Answer
$$
\begin{aligned}
I_{x}&=\frac{4}{9} \rho,
\end{aligned}
$$
$$
\begin{aligned}
I_{y} &=\rho\left(\pi^{2}-4\right),
\end{aligned}
$$
$$
\overline{\bar{y}}=\frac{\sqrt 2}{3}
$$
and
$$
\overline{\bar{x}}=\sqrt {\frac{\left(\pi^{2}-4\right)}{ 2 } }.
$$
Work Step by Step
A lamina with constant density $\rho(x, y)= \rho $ occupies the given region $D$ :
$$ D=\left\{(x, y ) | \quad y \geq 0 ,\, \quad 0 \leq x\leq \pi ,\, y= \sin x \right\} $$
So, the moment of inertia of the lamina about the $x$-axis is:
$$
\begin{aligned}
I_{x}&=\int_{0}^{\pi} \int_{0}^{\sin x} \rho y^{2} d y d x \\
&=\frac{1}{3} \rho \int_{0}^{\pi} \sin ^{3} x d x \\
&=\frac{1}{3} \rho \int_{0}^{\pi}\left(1-\cos ^{2} x\right) \sin x d x \\
&=\frac{1}{3} \rho\left[-\cos \theta+\frac{1}{3} \cos ^{3} \theta\right]_{0}^{\pi} \\
&=\frac{4}{9} \rho
\end{aligned}
$$
and the moment of inertia of the lamina about the $y$-axis is:
$$
\begin{aligned}
I_{y} &=\int_{0}^{\pi} \int_{0}^{\sin x} \rho x^{2} d y d x \\
&=\rho \int_{0}^{\pi} x^{2} \sin x d x \\
&\quad\quad\quad \text {use integration by parts with}\\
& \quad\quad\quad \left[\begin{array}{c}{u=x^{2}, \quad\quad dv= \sin x d x} \\ {d u= 2x dx, \quad\quad v=-\cos x }\end{array}\right], \text { then }\\
&=\rho[-x^{2}\cos x+2 \int x \cos x d x \\
&\quad\quad\quad \text {use integration by parts with}\\
& \quad\quad\quad \left[\begin{array}{c}{u=x, \quad\quad dv= \cos x d x} \\ {d u= dx, \quad\quad v=\sin x }\end{array}\right], \text { then we get }\\
&=\rho\left[-x^{2} \cos x+2 x \sin x+2 \cos x\right]_{0}^{\pi} \\
&=\rho\left(\pi^{2}-4\right)
\end{aligned}
$$
Notice that the mass of the region $D$ is given by:
$$
\begin{aligned}
m&=\iint_{D} \rho(x, y) d A \\
&=\int_{0}^{\pi} \int_{0}^{\sin x} \rho d y d x\\
&=\rho \int_{0}^{\pi} \sin x d x \\
&=\rho[-\cos x]_{0}^{\pi} \\
&=2 \rho
\end{aligned}
$$
So, the radius of gyration $\overline{\bar{y}}$ with respect to the $x$-axis is given by:
$$
\overline{\bar{y}}^{2}=\frac{I_{x}}{m}=\frac{\frac{4}{9} \rho}{2 \rho}=\frac{2}{9} \Rightarrow \overline{\bar{y}}=\frac{\sqrt 2}{3},
$$
and the radius of gyration $\overline{\bar{x}}$ with respect to the $y$-axis is given by:
$$
\overline{\bar{x}} ^{2}=\frac{I_{y}}{m}=\frac{\rho\left(\pi^{2}-4\right)}{ 2 \rho}=\frac{\left(\pi^{2}-4\right)}{ 2 } \Rightarrow \overline{\bar{x}}=\sqrt {\frac{\left(\pi^{2}-4\right)}{ 2 } }.
$$
