## Thomas' Calculus 13th Edition

$$\bar{x}=\frac{M_{y z}}{M}=\frac{3}{\pi},\ \ \bar{y}=\frac{M_{x z}}{M}=\frac{3}{\pi},\ \ \bar{z}=\frac{M_{x y}}{M}=\frac{3}{4}$$
Since \begin{align*} M&=\int_{0}^{\pi / 2} \int_{0}^{2} \int_{0}^{r} d z r d r d r d \theta\\ &=\int_{0}^{\pi / 2} \int_{0}^{2} r^{2} d r d \theta\\ &=\frac{8}{3} \int_{0}^{\pi / 2} d \theta\\ &=\frac{4 \pi}{3} \\ M_{y z}&=\int_{0}^{\pi / 2} \int_{0}^{2} \int_{0}^{r} x d z r d r d \theta\\ &=\int_{0}^{\pi / 2} \int_{0}^{2} r^{3} \cos \theta d r d \theta\\ &=4 \int_{0}^{\pi / 2} \cos \theta d \theta\\ &=4\\ M_{x z}&=\int_{0}^{\pi / 2} \int_{0}^{2} y d z r d r d \theta\\ &=\int_{0}^{\pi / 2} \int_{0}^{2} r^{3} \sin \theta d r d \theta\\ &=4 \int_{0}^{\pi / 2} \sin \theta d \theta=4\\ M_{x y}&=\int_{0}^{\pi / 2} \int_{0}^{2} \int_{0}^{r} z d z r d r d \theta\\ &=\frac{1}{2} \int_{0}^{\pi / 2} \int_{0}^{2} r^{3} d r d r d \theta\\ &=2 \int_{0}^{\pi / 2} d \theta=\pi \end{align*} $$\bar{x}=\frac{M_{y z}}{M}=\frac{3}{\pi},\ \ \bar{y}=\frac{M_{x z}}{M}=\frac{3}{\pi},\ \ \bar{z}=\frac{M_{x y}}{M}=\frac{3}{4}$$