Finding the Moment of Inertia of a Solid

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 \[ 0\leq \theta \leq 2\pi\qquad 0\leq r\leq 1\qquad \qquad 0\leq z\leq 1-r^{2} \]

954

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