Answer
$56 \pi$
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, $R_{outside}=2-(-3)=5$ and $ R_{inside}=\sqrt{y-2}-(-3)=\sqrt {y-2}+3$
Now, $V=\pi \int_0^2 [(5)^2-(3)^2] \ dy +\pi \int_2^6 [(5)^2-( \sqrt {y-2}+3)^2 ] \ dy \\=\pi \int_0^2 16 \ dy + \pi \int_2^6 (25-4+2-6\sqrt {y-2} -9) \ dy \\=\pi [16 y]_0^2 +\pi [18y -\dfrac{y^2}{2}-4(y-2)^{3/2}]_2^6 \\=32 \pi +\pi (58-34) \\=56 \pi$
