Answer
$$
\begin{aligned}
I_{x} &=\frac{1}{16} \rho a^{4} \pi ,
\end{aligned}
$$
$$
\begin{aligned}
I_{y} &=\frac{1}{16} \rho a^{4} \pi,
\end{aligned}
$$
$$
\overline{\bar{x}}=\overline{\bar{y}}=\frac{a}{2}.
$$
Work Step by Step
A lamina with constant density $\rho(x, y)= \rho $ occupies the given disk $D$ in the first quadrant:
$$ D=\left\{(x, y ) | \quad x \geq 0,\, y \geq 0 , \quad x^{2}+y^{2}\leq a^{2} \right\} $$
In polar coordinates, the region $D$ is
$$ D=\left\{(r, \theta) | 0 \leq r \leq a, \quad 0 \leq \theta \leq \frac{\pi}{2} \right\} $$
So, the moment of inertia of the lamina about the $x$-axis is:
$$
\begin{aligned}
I_{x} &=\iint_{D} y^{2} \rho d A\\
&=\int_{0}^{\pi / 2} \int_{0}^{a} \rho(r \sin \theta)^{2} r d r d \theta \\
&=\rho \int_{0}^{\pi / 2} \sin ^{2} d \theta \int_{0}^{a} r^{3} d r \\
&=\rho\left[\frac{1}{2} \theta-\frac{1}{4} \sin 2 \theta\right]_{0}^{\pi / 2}\left[\frac{1}{4} r^{4}\right]_{0}^{a}\\
&=\rho\left(\frac{\pi}{4}\right)\left(\frac{1}{4} a^{4}\right)\\
&=\frac{1}{16} \rho a^{4} \pi ,
\end{aligned}
$$
and the moment of inertia of the lamina about the $y$-axis is:
$$
\begin{aligned}
I_{y} &=\iint_{D} x^{2} \rho d A\\
&=\int_{0}^{\pi / 2} \int_{0}^{a} \rho(r \cos \theta)^{2} r d r d \theta \\
&=\rho \int_{0}^{\pi / 2} \cos ^{2} d \theta \int_{0}^{a} r^{3} d r \\
&=\rho\left[\frac{1}{2} \theta+\frac{1}{4} \sin 2 \theta\right]_{0}^{\pi / 2}\left[\frac{1}{4} r^{4}\right]_{0}^{a} \\
&=\rho\left(\frac{\pi}{4}\right)\left(\frac{1}{4} a^{4}\right) \\
&=\frac{1}{16} \rho a^{4} \pi.
\end{aligned}
$$
Notice that the mass of the region $D$ is
$m =\text { density * area }= \rho* \frac{1}{4} \pi a^{2} $.
So, the radius of gyration $\overline{\bar{y}}$ with respect to the $x$-axis is given by:
$$
\overline{\bar{y}}^{2}=\frac{I_{x}}{m}=\frac{\frac{1}{16} \rho a^{4} \pi}{ \rho. \frac{1}{4} \pi a^{2}}=\frac{a^{2}}{4} \Rightarrow \overline{\bar{y}}=\frac{a}{2},
$$
and the radius of gyration $\overline{\bar{x}}$ with respect to the $y$-axis is given by:
$$
\overline{\bar{x}} ^{2}=\frac{I_{y}}{m}=\frac{\frac{1}{16} \rho a^{4} \pi}{ \rho. \frac{1}{4} \pi a^{2}}=\frac{a^{2}}{4} \Rightarrow \overline{\bar{x}}=\frac{a}{2}.
$$
