Answer
$\dfrac{32 \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_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 (4-4\sqrt {y-2} +y-2) \ dy \\=\pi [4y]_0^2 +\pi [\dfrac{y^2}{2}+2y-\dfrac{8(y-2)^{3/2}}{3}]_2^6 \\=8 \pi +\pi (\dfrac{26}{3}-6) \\=\dfrac{32 \pi}{3} $
