Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.3 Volumes By Cylindrical Shells - Exercises Set 5.3 - Page 370: 9

$$V = \frac{{20}}{3}\pi $$

$$\eqalign{ & {\text{We have }}y = 2x - 1,\,\,\,y = - 2x + 3,\,\,\,x = 2 \cr & {\text{Calculate the intersection point}}{\text{. Let }}y = y \cr & 2x - 1 = - 2x + 3 \cr & 4x = 4 \cr & x = 1 \cr & 2x - 1 \geqslant x{\text{ on the interval }}1 \leqslant x \leqslant 2 \cr & \cr & {\text{The volume of the solid can be calculated using cylindrical shells}} \cr & V = \int_a^b {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]dx} \cr & {\text{Let }}f\left( x \right) = 2x - 1{\text{ and }}g\left( x \right) = - 2x + 3 \cr & V = \int_1^2 {2\pi x\left[ {\left( {2x - 1} \right) - \left( { - 2x + 3} \right)} \right]dx} \cr & V = 2\pi \int_1^2 {x\left( {2x - 1 + 2x - 3} \right)dx} \cr & V = 2\pi \int_1^2 {x\left( {4x - 4} \right)dx} \cr & V = 2\pi \int_1^2 {\left( {4{x^2} - 4x} \right)dx} \cr & {\text{Integrating}} \cr & V = 2\pi \left[ {\frac{4}{3}{x^3} - 2{x^2}} \right]_1^2 \cr & V = 2\pi \left[ {\frac{4}{3}{{\left( 2 \right)}^3} - 2{{\left( 2 \right)}^2}} \right] - 2\pi \left[ {\frac{4}{3}{{\left( 1 \right)}^3} - 2{{\left( 1 \right)}^2}} \right] \cr & V = 2\pi \left( {\frac{8}{3}} \right) - 2\pi \left( { - \frac{2}{3}} \right) \cr & V = \frac{{20}}{3}\pi \cr} $$