr$
$=\int_1^2\frac{r\theta^2}{2}]_0^{\pi/4}dr$
$=\int_1^2\frac{\pi^2r}{32}dr$
$=\frac{\pi^2r^2}{64}]_1^2$
$=\frac{3\pi^2}{64}$
