Answer
$V = \frac{{81}}{2}\pi $
Work Step by Step
$$\eqalign{
& y = 9 - {x^2},{\text{ }}y = 0 \cr
& {\text{Applying the shell method about the }}y{\text{ - axis}} \cr
& V = 2\pi \int_a^b {x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& {\text{From the graph we can see that the volume revolving can be}} \cr
& {\text{considered revolving the curve }}9 - {x^2}{\text{ from the interval }}\left[ {0,3} \right] \cr
& {\text{Let }}f\left( x \right) = 9 - {x^2}{\text{ and }}g\left( x \right) = 0{\text{ on the interval }}\left[ {0,3} \right],{\text{ then}} \cr
& V = 2\pi \int_0^3 {x\left[ {9 - {x^2} - \left( 0 \right)} \right]} dx \cr
& V = 2\pi \int_0^3 {\left( {9x - {x^3}} \right)} dx \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {\frac{9}{2}{x^2} - \frac{1}{4}{x^4}} \right]_0^3 \cr
& V = 2\pi \left[ {\frac{9}{2}{{\left( 3 \right)}^2} - \frac{1}{4}{{\left( 3 \right)}^4}} \right] - 2\pi \left[ {\frac{9}{2}{{\left( 0 \right)}^2} - \frac{1}{4}{{\left( 0 \right)}^4}} \right] \cr
& V = 2\pi \left[ {\frac{{81}}{4}} \right] - 2\pi \left[ 0 \right] \cr
& V = \frac{{81}}{2}\pi \cr} $$
