Finding the Moment of Inertia of a Sphere

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

1. Set up the integral to find the moment of inertia of the sphere about a diameter.
2. Use a CAS to find the moment of inertia from (a).

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 \[ I_{z}=\rho \int_{0}^{2\pi }\int_{0}^{2}\int_{-\sqrt{4-r^{2}}}^{\sqrt{4-r^{2}} }r^{3}\,dz\,dr\,d\theta =\dfrac{256\pi \rho }{15} \]