Answer
$\dfrac{128 \pi}{15}$
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, $V=2\pi \int_{0}^{2} (x)(4x-x^3) \ dx\\= 2\pi \int_{0}^{2} (4x^2-x^4) dx \\=2 \pi [\dfrac{4 x^3}{3}-\dfrac{x^5}{5}]_{0}^{2} \\=2 \pi [\dfrac{4 (2^3)}{3}-\dfrac{(2^5)}{5}] \\=\dfrac{128 \pi}{15}$
