## Thomas' Calculus 13th Edition

$\frac{3\pi}{16}$
We calculate the average as: average=$\frac{1}{(\frac{4\pi}{3})}\int^{2\pi}_0 \int^1_0 \int^{\sqrt{1-r^2}}_{-\sqrt{1-r^2}}r^2$ dz dr d $\theta$ =$\frac{3}{4\pi} \int^{2\pi}_0 \int^1_0 2r^2 \sqrt{1-r^2}$ dr d $\theta$ =$\frac{3}{2\pi} \int^{2\pi}_0 [\frac{1}{8}sin^{-1}r-\frac{1}{8}r\sqrt{1-r^2}(1-2r^2)]^1_0d\theta$ =$\frac{3}{16\pi}\int^{2\pi}_0 (\frac{\pi}{2}+0)$ d $\theta$ =$\frac{3}{32}\int^{2\pi}_0 d\theta$ =$(\frac{3}{32})(2\pi)$ =$\frac{3\pi}{16}$