Answer
$$\frac{50}{3}$$
Work Step by Step
Using the rules of integrals, we obtain:
$\int_{0}^{2} \int_{y^{2}}^{6-y} x y d x d y=\iint_{R} x y d A$
$=\int_{0}^{2}\left[\frac{x^{2} y}{2}\right]_{y^{2}}^{6-y} d y$
$=\int_{0}^{2} \frac{(6-y)^{2} y}{2}-\frac{\left(y^{2}\right)^{2} y}{2} d y$
$=\int_{0}^{2} \frac{y\left(y^{2}-12 y+36\right) }{2}-\frac{y^{5}}{2} d y$
$=\frac{1}{2} \int_{0}^{2} -y^{5} +y^{3}-12 y^{2}+36 yd y$
$=\frac{1}{2}\left[\frac{y^{4}}{4}-4 y^{3}+18 y^{2}-\frac{y^{6}}{6}\right]_{0}^{2}$
$\frac{50}{3}=\left[\frac{2^{4}}{4}-4 \cdot 2^{3}+18 \cdot 2^{2}-\frac{2^{6}}{6}\right]\frac{1}{2}$
