Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.3 Volumes By Cylindrical Shells - Exercises Set 5.3 - Page 370: 10

$$V = \frac{8}{3}\pi $$

$$\eqalign{ & {\text{We have }}y = 2x - {x^2},\,\,\,y = 0 \cr & {\text{Calculate the intersection point}}{\text{. Let }}y = y \cr & 2x - {x^2} = 0 \cr & x\left( {2 - x} \right) \cr & {x_1} = 0{\text{ and }}{x_2} = 2 \cr & 2x - {x^2} \geqslant 0{\text{ on the interval 0}} \leqslant x \leqslant 2 \cr & \cr & {\text{The volume of the solid can be calculated using cylindrical shells}} \cr & V = \int_a^b {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]dx} \cr & {\text{Let }}f\left( x \right) = 2x - {x^2}{\text{ and }}g\left( x \right) = 0 \cr & V = \int_0^2 {2\pi x\left[ {\left( {2x - {x^2}} \right) - \left( 0 \right)} \right]dx} \cr & V = 2\pi \int_0^2 {\left( {2{x^2} - {x^3}} \right)dx} \cr & V = 2\pi \left[ {\frac{2}{3}{x^3} - \frac{1}{4}{x^4}} \right]_0^2 \cr & {\text{Integrating}} \cr & V = 2\pi \left[ {\frac{2}{3}{{\left( 2 \right)}^3} - \frac{1}{4}{{\left( 2 \right)}^4}} \right] - 2\pi \left[ {\frac{2}{3}{{\left( 0 \right)}^3} - \frac{1}{4}{{\left( 0 \right)}^4}} \right] \cr & V = 2\pi \left[ {\frac{4}{3}} \right] - 0 \cr & V = \frac{8}{3}\pi \cr} $$