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: 13

$$V = \frac{{2048}}{{15}}\pi $$

$$\eqalign{ & {\text{We have that }}x = \sqrt y ,\,\,\,x = \frac{y}{4},\, \cr & x = \sqrt y \Rightarrow y = {x^2},\,\,\,\,x \geqslant 0 \cr & x = \frac{y}{4} \Rightarrow y = 4x \cr & {\text{Find the intersection points set }}f\left( x \right) = g\left( x \right) \cr & {x^2} = 4x \cr & {x^2} - 4x = 0 \cr & x\left( {x - 4} \right) = 0 \cr & {x_1} = 0{\text{ and }}{x_2} = 4 \cr & {\text{The intersection points are }}\left( {0,0} \right){\text{ and }}\left( {4,4} \right) \cr & {\text{Where }} \cr & 4x \geqslant {x^2}{\text{ on the interval 0}} \leqslant x \leqslant 4 \cr & \cr & {\text{The volume of the solid can be calculated using}} \cr & V = \int_0^4 {\pi \left( {{{\left[ {4x} \right]}^2} - {{\left[ {{x^2}} \right]}^2}} \right)dx} \cr & {\text{Thus}}{\text{,}} \cr & V = \pi \int_0^4 {\left( {16{x^2} - {x^4}} \right)dx} \cr & {\text{Integrating}} \cr & V = \pi \left[ {\frac{{16}}{3}{x^3} - \frac{1}{5}{x^5}} \right]_0^4 \cr & V = \pi \left[ {\frac{{16}}{3}{{\left( 4 \right)}^3} - \frac{1}{5}{{\left( 4 \right)}^5}} \right] - \pi \left[ 0 \right] \cr & V = \pi \left( {\frac{{2048}}{{15}}} \right) \cr & V = \frac{{2048}}{{15}}\pi \cr} $$