Answer
$\dfrac{2}{15}$
Work Step by Step
Given $$ \iint_{D} xy^2 d A$$
$$D=\{(x, y) |-1\leq x \leq 1,0\leq y \leq \sqrt{1-y^2} \}$$
So, we have
$A=\int_{-1}^{1} \int_{0}^{\sqrt{1-y^2}} xy^2 dx \ d y\\
=\int_{-1}^{1} y^2 [ \dfrac{x^2}{2}]_{0}^{\sqrt{1-y^2}} dy \\ = \dfrac{1}{2} \int_{-1}^1 y^2 (1-y^2) \ dy \\=\dfrac{1}{2} \int_{-1}^1 (y^2-y^4) \ dy \\ = \dfrac{1}{2} [\dfrac{y^3}{3}-\dfrac{y^5}{5}]_{-1}^1 \\= \dfrac{1}{2} [ (\dfrac{1}{3} - \dfrac{1}{5})+(\dfrac{1}{3}-\dfrac{1}{5})] \\=\dfrac{2}{15}$
