Answer
$\dfrac{40 \pi}{3}$
Work Step by Step
The volume of a revolution can be calculated as:
$V=\pi \int_{m}^{n} (R^2_{outside}-r^2_{inside}) \ dy$
where, $R_{outside}=2-0=2$ and $ R_{inside}=2-\sqrt{y-2}$
Now, $V=\pi \int_2^6 [(2)^2-(2- \sqrt {y-2})^2 ] \ dy \\=\pi \int_2^6 (2-y+4 \sqrt {y-2}) \ dy\\=\pi [2y-\dfrac{y^2}{2}+\dfrac{8 (y-2)^{3/2}}{3}]_2^6 \\=\pi (\dfrac{46}{3}-2) \\=\dfrac{40 \pi}{3}$
