[m]=ab\int^{1}_{0}\int^{2\pi}_{0} \sqrt{1-r^2} rdrd\varphi=ab \int^{2\pi}_{0}d\varphi (-\frac{1}{2}\int^{1}_{0}\sqrt{1-r^2} (-2rdr))=[/m]
=[m]2\pi ab\cdot (-\frac{1}{2})\int^{1}_{0}\sqrt{1-r^2}d(1-r^2)=-\pi ab\cdot\frac{{} (1-r^{2})^{\frac{3}{2}}}{\frac{3}{2}}|^{1}_{0}=\frac{2}{3}\pi ab[/m]