## Thomas' Calculus 13th Edition

$\overline{x}=\dfrac{13}{3\pi} ; \\ \overline{y}=\dfrac{13}{3\pi}$
$$M=\int_{0}^{\pi/2} \int_{1}^3 r \ dr \ d \theta \\=4 \int_{0}^{\pi/2} d \theta \\= 2 \pi$$ $$M_y=\int_{0}^{\pi/2} \int_{1}^3 r^2 \cos \theta \ dr \ d \theta \\=(26/3) \int_{0}^{\pi/2} \cos \theta d \theta \\= \dfrac{26}{3} \\ M_x =\int_{0}^{\pi/2} \int_{1}^3 [r^2 \cos \theta] \ dr \ d \theta \\= \int_{0}^{\pi/2} (\dfrac{26}{3}) \times \cos \theta d \theta \\= \dfrac{26}{3}$$ So, $\overline{x}=\dfrac{13}{3\pi} ; \\ \overline{y}=\dfrac{13}{3\pi}$ (By symmetry )