Chapter 6 - Applications of Integration - 6.4 Volume by Shells - 6.4 Exercises - Page 443: 42

$$V = \frac{{32}}{{15}}\pi $$

$$\eqalign{ & y = {x^2},{\text{ }}y = 2 - x,{\text{ and }}x = 0,{\text{ revolved about the }}x{\text{ - axis}} \cr & {\text{Graph of the region shown below}} \cr & \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 & 2 - x \geqslant {x^2}{\text{ on the interval }}\left[ {0,1} \right]{\text{ }} \cr & {\text{Let }}f\left( x \right) = 2 - x{\text{ and }}g\left( x \right) = {x^2} \cr & {\text{Therefore}}{\text{,}} \cr & V = \int_0^1 {\pi \left[ {{{\left( {2 - x} \right)}^2} - {{\left( {{x^2}} \right)}^2}} \right]} dx \cr & V = \pi \int_0^1 {\left( {4 - 4x + {x^2} - {x^4}} \right)} dx \cr & {\text{Integrating}} \cr & V = \pi \left[ {4x - 2{x^2} + \frac{1}{3}{x^3} - \frac{1}{5}{x^5}} \right]_0^1 \cr & V = \pi \left[ {4\left( 1 \right) - 2{{\left( 1 \right)}^2} + \frac{1}{3}{{\left( 1 \right)}^3} - \frac{1}{5}{{\left( 1 \right)}^5}} \right] - \pi \left[ 0 \right] \cr & V = \pi \left( {\frac{{32}}{{15}}} \right) \cr & V = \frac{{32}}{{15}}\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 = {x^2} \to x = \sqrt y \cr & y = 2 - x \to x = 2 - y \cr & \sqrt y \geqslant 0{\text{ on the interval }}\left[ {0,1} \right]{\text{ and }}2 - y \geqslant 0{\text{ on }}\left[ {1,2} \right] \cr & {\text{Therefore}}{\text{,}} \cr & V = \int_0^1 {2\pi y\left( {\sqrt y - 0} \right)} dy + \int_1^2 {2\pi y\left( {2 - y - 0} \right)} dy \cr & V = 2\pi \int_0^1 {{y^{3/2}}} dy + 2\pi \int_1^2 {\left( {2y - {y^2}} \right)} dy \cr & {\text{Integrating}} \cr & V = 2\pi \left[ {\frac{{{y^{5/2}}}}{{5/2}}} \right]_0^1 + 2\pi \left[ {{y^2} - \frac{1}{3}{y^3}} \right]_1^2 \cr & V = \frac{4}{5}\pi \left( 1 \right) + 2\pi \left[ {{{\left( 2 \right)}^2} - \frac{1}{3}{{\left( 2 \right)}^3}} \right] - 2\pi \left[ {{{\left( 1 \right)}^2} - \frac{1}{3}{{\left( 1 \right)}^3}} \right] \cr & V = \frac{4}{5}\pi + \frac{8}{3}\pi - \frac{4}{3}\pi \cr & V = \frac{{32}}{{15}}\pi \cr & {\text{Washer method is easier to apply}} \cr} $$