Chapter 6 - Applications of Integration - Review Exercises - Page 508: 27

\[V = \frac{8}{5}\pi \]

\[\begin{gathered} y = 1 + \sqrt x {\text{ and }}y = 1 - \sqrt x ,{\text{ }}x = 1 \hfill \\ \hfill \\ \left. {\text{a}} \right){\text{Integrating with respect to }}x \hfill \\ {\text{From the graph of the region shown below}} \hfill \\ 1 + \sqrt x > 1 - \sqrt x {\text{ on the interval }}\left[ {0,1} \right]{\text{ }} \hfill \\ {\text{Using the shell method about the }}y{\text{ - axis}} \hfill \\ V = \int_a^b {2\pi x\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \hfill \\ {\text{Let }}f\left( x \right) = 1 + \sqrt x {\text{ and }}g\left( x \right) = 1 - \sqrt x \hfill \\ V = \int_0^1 {2\pi x\left[ {\left( {1 + \sqrt x } \right) - \left( {1 - \sqrt x } \right)} \right]} dx \hfill \\ V = 2\pi \int_0^1 {x\left( {1 + \sqrt x - 1 + \sqrt x } \right)} dx \hfill \\ V = 2\pi \int_0^1 {\left( {2{x^{3/2}}} \right)} dx \hfill \\ V = 4\pi \int_0^1 {{x^{3/2}}} dx \hfill \\ {\text{Integrating}} \hfill \\ V = 4\pi \left[ {\frac{{{x^{5/2}}}}{{5/2}}} \right]_0^1 \hfill \\ V = \frac{{8\pi }}{5}\left[ {{x^{5/2}}} \right]_0^1 \hfill \\ V = \frac{{8\pi }}{5}\left[ {{{\left( 1 \right)}^{5/2}} - {{\left( 0 \right)}^{5/2}}} \right] \hfill \\ {\text{Simplifying}} \hfill \\ V = \frac{{8\pi }}{5} \hfill \\ \hfill \\ \left. {\text{b}} \right){\text{Integrating with respect to }}y \hfill \\ y = 1 + \sqrt x {\text{ and }}y = 1 - \sqrt x \Rightarrow x = {\left( {y - 1} \right)^2} \hfill \\ {\text{Using the washer method about the }}y{\text{ - axis}} \hfill \\ V = \int_c^d {\pi \left[ {p{{\left( y \right)}^2} - q{{\left( y \right)}^2}} \right]} dy \hfill \\ 1 \geqslant {\left( {y - 1} \right)^2}{\text{ on the interval }}\left[ {0,2} \right],{\text{ then}} \hfill \\ V = \int_0^2 {\pi \left[ {{{\left( 1 \right)}^2} - {{\left( {{{\left( {y - 1} \right)}^2}} \right)}^2}} \right]} dy \hfill \\ V = \pi \int_0^2 {\left( {1 - {{\left( {y - 1} \right)}^4}} \right)} dy \hfill \\ {\text{Integrating}} \hfill \\ V = \pi \left[ {y - \frac{{{{\left( {y - 1} \right)}^5}}}{5}} \right]_0^2 \hfill \\ V = \pi \left[ {2 - \frac{{{{\left( {2 - 1} \right)}^5}}}{5}} \right] - \pi \left[ {0 - \frac{{{{\left( {0 - 1} \right)}^5}}}{5}} \right] \hfill \\ {\text{Simplifying}} \hfill \\ V = \pi \left( {\frac{9}{5}} \right) - \pi \left( {\frac{1}{5}} \right) \hfill \\ V = \frac{8}{5}\pi \hfill \\ \end{gathered} \]