Answer
$$I_{o}=I_{x}+I_{y}=8 \pi$$
Work Step by Step
Since
\begin{align*}
I_{x}&=\int_{-2}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} y^{2} d y d x\\
&=\int_{-2}^{2}\left[\frac{y^{3}}{3}\right]_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} d x\\
&=\frac{2}{3} \int_{-2}^{2}\left(4-x^{2}\right)^{3 / 2} d x\\
&=\frac{4}{3} \int_{0}^{2}\left(4-x^{2}\right)^{3 / 2} d x \\
& \text{Let } x=2\sin u: \\
&=\frac{8}{3} \int_{0}^{\pi/2}\left(4-4\sin^{2}u\right)^{3 / 2} \cos ud u\\
&=\frac{64}{3}\int_{0}^{\pi/2} \cos^4udu\\
&=4\pi
\end{align*}
and by symmetry
$$ I_{y}=4 \pi,$$
then
$$I_{o}=I_{x}+I_{y}=8 \pi$$
