Answer
$\dfrac{704 \pi}{15}$
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:
$V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$
$V=\pi \int_0^2 [(6)^2 -(x^2+2)^2] \ dx \\ = \pi \int_0^2 [36 -x^4 -4-4x^2 ] \ dx \\=\pi \int_0^2[-x^4 -4x^2+32] \ dx \\=\pi [-\dfrac{x^5}{5} -\dfrac{4 x^3}{3}+32 x]_0^2 \\=\dfrac{704 \pi}{15}$
