1. The problem statement, all variables and given/known data Given that the conical shell has uniform density and thickness is made of one sheet, has mass M, height h, and base radius R, derive the moment of inertia about its axis of symmetry. 2. Relevant equations I = MR^2 for a ring about its central axis. I = ∫dI - my approach 3. The attempt at a solution For a conical shell, I am equating the change in the moment of inertia as a series of rings, where dI = dm(R)^2. R = y, and since y/z = r/h, y = (rz)/h = R. dm = σdA = M(2∏(rz)/h)dz/(∏r^2 + √(r^2+h^2)) Therefore I = ∫[[M(2∏(rz)/h)[(rz)/h]^2]dz]/(∏r^2 + √(r^2+h^2)). I know that is not the correct dm, as it is not including the base, while the base is included within the density function. I am able to derive the moment of inertia for a conical shell with an open base just fine (just take out the ∏r^2 from the denominator in the integral), and have found it to be [MR^2]/√8. I am aware that the moment of inertia for a conical shell is [MR^2]/2, yet I do not know quite how to approach this. It would be easier if I were able to set up a density function, and utilize a triple integral for Izz, but I had issues with that approach as well. How else could I approach this?