Answer
$\dfrac{96 \pi }{5}$
Work Step by Step
The Washer method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq g(x) \geq 0$ on the interval $[m,n]$, then the volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by:
$V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$
where, $R_{outside}=(y+2)^2 $ and $ R_{inside}=\dfrac{y^2}{2}+2$
Now, $V=\pi \int_{0}^{4} [(y+2)^2-(\dfrac{y^2}{2}+2)^2] \ dy \\ = \pi (y^2+4y+4) -(\dfrac{y^4}{16}+y^2+4)] \ dy \\=\pi \int_0^4 [4y-\dfrac{y^4}{16}] \ dy \\=\dfrac{96 \pi }{5}$
