Chapter 6 - Applications of Integration - 6.3 Volume by Slicing - 6.3 Exercises - Page 432: 44

$$\eqalign{ & {\text{ The volume generated revolving about the }}x{\text{ - axis}} \cr & {\text{is greater than the volume generated revolving about the }}y{\text{ - axis}} \cr} $$

$$\eqalign{ & {\text{Let the functions}}{\text{, }}y = {x^2},{\text{ }}y = \sqrt {8x} \cr & {\text{The graph of the region }}R{\text{ is shown below}}{\text{.}} \cr & \cr & {\text{Revolving the region about the }}x{\text{ - axis}}{\text{, using the}} \cr & {\text{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{Let }}f\left( x \right) = \sqrt {8x} {\text{ and }}g\left( x \right) = {x^2}{\text{ on the interval }}\left[ {0,2} \right] \cr & V = \int_0^2 {\pi \left[ {{{\left( {\sqrt {8x} } \right)}^2} - {{\left( {{x^2}} \right)}^2}} \right]} dx \cr & V = \pi \int_0^2 {\left( {8x - {x^4}} \right)} dx \cr & {\text{Integrating}} \cr & V = \pi \left[ {4{x^2} - \frac{1}{5}{x^5}} \right]_0^2 \cr & V = \pi \left[ {4{{\left( 2 \right)}^2} - \frac{1}{5}{{\left( 2 \right)}^5}} \right] - \pi \left[ {4{{\left( 0 \right)}^2} - \frac{1}{5}{{\left( 0 \right)}^5}} \right] \cr & V = \pi \left( {\frac{{48}}{5}} \right) - 0 \cr & V = \frac{{48\pi }}{5} \cr & \cr & {\text{Revolving the region about the }}y{\text{ - axis}}{\text{, using the}} \cr & {\text{Washer Method about the }}y{\text{ - axis }} \cr & V = \int_c^d {\pi \left[ {p{{\left( y \right)}^2} - q{{\left( y \right)}^2}} \right]} dy \cr & y = {x^2}{\text{ }} \to x = \sqrt y {\text{, }}p\left( y \right) = \sqrt y \cr & y = \sqrt {8x} {\text{ }} \to x = \frac{1}{8}{y^2},{\text{ }}q\left( y \right) = \frac{1}{8}{y^2} \cr & \sqrt y \geqslant \frac{1}{8}{y^2}{\text{ on the interval }}\left[ {0,4} \right] \cr & {\text{Therefore}} \cr & V = \int_0^4 {\pi \left[ {{{\left( {\sqrt y } \right)}^2} - {{\left( {\frac{1}{8}{y^2}} \right)}^2}} \right]} dy \cr & V = \pi \int_0^4 {\left( {y - \frac{1}{{64}}{y^4}} \right)} dy \cr & {\text{Integrating}} \cr & V = \pi \left[ {\frac{1}{2}{y^2} - \frac{1}{{256}}{y^5}} \right]_0^4 \cr & V = \pi \left[ {\frac{1}{2}{{\left( 4 \right)}^2} - \frac{1}{{256}}{{\left( 4 \right)}^5}} \right] - \pi \left[ {\frac{1}{2}{{\left( 0 \right)}^2} - \frac{1}{{256}}{{\left( 0 \right)}^5}} \right] \cr & V = \pi \left( 4 \right) - 0 \cr & V = 4\pi \cr & \cr & \frac{{48\pi }}{5} > 4\pi \cr & {\text{Therefore}}{\text{, }} \cr & {\text{Therefore}}{\text{, the volume generated revolving about the }}x{\text{ - axis}} \cr & {\text{is greater than the volume generated revolving about the }}y{\text{ - axis}} \cr} $$