Let x=rcos(θ) and y=rsin(θ). The upper bound of the solid is z=16-4(x^2+y^2) = 16 - 4r^2 and the lower bound of the solid is z=0. That is, 0<=z<=16-4r^2. Furthermore, 0=16-4(x^2+y^2) yields x^2+y^2=4 which indicates that the projection of the solid onto the xy- plane is the circular region with radius 2, that is, 0<=r<=2 and 0<=θ<=2pi. Therefore, the triple integral can be written into
\int_0^{2π} \int _0^2 \int_0^{16-4r^2} r*rdzdrdθ = \int_0^{2π} \int_0^2 r^2(16-4r^2) drdθ
= \int_0^{2ππ} (256/15) dθ
= (512π)/15.
