Answer
$$V = \frac{5}{2}\pi $$
Work Step by Step
$$\eqalign{
& y = 9 - {x^2},{\text{ }}x = \sqrt {9 - y} \cr
& {\text{Apply the washer method}} \cr
& V = \pi \int_c^d {\left( {{{\left[ {R\left( y \right)} \right]}^2} - {{\left[ {r\left( y \right)} \right]}^2}} \right)} dy \cr
& {\text{So}},{\text{ the volume of the solid of revolution is}} \cr
& V = \pi \int_2^3 {\left( {{{\left[ {\sqrt {9 - y} } \right]}^2} - {{\left[ 2 \right]}^2}} \right)} dy \cr
& V = \pi \int_2^3 {\left( {9 - y - 4} \right)} dy \cr
& V = \pi \int_2^3 {\left( {5 - y} \right)} dy \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {\frac{{{{\left( {5 - y} \right)}^2}}}{2}} \right]_3^2 \cr
& V = \pi \left[ {\frac{{{{\left( {5 - 2} \right)}^2}}}{2} - \frac{{{{\left( {5 - 3} \right)}^2}}}{2}} \right] \cr
& V = \pi \left( {\frac{9}{2} - 2} \right) \cr
& V = \frac{5}{2}\pi \cr} $$
