Chapter 6 - Applications of Integration - 6.4 Volume by Shells - 6.4 Exercises - Page 443: 46

$$V = \frac{1}{3}\pi $$

$$\eqalign{ & y = \frac{6}{{x + 3}},{\text{ }}y = 2 - x \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 & 2 - x \geqslant \frac{6}{{x + 3}}{\text{ on the interval }}\left[ { - 1,0} \right]{\text{ }} \cr & {\text{Therefore}}{\text{,}} \cr & V = \int_{ - 1}^0 {\pi \left[ {{{\left( {2 - x} \right)}^2} - {{\left( {\frac{6}{{x + 3}}} \right)}^2}} \right]} dx \cr & V = \int_{ - 1}^0 {\pi \left( {4 - 4x + {x^2} - \frac{{36}}{{{{\left( {x + 3} \right)}^2}}}} \right)} dx \cr & {\text{Integrating}} \cr & V = \pi \left[ {4x - 2{x^2} + \frac{1}{3}{x^3} + \frac{{36}}{{x + 3}}} \right]_{ - 1}^0 \cr & V = \pi \left[ {4\left( 0 \right) - 2{{\left( 0 \right)}^2} + \frac{1}{3}{{\left( 0 \right)}^3} + \frac{{36}}{{\left( 0 \right) + 3}}} \right] \cr & - \pi \left[ {4\left( { - 1} \right) - 2{{\left( { - 1} \right)}^2} + \frac{1}{3}{{\left( { - 1} \right)}^3} + \frac{{36}}{{\left( { - 1} \right) + 3}}} \right] \cr & {\text{Simplifying}} \cr & V = \pi \left( {12} \right) - \pi \left( {\frac{{35}}{3}} \right) \cr & V = \frac{1}{3}\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{6}{{x + 3}} \to x = \frac{6}{y} - 3 \cr & {\text{ }}y = 2 - x \to x = 2 - y \cr & 2 - y \geqslant \frac{6}{y} - 3{\text{ on the interval }}\left[ { - 2,3} \right]{\text{ }} \cr & {\text{Therefore}}{\text{,}} \cr & V = \int_2^3 {2\pi y\left( {2 - y - \frac{6}{y} + 3} \right)} dy \cr & V = 2\pi \int_2^3 {\left( {5y - {y^2} - 6} \right)} dy \cr & {\text{Integrating}} \cr & V = 2\pi \left[ {\frac{5}{2}{y^2} - \frac{1}{3}{y^3} - 6y} \right]_2^3 \cr & V = 2\pi \left[ {\frac{5}{2}{{\left( 3 \right)}^2} - \frac{1}{3}{{\left( 3 \right)}^3} - 6\left( 3 \right)} \right] - 2\pi \left[ {\frac{5}{2}{{\left( 2 \right)}^2} - \frac{1}{3}{{\left( 2 \right)}^3} - 6\left( 2 \right)} \right] \cr & V = 2\pi \left[ { - \frac{9}{2}} \right] - 2\pi \left[ { - \frac{{14}}{3}} \right] \cr & V = - 9\pi + \frac{{28}}{3}\pi \cr & V = \frac{1}{3} \cr & {\text{Shell method is easier to apply}} \cr} $$