Answer
$$\Rightarrow \bar{x}=\frac{64}{35} \ \text{ and }\ \bar{y}=\frac{5}{7}$$
Work Step by Step
Since
\begin{align*}
M&=\int_{0}^{2} \int_{y^{2} / 2}^{4-y} d x d y\\
&=\int_{0}^{2}\left(4-y-\frac{y^{2}}{2}\right) d y\\
&=\frac{14}{3}
\end{align*}
and
\begin{align*}
M_{y}&=\int_{0}^{2} \int_{y^{2} / 2}^{4-y} x d x d y\\
&=\frac{1}{2} \int_{0}^{2}\left[x^{2}\right]_{y^{2} / 2}^{4-y} d y\\
& =\frac{1}{2} \int_{0}^{2}\left(16-8 y+y^{2}-\frac{y^{4}}{4}\right) dy\\ &=\frac{128}{3} \\
M_{x}&=\int_{0}^{2} \int_{y^{2} / 2}^{4-y} y d x d y\\
&=\int_{0}^{2}\left(4 y-y^{2}-\frac{y^{3}}{2}\right) d y\\
&=\frac{10}{3}
\end{align*}
Then
$$\Rightarrow \bar{x}=\frac{64}{35} \ \text{ and }\ \bar{y}=\frac{5}{7}$$
