Answer
$$ \bar{x}=0,\ \ \ \bar{y}=\frac{9}{14}$$
$$I_{y} =\frac{16}{35}$$
Work Step by Step
Since
\begin{align*}
M&=\int_{-1}^{1} \int_{x^{2}}^{1}(y+1) d y d x\\
&=-\int_{-1}^{1}\left(\frac{x^{4}}{2}+x^{2}-\frac{3}{2}\right) d x\\
&=-\left(\frac{x^{5}}{10}+\frac{1}{3}x^{3}-\frac{3}{2}x\right) \bigg|_{-1}^{1}\\
&=\frac{32}{15} \\
M_{x}&=\int_{-1}^{1} \int_{x^{2}}^{1} y(y+1) d y d x\\
&=\int_{-1}^{1}\left(\frac{5}{6}-\frac{x^{6}}{3}-\frac{x^{4}}{2}\right) d x\\
&=\left(\frac{5}{6}x-\frac{x^{7}}{21}-\frac{x^{5}}{10}\right)\bigg|_{-1}^{1}\\
&=\frac{48}{35}\\
M_{y}&=\int_{-1}^{1} \int_{x^{2}}^{1} x(y+1) d y d x\\
&=\int_{-1}^{1}\left(\frac{3 x}{2}-\frac{x^{5}}{2}-x^{3}\right) d x=0
\end{align*}
Then
$$ \bar{x}=0,\ \ \ \bar{y}=\frac{9}{14}$$
and
\begin{align*}
I_{y}&=\int_{-1}^{1} \int_{x^{2}}^{1} x^{2}(y+1) d y d x\\
&=\int_{-1}^{1}\left(\frac{3 x^{2}}{2}-\frac{x^{6}}{2}-x^{4}\right) d x\\
&=\left(\frac{3 x^{3}}{6}-\frac{x^{7}}{14}-\frac{1}{5}x^{5}\right)\bigg|_{-1}^{1}\\
&=\frac{16}{35}
\end{align*}
