## Thomas' Calculus 13th Edition

$\bar x=\bar y=0$
We calculate the center of mass as follows: First calculate the mass: $M=4\int^{\pi/2}_0 \int^1_0 \int^r_0$ dz r dr $d\theta$ =$4\int^{\pi/2}_0 \int^1_0 r^2$ dr $d\theta$ =$\frac{4}{3}\int^{\pi/2}_0 d\theta$ =$\frac{2\pi}{3}$ Now calculate the moment: $M_{xy}$= $\int^{2\pi}_0 \int^1_0 \int^r_0$ z dz r dr $d\theta$ =$\frac{1}{2}\int^{2\pi}_0 \int^1_0 r^3 dr d\theta$ =$\frac{1}{8} \int^{2\pi}_0 d\theta$ =$\frac{\pi}{4}$ And take the ratio: $\frac{M_{xy}}{M}=(\frac{\pi}{4})(\frac{3}{2\pi})=\frac{3}{8}$ Thus, we have (by symmetry): $\bar x=\bar y=0$