Chapter 6 - Applications of Integration - 6.4 Volume by Shells - 6.4 Exercises - Page 443: 43

$$V = \frac{8}{{27}}\pi $$

$$\eqalign{ & y = \frac{1}{{x + 1}},{\text{ }}y = 1 - \frac{x}{3},{\text{ revolved about the }}x{\text{ - axis}} \cr & {\text{Using the washer method about the }}x{\text{ - axis}} \cr & V = \int_a^b {\pi \left[ {f{{\left( x \right)}^2} - g{{\left( x \right)}^2}} \right]} dx \cr & {\text{From the graph of the region shown below}} \cr & 1 - \frac{x}{3} \geqslant \frac{1}{{x + 1}}{\text{ on the interval }}\left[ {0,2} \right]{\text{ }} \cr & {\text{Let }}f\left( x \right) = 1 - \frac{x}{3}{\text{ and }}g\left( x \right) = \frac{1}{{x + 1}} \cr & {\text{Therefore}}{\text{,}} \cr & V = \int_0^2 {\pi \left[ {{{\left( {1 - \frac{x}{3}} \right)}^2} - {{\left( {\frac{1}{{x + 1}}} \right)}^2}} \right]} dx \cr & V = \pi \int_0^2 {\left( {1 - \frac{{2x}}{3} + \frac{{{x^2}}}{9} - \frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right)} dx \cr & {\text{Integrating}} \cr & V = \pi \left[ {x - \frac{{{x^2}}}{3} + \frac{{{x^3}}}{{27}} + \frac{1}{{x + 1}}} \right]_0^2 \cr & V = \pi \left[ {2 - \frac{{{2^2}}}{3} + \frac{{{2^3}}}{{27}} + \frac{1}{{2 + 1}}} \right] - \pi \left[ {0 + \frac{1}{{0 + 1}}} \right] \cr & V = \pi \left( {\frac{{35}}{{27}}} \right) - \pi \cr & V = \frac{8}{{27}}\pi \cr & \cr & {\text{Using the shell method about the }}x{\text{ - axis}} \cr & V = \int_c^d {2\pi y\left[ {p\left( y \right) - q\left( y \right)} \right]} dy \cr & y = \frac{1}{{x + 1}} \to x = \frac{1}{y} - 1 \cr & {\text{ }}y = 1 - \frac{x}{3} \to x = 3 - 3y \cr & {\text{Find the intersection points}}{\text{, let }}y = y \cr & \frac{1}{y} - 1 = 3 - 3y \cr & y = 1{\text{ and }}y = \frac{1}{3} \cr & 3 - 3y \geqslant \frac{1}{y} - 1{\text{ on the interval }}\left[ {\frac{1}{3},1} \right]{\text{ }} \cr & {\text{Therefore}}{\text{,}} \cr & V = \int_{1/3}^1 {2\pi y\left[ {\left( {3 - 3y} \right) - \left( {\frac{1}{y} - 1} \right)} \right]} dy \cr & V = 2\pi \int_{1/3}^1 {y\left( {3 - 3y - \frac{1}{y} + 1} \right)} dy \cr & V = 2\pi \int_{1/3}^1 {y\left( {4 - 3y - \frac{1}{y}} \right)} dy \cr & V = 2\pi \int_{1/3}^1 {\left( {4y - 3{y^2} - 1} \right)} dy \cr & {\text{Integrating}} \cr & V = 2\pi \left[ {2{y^2} - {y^3} - y} \right]_{1/3}^1 \cr & V = 2\pi \left[ {2{{\left( 1 \right)}^2} - {{\left( 1 \right)}^3} - \left( 1 \right)} \right] - 2\pi \left[ {2{{\left( {\frac{1}{3}} \right)}^2} - {{\left( {\frac{1}{3}} \right)}^3} - \left( {\frac{1}{3}} \right)} \right] \cr & V = 2\pi \left( 0 \right) - 2\pi \left( { - \frac{4}{{54}}} \right) \cr & V = \frac{8}{{27}}\pi \cr & {\text{Shell method is easier to apply}} \cr} $$