Answer
$\dfrac{32 \pi}{3}$
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, $V=2\pi \int_{0}^{4} y (2-\dfrac{y}{2}) \ dy\\= 2\pi \int_{0}^{4} (2y-\dfrac{y^2}{2}) \ dy \\=2 \pi [y^2-\dfrac{y^3}{6}]_{0}^{4} \\= 2\pi [16-\dfrac{64}{6}] \\=\dfrac{32 \pi}{3}$
