Answer
$\dfrac{128 \pi}{5}$
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}=x^2+2-2=x^2 $ and $ R_{inside}=6-2=4$
Now, $V=\pi \int_0^2 [(4)^2 -(x^2)^2] \ dx \\ = \pi \int_0^2 [16 -x^4 ] \ dx \\=\pi [16x -\dfrac{x^5}{5} ]_0^2 \ dx \\=\pi (32 -\dfrac{2^5}{5} ) \ dx \\=\dfrac{128 \pi}{5}$
