# Determining Moment of Inertia of a sphere.

by aFk-Al
Tags: determining, inertia, moment, sphere
 P: 19 I'm having some troubles determining the moment of inertia of a sphere about it's central axis. My original question was to calculate it for a cylinder, which I've done, but I'd like to know how to find it for a sphere. Here is the problem solved for a cylinder: Problem: A uniform solid cylinder has radius R, mass M, and length L. Calculate its moment of inertia about its central axis (the z axis). Solution: I divided the cylinder into infinitesimally small layers because I knew that $$dV = (2\pi*r*dr)*L$$. From here I calculated the integral $$I = \int \rho*r^2 dV = \int_{0}^R \rho*r^2*(2\pi*r*L)dr = 2*\pi*\rho*L*R^4$$ I substituted $$\frac {M}{\pi*R^2*L}$$ (or $$\frac{M}{V}$$) for $$\rho$$ into the equation to get $$I = \frac{1}{2}*\pi*(\frac {M}{\pi*R^2*L})*L*R^4 = \frac{1}{2}*M*R^2$$ I understand this, but when I tried to get it as a sphere I ended up getting the wrong answer. Could anyone please show me how to start the problem with a sphere?
 P: 19 Couldn't I divide the sphere into a hemisphere and multiply it by 2? E.G. $$2*\pi*\int_{0}^R (\sqrt{R^2-r^2}*r^2)*r^2 dr$$ Where R is the radius of the sphere. So essentially I'm adding a bunch of $$\pi*r^2$$ (circles) to get a half of a sphere. Then multiplying the whole thing times two.