Chapter 6 - Applications of the Integral - 6.3 Volumes of Revolution - Exercises - Page 305: 31

$40 \pi$

The volume of a revolution can be calculated as: $V=\pi \int_{m}^{n} (R^2_{outside}-r^2_{inside}) \ dy$ where, $R_{outside}=\sqrt {y-2}-(-3)=\sqrt {y-2}+3$ and $ R_{inside}=0-(-3)=3$ Now, $V=\pi \int_2^6 (\sqrt {y-2}+3)^2 -(3)^2 \ dy \\=\pi \int_2^6 (y+6\sqrt {y-2}-2) \ dy\\=\pi [\dfrac{y^2}{2}+4 (y-2)^{3/2} -2y]_2^6 \\=\pi (38+2) \\=40 \pi$