Answer
$\overline{x}=\dfrac{13}{3\pi} $ and $\overline{y}=\dfrac{13}{3\pi}$
Work Step by Step
$M=\int_{0}^{\pi/2} \int_{1}^3 r \ dr \ d \theta=4 \int_{0}^{\pi/2} d \theta = 2 \pi$
Now, $M_y=\int_{0}^{\pi/2} \int_{1}^3 r^2 \cos \theta \ dr \ d \theta=(26/3) \int_{0}^{\pi/2} \cos \theta d \theta = \dfrac{26}{3}$ and $M_x=\int_{0}^{\pi/2} \int_{1}^3 r^2 \cos \theta \ dr \ d \theta=(26/3) \int_{0}^{\pi/2} \cos \theta d \theta = \dfrac{26}{3}$
So, by symmetry $\overline{x}=\dfrac{13}{3\pi} $ and $\overline{y}=\dfrac{13}{3\pi}$
