Chapter 15: Multiple Integrals - Section 15.6 - Moments and Centers of Mass - Exercises 15.6 - Page 908: 5

$ \bar{x}=\bar{y}=\frac{4 a}{3 \pi}$

Since \begin{align*} M&=\int_{0}^{a} \int_{0}^{\sqrt{a^{2}-x^{2}}} d y d x\\ &=\frac{\pi a^{2}}{4} ; \\ M_{y}&=\int_{0}^{a} \int_{0}^{\sqrt{a^{2}-x^{2}}} x d y d x\\ &=\int_{0}^{a}[x y]_{0}^{\sqrt{a^{2}-x^{2}}} d x\\ &=\int_{0}^{a} x \sqrt{a^{2}-x^{2}} d x\\ &=\frac{a^{3}}{3} \end{align*} Then by symmetry $ \bar{x}=\bar{y}=\frac{4 a}{3 \pi}$