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:
$V=2 \pi \int_{m}^{n} (Radius) \times (height \ of \ the \ shell) \ dy=2 \pi \int_{m}^{n} (y) \times f(y) \ dy$
Now, $V=2\pi \int_{0}^{2} (y) (4y-y^3) \ dy \\= 2\pi \int_{0}^{2} (4y^2-y^4) \ dy \\=2 \pi \times [\dfrac{4y^3}{3}-\dfrac{y^5}{5}]_0^2 \\=2 \pi \times [\dfrac{4(2^3)}{3}-\dfrac{(2^5)}{5} ] \\=\dfrac{128 \pi}{15}$
