Find the moment of inertia of a homogeneous solid $E$ in the shape of a sphere of radius 2 about a diameter.

Solution (a) Let $\rho$ denote the constant mass density of the sphere of radius $2.$ We position the sphere so that its center is at the origin. The equation of the sphere is then $x^{2}+y^{2}+z^{2}=4$, or, in cylindrical coordinates, $r^{2}+z^{2}=4$. If we use the $z$-axis as the diameter, the moment of inertia about a diameter is given by the moment of inertia $I_{z}$ about the $z$-axis. Then $$I_{z}=\iiint\limits_{\kern-3ptE}r^{2}\rho \,dV=\rho \int_{0}^{2\pi }\int_{0}^{2}\int_{-\sqrt{4-r^{2}}}^{\sqrt{4-r^{2}}}r^{3}\,dz\,dr\,d\theta$$

(b) We use a CAS to find $$I_{z}=\rho \int_{0}^{2\pi }\int_{0}^{2}\int_{-\sqrt{4-r^{2}}}^{\sqrt{4-r^{2}} }r^{3}\,dz\,dr\,d\theta =\dfrac{256\pi \rho }{15}$$