Answer
$\dfrac{32 \pi}{3}$
Work Step by Step
The shell method to compute the volume of a region: The volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the y-axis is given by:
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (y) \times f(y) \ dy$
Now, $V=2\pi \int_{0}^{2} (2-x) (x^2+2) \ dx\\= 2 \pi \int_0^2 (-x^3-2x+2x^2+4) \ dx \\= 2\pi [-\dfrac{x^4}{4}-x^2+\dfrac{2x^3}{3}+4x]_0^2 \\=2 \pi [-\dfrac{16}{4}-4+\dfrac{16}{3}+8-0] \\=\dfrac{32 \pi}{3}$
