Chapter 6 - Applications of Integration - 6.3 Volume by Slicing - 6.3 Exercises - Page 432: 43

$$x{\text{ - axis}}$$

$$\eqalign{ & {\text{Let the graphs }}y = 1 - {x^3}{\text{ and }}y = 0{\text{ }}\left( {x{\text{ - axis}}} \right),x = 0{\text{ }}\left( {y{\text{ - axis}}} \right) \cr & \cr & *{\text{ Revolving the region about the }}x{\text{ - axis:}} \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 the interval of integration }}\left[ {a,b} \right]{\text{ is }}\left[ {0,1} \right] \cr & 1 - {x^3} \geqslant 0,{\text{ for }}\left[ {0,1} \right] \to f\left( x \right) = 1 - {x^3}{\text{ and }}g\left( x \right) = 0{\text{ }} \cr & {\text{We can represent the volume as:}} \cr & V = \int_0^1 {\pi \left[ {{{\left( {1 - {x^3}} \right)}^2} - {{\left( 0 \right)}^2}} \right]} dx \cr & V = \pi \int_0^1 {{{\left( {1 - {x^3}} \right)}^2}} dx \cr & {\text{Integrating we obtain}} \cr & V = \frac{3}{4}\pi \cr & \cr & *{\text{ Revolving the region about the }}y{\text{ - axis:}} \cr & {\text{Using the Washer Method about the }}y{\text{ - axis:}} \cr & V = \int_c^d {\pi \left[ {p{{\left( y \right)}^2} - g{{\left( y \right)}^2}} \right]} dy \cr & {\text{From the graph the interval of integration }}\left[ {c,d} \right]{\text{ is }}\left[ {0,1} \right] \cr & y = 1 - {x^3} \to x = \root 3 \of {1 - y} {\text{ and }}x = 0 \cr & \root 3 \of {1 - y} \geqslant 0{\text{ for }}\left[ {0,1} \right] \to p\left( y \right) = \root 3 \of {1 - y} {\text{ and }}q\left( y \right) = 0 \cr & {\text{We can represent the volume as:}} \cr & V = \int_0^1 {\pi \left[ {{{\left( {\root 3 \of {1 - y} } \right)}^2} - {{\left( 0 \right)}^2}} \right]} dy \cr & V = \int_0^1 {\pi {{\left( {1 - y} \right)}^{2/3}}} dy \cr & {\text{Integrating we obtain}} \cr & V = - \frac{3}{5}\pi \left[ {{{\left( {1 - y} \right)}^{5/3}}} \right]_0^1 \cr & V = - \frac{3}{5}\pi \left[ {{{\left( {1 - 1} \right)}^{5/3}} - {{\left( {1 - 0} \right)}^{5/3}}} \right] \cr & V = - \frac{3}{5}\pi \left[ {{{\left( 0 \right)}^{5/3}} - {{\left( 1 \right)}^{5/3}}} \right] \cr & V = \frac{3}{5}\pi \cr & \cr & \frac{3}{4}\pi > \frac{3}{5}\pi \cr & \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 & \cr & {\text{Graph}} \cr} $$