Answer
$$\bar{x}=0,\ \bar{y}=\frac{13}{31},\ \ I_y=\frac{7}{5} $$
Work Step by Step
Since
\begin{align*}
M&=\int_{-1}^{1} \int_{0}^{x^{2}}(7 y+1) d y d x\\
&=\int_{-1}^{1}\left(\frac{7 x^{4}}{2}+x^{2}\right) d x\\
&=\left(\frac{7 x^{5}}{10}+\frac{1}{3}x^{3}\right) \bigg|_{-1}^{1}\\
&=\frac{31}{15}\\
M_{x}&=\int_{-1}^{1} \int_{0}^{x^{2}} y(7 y+1) d y d x\\
&=\int_{-1}^{1}\left(\frac{7 x^{6}}{3}+\frac{x^{4}}{2}\right) d x\\
&=\frac{13}{15}\\
M_{y}&=\int_{-1}^{1} \int_{0}^{x^{2}} x(7 y+1) d y d x\\
&=\int_{-1}^{1}\left(\frac{7 x^{5}}{2}+x^{3}\right) d x=0
\end{align*}
Then $$\bar{x}=0,\ \bar{y}=\frac{13}{31}$$
and \begin{align*}
I_{y}&=\int_{-1}^{1} \int_{0}^{x^{2}} x^{2}(7 y+1) d y d x\\
&=\int_{-1}^{1}\left(\frac{7 x^{6}}{2}+x^{4}\right) d x\\
&=\left(\frac{7 x^{7}}{14}+\frac{1}{5}x^{5}\right)\bigg|_{-1}^{1} \\
&=\frac{7}{5}
\end{align*}
