Answer
$$ \bar{x}=\frac{8}{15}\ \text{ and }\ \bar{y}=\frac{8}{15}$$
$$I_{x} =\frac{1}{6}$$
Work Step by Step
Since
\begin{align*}
M&=\int_{0}^{1} \int_{y^{2}}^{2 y-y^{2}}(y+1) d x d y\\
&=\int_{0}^{1}\left(2 y-2 y^{3}\right) d y\\
&=\frac{1}{2} \\
M_{x}&=\int_{0}^{1} \int_{y^{2}}^{2 y-y^{2}} y(y+1) d x d y\\
&=\int_{0}^{1}\left(2 y^{2}-2 y^{4}\right) d y\\
&=\left(\frac{2}{3} y^{3}-\frac{2}{5} y^{5}\right)\bigg|_{0}^{1} \\
&=\frac{4}{15}\\
M_{y}&=\int_{0}^{1} \int_{y^{2}}^{2 y-y^{2}} x(y+1) d x d y\\
&=\int_{0}^{1}\left(2 y^{2}-2 y^{4}\right) d y\\
&=\left(\frac{2}{3} y^{3}-\frac{2}{5} y^{5}\right)\bigg|_{0}^{1} \\
&=\frac{4}{15}
\end{align*}
Then
$$ \bar{x}=\frac{8}{15}\ \text{ and }\ \bar{y}=\frac{8}{15}$$
and
\begin{align*}
I_{x}&=\int_{0}^{1} \int_{y^{2}}^{2 y-y^{2}} y^{2}(y+1) d x d y\\
&=2 \int_{0}^{1}\left(y^{3}-y^{5}\right) d y\\
&=2\left(\frac{1}{4}y^{4}-\frac{1}{6}y^{6}\right) \bigg|_{0}^{1}\\
&=\frac{1}{6}
\end{align*}
