Answer
$8 \pi$
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, $V=\pi \int_2^6 [R(y)^2] \ dy \\ = \pi \int_2^6 [y-2] \ dy \\=\pi [\dfrac{y^2}{2}-2y ]_2^6 \\=\pi[(\dfrac{6^2}{2}-\dfrac{2^2}{2})-2(6-2)] \\=8 \pi$
