Answer
$$\int_{D} y^{2} d A=\frac{4}{3}$$,
Work Step by Step
Given $$\int_{D} y^{2} d A$$
$$D=\{(x, y) |-1 \leq y \leq 1,-y-2 \leq x \leq y\}$$
So, we have
\begin{aligned} A&=\int_{D} y^{2} d A=\int_{-1}^{1} \int_{-y-2}^{y} y^{2} d x d y\\
& =\int_{-1}^{1}\left[x y^{2}\right]_{-y-2}^{y} d y\\
&=\int_{-1}^{1}(y) y^{2}-(-y-2) y^{2} d y\\
&=\int_{-1}^{1} y^{3}-\left(-y^{3}-2 y^{2}\right) d y\\
& =\int_{-1}^{1} 2 y^{3}+2 y^{2} d y\\
&=\left[\frac{y^{4}}{2}+\frac{2 y^{3}}{3}\right]_{-1}^{1}
\\
&=\left[\frac{1}{2}+\frac{2}{3}\right]-\left[\frac{1}{2}-\frac{2}{3}\right]\\
&=\frac{4}{3}\end{aligned}
