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Locate the centroid  $\overline{y}$  of the homogeneous solid  formed by revolving the shaded area about the  $y$  axis.

$d v=\pi z^{2} d y$

$z=a-\sqrt{a^{2}-y^{2}}$

$d v=\pi\left(a-\sqrt{a^{2}-y^{2}}\right)^{2} d y$

$=\pi\left(2 a^{2}-y^{2}-2 a \sqrt{a^{2}-y^{2}}\right) dy$

$\tilde{y}=y \quad \longrightarrow\quad \overline{y}= \frac{\int \tilde{y} d v}{\int d v}= \frac{\int_{0}^{a} y\left(\pi\left[2 a^{2}-y^{2}-2 a \sqrt{a^{2}-y^{2}}\right]\right) d y}{\int_{0}^{a} \pi\left(2 a^{2}-y^{2}-2 a \sqrt{a^{2}-y^{2}}\right) d y}$

$=\frac{\pi \int_{0}^{a}\left(2 a^{2} y-y^{3}-2 a y \sqrt{a^{2}-y^{2}}\right) d y}{\pi \int_{0}^{a}\left(2 a^{2}-y^{2}-2 a \sqrt{a^{2}-y^{2}}\right) d y} =\frac{\pi\left(a^{2} y^{2}-\frac{y^{4}}{4}+\frac{2a}{3} \sqrt{(a^{2}-y^{2})^{3}}\right)|_{0}^{a}}{\pi\left(2 a^{2} y-\frac{y^{3}}{3}-a\left(y \sqrt{a^{2}-y^{2}}\right)+a^{2}\sin^{-1}\frac{y}{a})\right)|_{0}^{a}}$

$=\frac{\frac{1}{12} a^{4}}{\left(\frac{10-3 \pi}{6}\right)a^{3}}=\frac{a}{2(10-3 \pi)}$