Answer
$\dfrac{824 \pi}{15}$
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, $R_{outside}=6-(x^2+2)=4-x^2$ and $ R_{inside}= 6$
Now, $V=\pi \int_0^2 [(6)^2 - (4-x^2)^2] \ dx \\ = \pi \int_0^2 [36-x^4 -16+8x^2] \ dx \\=\pi \int_0^2 [-x^4+8x^2+20] \ dx \\=\pi (-\dfrac{x^5}{5}+\dfrac{8x^3}{3}+20 x )_0^2 \ dx \\=\dfrac{824 \pi}{15}$
