1. The problem statement, all variables and given/known data The density of Earth, at any distance r from it's center is approximately p(r) = [14.2 - 11.6 (r/R)] * 10^3 kg*m^-3 where R is the radius of the Earth. Show that the density leads to a moment of inertia I = 0.330 MR^2 about an axis through the center, where M is the mass of the Earth. 2. Relevant equations I = [tex]\int[/tex] r^2 dm I of a hollow sphere = 2/3MR^2 I = [tex]\int[/tex] dI 3. The attempt at a solution So what I believe is right is to sum up the contributions of the moments of inertia of infinitesimally thin hollow spheres, with limits of integration 0 to R. So, I have a hollow sphere with thickness dr a distance r from the center of the Earth, whose moment is given by 2/3MR^2 so I have: I = 2/3 [tex]\int[/tex] [R,0] r^2 dm dm would be p(r) dV dV should be 4/3[tex]\pi[/tex](r+dr)^3 - 4/3[tex]\pi[/tex]r^3 because the all the mass is concentrated at the outside of the shell of thickness dr. This equals 3r^2dr. So. 2 [tex]\int[/tex] p(r) * 3r^2 * r^2 dr = I integrating with limits [R,0] and substituting in the actual values i got about 2 * 10^37 kg * m^2, but the answer is 8 * 10^37 kgm^2. What did I do incorrectly?