Answer
$16 \pi $
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}=\sqrt{y-2}$
Now, $V=\pi \int_0^2 (2)^2 \ dy +\pi \int_2^6 [(2)^2-( \sqrt {y-2})^2 ] \ dy \\=\pi \int_0^2 4 \ dy + \pi \int_2^6 (6-y) \ dy \\=\pi [4y]_0^2 +\pi [6y -\dfrac{y^2}{2}]_2^6 \\=8 \pi +\pi (18-10) \\=16 \pi $
