Answer
$8 \pi$
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, $V=2\pi \int_{0}^{2} (x+2) (2x-x^2) \ dx \\= 2\pi \int_{0}^{2} (2x^2-x^3+4x-2x^2) \ dx \\= 2 \pi \int_0^2 (4x-x^3) \ dx \\=2 \pi \times [2x^2 -\dfrac{x^4}{4}]_{0}^{2} \\=2 \pi [(2)(4)-\dfrac{(2^4)}{4}] \\= 8 \pi$
