Question

Find the moment of inertia I₂ for the solid above the x-y plane bounded by the paraboloid z = x² + y² and the cylinder x² + y² = 9, assuming the mean density to be constant C.

Answer 1

There are three solid objects here. 

1. Solid cylinder of radius R = 3 units.  R² = 9 (found from x²+y²=9)
    Enclosed between two planes z=0 and z=9=Z (as z=x²+y²=R²=9)
    M1 = π R² * Z *C = π R² Z C           (Z = height of cylinder)
    MOI about z axis = I₂ = M1 * R²/2 = πC R⁴ Z / 2

2. Solid enclosed inside Paraboloid: between z=0, z = 9 (=Z), z = x²+y²
    Consider a thin disc dm of radius r,  r² = x²+y² and height = dz
    MOI (I₂) about z-axis: = dm r²/2
              = (π r² dz) C* r²/2 = π C/2 * (x²+y²)² dz = πC/2 * z² dz
    MOI about z axis = integral from z= 0 to Z,  [πC/2 * z²] dz
             =πC Z³/6

3. MOI (I₂) of the solid between (outside and below) the paraboloid
    and axes z = 0, z =Z  : (like the open cup shape).
    
     MOI = MOI of full cylinder - MOI of solid paraboloid 
             = π C R⁴ Z/2  -  π C Z³ / 6 = π C Z [3R⁴ - Z²] /6
             = π C Z³ / 3         (as  R² = 9 = Z  in this problem.)
             = π C 243