Answer
$$
\begin{aligned}
\iint_{R} \frac{y^{2}}{x^{2}+y^{2}} d A =\frac{\pi}{2}\left(b^{2}-a^{2}\right)
\end{aligned}
$$
Work Step by Step
$$
\iint_{R} \frac{y^{2}}{x^{2}+y^{2}} d A
$$
The region R can be described as
$$
R=\left\{(x, y) | a \leqslant x^{2}+y^{2} \leqslant b \right\}
$$
and in polar coordinates it is given by
$$
R=\left\{(r, \theta) | a \leq r \leq b , \quad 0 \leq \theta \leq 2\pi \right\}
$$
Therefore,
$$
\begin{aligned}
\iint_{R} \frac{y^{2}}{x^{2}+y^{2}} d A &=\int_{0}^{2 \pi} \int_{a}^{b} \frac{(r \sin \theta)^{2}}{r^{2}} r d r d \theta \\
&=\left(\int_{0}^{2 \pi} \sin ^{2} \theta d \theta\right)\left(\int_{a}^{b} r d r\right) \\
&=\int_{0}^{2 \pi} \frac{1}{2}(1-\cos 2 \theta) d \theta \int_{a}^{b} r d r \\
& =\frac{1}{2}\left[\theta-\frac{1}{2} \sin 2 \theta\right]_{0}^{2 \pi}\left[\frac{1}{2} r^{2}\right]_{a}^{b} \\
&=\frac{1}{2}(2 \pi-0-0)\left[\frac{1}{2}\left(b^{2}-a^{2}\right)\right] \\
&=\frac{\pi}{2}\left(b^{2}-a^{2}\right).
\end{aligned}
$$