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