Thomas' Calculus 13th Edition

$$\overline{x}=\dfrac{3\sqrt 3}{\pi} ; \\ \overline{y}=0$$
$M=\int_{-\pi/3}^{\pi/3} \int_{0}^3 r \ dr \ d \theta \\=\dfrac{9}{2} \times \int_{-\pi/3}^{\pi/3} d \theta \\= 3 \pi$ Now, $$M_y=\int_{-\pi/3}^{\pi/3} \int_{0}^3 r^2 \cos \theta \ dr \ d \theta \\= \int_{-\pi/3}^{\pi/3} 9 \cos \theta d \theta \\= 9 \sqrt 3$$ and $M_x=0$ We have: $\overline{x}=\dfrac{3\sqrt 3}{\pi}$ and $\overline{y}=0$