Answer
$\dfrac{408 \pi}{5} $
Work Step by Step
The volume of a revolution can be calculated as:
$V=\pi \int_{m}^{n} (R^2_{outer}-r^2_{inner}) \ dy$
where, $R_{outer}=8-0=8$ and $ R_{inner}=8-(x^2+2)=6-x^2$
Now, $V=\pi \int_0^2 [(8)^2-(6-x^2)^2] \ dx \\=\pi \int_0^2 (64-36+12x^2-x^4) \ dx \\=\pi \int_0^2 (28+12x^2-x^4) \ dx \\= \pi (28 x+4x^3-\dfrac{x^5}{5}]_0^2 \\=\dfrac{408 \pi}{5} $
