Chapter 6 - Applications of Integration - 6.4 Volume by Shells - 6.4 Exercises - Page 443: 41

$$V = \frac{4}{{15}}\pi $$

$$\eqalign{ & y = x,{\text{ }}y = {x^{1/3}},{\text{ in the first quadrant}}{\text{, revolved about }}x{\text{ - axis}} \cr & {\text{From the 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 & {x^{1/3}} \geqslant x{\text{ on the interval }}\left[ {0,1} \right]{\text{ }} \cr & {\text{Let }}f\left( x \right) = {x^{1/3}}{\text{ and }}g\left( x \right) = x \cr & {\text{Therefore}}{\text{,}} \cr & V = \int_0^1 {\pi \left[ {{{\left( {{x^{1/3}}} \right)}^2} - {{\left( x \right)}^2}} \right]} dx \cr & V = \pi \int_0^1 {\left( {{x^{2/3}} - {x^2}} \right)} dx \cr & {\text{Integrating}} \cr & V = \pi \left[ {\frac{{{x^{5/3}}}}{{5/3}} - \frac{{{x^3}}}{3}} \right]_0^1 \cr & V = \pi \left[ {\frac{3}{5}{{\left( 1 \right)}^{5/3}} - \frac{1}{3}{{\left( 1 \right)}^3}} \right] - \pi \left[ {\frac{3}{5}{{\left( 0 \right)}^{5/3}} - \frac{1}{3}{{\left( 0 \right)}^3}} \right] \cr & V = \pi \left[ {\frac{4}{{15}}} \right] - \pi \left[ 0 \right] \cr & V = \frac{4}{{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 \to x = y \cr & y = {x^{1/3}} \to x = {y^3} \cr & y \geqslant {y^3}{\text{ on the interval }}\left[ {0,1} \right]{\text{ }} \cr & {\text{Let }}p\left( y \right) = y{\text{ and }}q\left( y \right) = {y^3} \cr & {\text{Therefore}}{\text{,}} \cr & V = \int_0^1 {2\pi y\left[ {y - {y^3}} \right]} dy \cr & V = 2\pi \int_0^1 {\left( {{y^2} - {y^4}} \right)} dy \cr & {\text{Integrating}} \cr & V = 2\pi \left[ {\frac{1}{3}{y^3} - \frac{1}{5}{y^5}} \right]_0^1 \cr & V = 2\pi \left[ {\frac{1}{3}{{\left( 1 \right)}^3} - \frac{1}{5}{{\left( 1 \right)}^5}} \right] - 2\pi \left[ {\frac{1}{3}{{\left( 0 \right)}^3} - \frac{1}{5}{{\left( 0 \right)}^5}} \right] \cr & V = 2\pi \left( {\frac{2}{{15}}} \right) - 2\pi \left( 0 \right) \cr & V = \frac{4}{{15}}\pi \cr & {\text{Shell method is easier to apply}} \cr} $$