Answer
$\dfrac{776 \pi}{15}$
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}=2+x^2+2=x^2+4 $ and $ R_{inside}=4$
Now, $V=\pi \int_0^2 [(x^2+4)^2-(2)^2] \ dx \\ = \pi \int_0^2 [x^4 +16+8x^2-4] \ dx \\=\pi \int_0^2 [x^4+8x^2+12] \ dx \\=\pi (\dfrac{x^5}{5}+\dfrac{8x^3}{3}+12x )_0^2 \ dx \\=\dfrac{776 \pi}{15}$
