How to do the moment of inertia of a cylinder

Determine the moment of inertia of a cylinder of radius 0.51 m, height 1.9 m and density (0.729 - 0.3645r + 0.371r^2) kg/m^3 about the center. Answer in units of kg*m^2.

I know I have to integrate using the density and I=int[r^2], but I don't know how to go from there.
Help would be greatly appreciated!!!

Remember your respective term in the integral...i.e. the d-whatever term.

In rotational inertia, that term is dm. An infinitesimal mass unit.

You will want your infinitesimal mass unit to be a thin hollow cylinder, at radial position r, of thickness dr, of mass dm, and of the same height as the cylinder.

EQUATE dm to the mass of this hollow cylinder in terms of H, r, dr, and density rho(r). You should get:
dm = 2*pi*rho(r)*H*r*dr

Multiply the circumference with dr, and that gives you differential cross sectional area.
Multiply that by height to get differential volume
Multiply by local density to get differential mass



Re-construct the integral. For convenience, pull out all coefficients:
I = 2*pi*H*integral(rho(r)*r^2 * r dr)

I = 2*pi*H*integral(rho(r) * r^3 dr)


I will leave the rest up to you.

Keep in mind, it may not actually be possible to evaluate the integral pencil-and-paper. You may need to consult a numeric integration technique. Many of these are embedded in computer mathematics software You can consult www.wolframalpha.com, if you don't have alternatives.

I don't really know how to integrate well, but I do know that the moment of inertial through the center of a cylinder is (1/2)MR^2

You can do the math (I'm lazy)

It makes sense if you already know how to find the moment of inertia for a circle.