## Thomas' Calculus 13th Edition

$$\bar{x}=\frac{M_{y z}}{M}=\frac{9 \sqrt{3}}{32},\ \bar{y}=\bar{z}=0,$$
Since \begin{align*} M&=\int_{-\pi / 3}^{\pi / 3} \int_{0}^{1} \int_{-\sqrt{1-r^{2}}}^{\sqrt{1-r^{2}}} d z r d r d \theta\\ &=\int_{-\pi / 3}^{\pi / 3} \int_{0}^{1} 2 r \sqrt{1-r^{2}} d r d \theta\\ &=\int_{-\pi / 3}^{\pi / 3}\left[-\frac{2}{3}\left(1-r^{2}\right)^{3 / 2}\right]_{0}^{1} d \theta\\ &=\frac{2}{3} \int_{-\pi / 3}^{\pi / 3} d \theta\\ &=\left(\frac{2}{3}\right)\left(\frac{2 \pi}{3}\right)\\ &=\frac{4 \pi}{9}\\ M_{y z}&=\int_{-\pi / 3}^{\pi / 3} \int_{0}^{1} \int_{-\sqrt{1-r^{2}}}^{\sqrt{1-r^{2}}} \cos \theta d z d r d \theta\\ &=2 \int_{-\pi / 3}^{\pi / 3} \int_{0}^{1} r^{2} \sqrt{1-r^{2}} \cos \theta d r d \theta\\ &=2 \int_{-\pi / 3}^{\pi / 3}\left[\frac{1}{8} \sin ^{-1} r-\frac{1}{8} r \sqrt{1-r^{2}}\left(1-2 r^{2}\right)\right]_{0}^{1} \cos \theta d \theta\\ &=\frac{\pi}{8} \int_{-\pi / 3}^{\pi / 3} \cos \theta d \theta\\ &=\frac{\pi}{8}[\sin \theta]_{-\pi / 3}^{\pi / 3}\\ &=\left(\frac{\pi}{8}\right)\left(2 \cdot \frac{\sqrt{3}}{2}\right)\\ &=\frac{\pi \sqrt{3}}{8} \end{align*} Then by symmetry $$\bar{x}=\frac{M_{y z}}{M}=\frac{9 \sqrt{3}}{32},\ \bar{y}=\bar{z}=0,$$