Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.2 Volumes By Slicing; Disks And Washers - Exercises Set 5.2 - Page 362: 12

$$V = \frac{{1296}}{5}\pi $$

$$\eqalign{ & {\text{We have that }}f\left( x \right) = 9 - {x^2},\,\,\,\,\,g\left( x \right) = 0 \cr & {\text{Find the intersection points set }}f\left( x \right) = g\left( x \right) \cr & 9 - {x^2} = 0 \cr & {x^2} = 9 \cr & x = \pm \sqrt 9 \cr & x = \pm 3 \cr & {\text{The volume of the solid can be calculated using}} \cr & V = \int_a^b {\pi \left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)dx} \cr & {\text{Thus}}{\text{,}} \cr & V = \pi \int_{ - 3}^3 {\left( {{{\left[ {9 - {x^2}} \right]}^2} - {{\left[ 0 \right]}^2}} \right)dx} \cr & V = \pi \int_{ - 3}^3 {\left( {81 - 18{x^2} + {x^4}} \right)dx} \cr & {\text{Integrating}} \cr & V = \pi \left[ {81x - 6{x^3} + \frac{1}{5}{x^5}} \right]_{ - 3}^3 \cr & V = \pi \left[ {81\left( 3 \right) - 6{{\left( 3 \right)}^3} + \frac{1}{5}{{\left( 3 \right)}^5}} \right] - \pi \left[ {81\left( { - 3} \right) - 6{{\left( { - 3} \right)}^3} + \frac{1}{5}{{\left( { - 3} \right)}^5}} \right] \cr & V = \pi \left( {\frac{{648}}{5}} \right) - \pi \left( { - \frac{{648}}{5}} \right) \cr & V = \frac{{1296}}{5}\pi \cr} $$