## Thomas' Calculus 13th Edition

$$\bar{z}=\frac{M_{x y}}{M}=\left(\frac{\pi a^{4}}{8}\right)\left[\frac{3}{\pi a^{3}(2-\sqrt{2})}\right]=\left(\frac{3 a}{8}\right)\left(\frac{2+\sqrt{2}}{2}\right)=\frac{3(2+\sqrt{2}) a}{16},\ \bar{x}=\bar{y}=0$$
Since \begin{align*} M&=\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{a} \rho^{2} \sin \phi d \rho d \phi d \theta\\ &=\frac{a^{3}}{3} \int_{0}^{2 \pi} \int_{0}^{\pi / 4} \sin \phi d \phi d \theta\\ &=\frac{a^{3}}{3} \int_{0}^{2 \pi} \frac{2-\sqrt{2}}{2} d \theta\\ &=\frac{\pi a^{3}(2-\sqrt{2})}{3}\\ M_{x y}&=\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{a} \rho^{3} \sin \phi \cos \phi d \rho d \phi d \theta\\ &=\frac{a^{4}}{4} \int_{0}^{2 \pi} \int_{0}^{\pi / 4} \sin \phi d \phi d \theta\\ &=\frac{a^{4}}{16} \int_{0}^{2 \pi} d \theta\\ &=\frac{\pi a^{4}}{8} \end{align*} Then by symmetry $$\bar{z}=\frac{M_{x y}}{M}=\left(\frac{\pi a^{4}}{8}\right)\left[\frac{3}{\pi a^{3}(2-\sqrt{2})}\right]=\left(\frac{3 a}{8}\right)\left(\frac{2+\sqrt{2}}{2}\right)=\frac{3(2+\sqrt{2}) a}{16},\ \bar{x}=\bar{y}=0$$