Find the moment of inertia about the $z$-axis of a homogeneous solid $E$ of mass density $\rho$ enclosed by the paraboloid $z=1-x^{2}-y^{2}$ and the $xy$ -plane.

Solution In cylindrical coordinates, the solid $E$ is enclosed on the top by the paraboloid $z=1-x^{2}-y^{2}=1-r^{2}$ and on the bottom by $z=0.$ In the $xy$-plane, the region is bounded by the circle $x^{2}+y^{2}=1.$ So, we have $$0\leq \theta \leq 2\pi\qquad 0\leq r\leq 1\qquad \qquad 0\leq z\leq 1-r^{2}$$

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The moment of inertia $I_{z}$ about the $z$-axis is $$\begin{eqnarray*} I_{z}& =&\iiint\limits_{\kern-3ptE}r^{2}\rho \,dV=\rho \iiint\limits_{\kern-3ptE}r^{2}r\,dr\,d\theta \,dz=\rho \int_{0}^{2\pi }\!\int_{0}^{1}\int_{0}^{1-r^{2}}r^{3}\,dz\,dr\,d\theta\\ & =&\rho \int_{0}^{2\pi }\int_{0}^{1}r^{3}(1-r^{2})\,dr\,d\theta =\rho \int_{0}^{2\pi }\left[ \dfrac{r^{4}}{4}-\dfrac{r^{6}}{6}\right] _{0}^{1}\,d\theta =\rho \int_{0}^{2\pi }\dfrac{1}{12}\,d\theta =\dfrac{\pi }{6} \rho \qquad \end{eqnarray*}$$