Answer
$$ \bar{x}=\frac{\pi}{2}\ \ \text{ and }\ \ \bar{y}=\frac{\pi}{8}$$
Work Step by Step
Since
\begin{align*}
M&=\int_{0}^{\pi} \int_{0}^{\sin x} d y d x\\
&=\int_{0}^{\pi} \sin x d x\\
&=2 ; \\
M_{x}&=\int_{0}^{\pi} \int_{0}^{\sin x} y d y d x\\
&=\frac{1}{2} \int_{0}^{\pi}\left[y^{2}\right]_{0}^{\sin x} d x\\
&=\frac{1}{2} \int_{0}^{\pi} \sin ^{2} x d x\\
&=\frac{1}{4} \int_{0}^{\pi}(1-\cos 2 x) d x\\
&=\frac{\pi}{4}
\end{align*}
Then
$$ \bar{x}=\frac{\pi}{2}\ \ \text{ and }\ \ \bar{y}=\frac{\pi}{8}$$
