Answer
$16 \pi$
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:
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (x) \times f(x) \ dx$
Now, $V=2\pi \int_{0}^{2} (x)(8-2x^2) \ dx\\= 2\pi \int_{0}^{2} (8x-2x^3) dx \\=2 \pi [4x^2-\dfrac{x^4}{2}]_{0}^{2} \\=2 \pi [4(2)^2-\dfrac{2^4}{2}] \\=16 \pi$
