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: 11

$$V = \frac{{256}}{3}\pi $$

$$\eqalign{ & {\text{We have that }}f\left( x \right) = \sqrt {25 - {x^2}} ,\,\,\,x = 0,\,\,\,\,g\left( x \right) = 3 \cr & {\text{Find the intersection points set }}f\left( x \right) = g\left( x \right) \cr & \sqrt {25 - {x^2}} = 3 \cr & 25 - {x^2} = 9 \cr & {x^2} = 16 \cr & x = \pm \sqrt {16} \cr & x = \pm 4 \cr & {\text{The volume of the solid can be calculated using}} \cr & V = \int_a^b {\pi \left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)dx} \cr & {\text{Thus}}{\text{,}} \cr & V = \pi \int_{ - 4}^4 {\left( {{{\left[ {\sqrt {25 - {x^2}} } \right]}^2} - {{\left[ 3 \right]}^2}} \right)dx} \cr & V = \pi \int_{ - 4}^4 {\left( {25 - {x^2} - 9} \right)dx} \cr & V = \pi \int_{ - 4}^4 {\left( {16 - {x^2}} \right)dx} \cr & {\text{Integrating}} \cr & V = \pi \left[ {16x - \frac{{{x^3}}}{3}} \right]_{ - 4}^4 \cr & V = \pi \left[ {16\left( 4 \right) - \frac{{{{\left( 4 \right)}^3}}}{3}} \right] - \pi \left[ {16\left( { - 4} \right) - \frac{{{{\left( { - 4} \right)}^3}}}{3}} \right] \cr & V = \pi \left( {\frac{{128}}{3}} \right) - \pi \left( { - \frac{{128}}{3}} \right) \cr & V = \frac{{256}}{3}\pi \cr} $$