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