Chapter 6 - Applications of Integration - 6.3 Volume by Slicing - 6.3 Exercises - Page 432: 42

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

$$\eqalign{ & {\text{Let the graphs }}y = 4 - 2x{\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[ {0,2} \right] \cr & 4 - 2x \geqslant 0,{\text{ for }}\left[ {0,2} \right] \to f\left( x \right) = 4 - 2x{\text{ and }}g\left( x \right) = 0{\text{ }} \cr & {\text{We can represent the volume as:}} \cr & V = \int_0^2 {\pi \left[ {{{\left( {4 - 2x} \right)}^2} - {{\left( 0 \right)}^2}} \right]} dx \cr & V = \pi \int_0^2 {{{\left( {4 - 2x} \right)}^2}} dx \cr & V = - \frac{1}{2}\pi \int_0^2 {{{\left( {4 - 2x} \right)}^2}\left( { - 2} \right)} dx \cr & {\text{Integrating we obtain}} \cr & V = - \frac{1}{2}\pi \left[ {\frac{{{{\left( {4 - 2x} \right)}^3}}}{3}} \right]_0^2 \cr & V = - \frac{1}{6}\pi \left[ {{{\left( {4 - 2\left( 2 \right)} \right)}^3} - {{\left( {4 - 2\left( 0 \right)} \right)}^3}} \right] \cr & V = - \frac{1}{6}\pi \left[ { - 64} \right] \cr & V = \frac{{32}}{3}\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[ {0,4} \right] \cr & y = 4 - 2x \to x = 2 - \frac{1}{2}y{\text{ and }}x = 0 \cr & 2 - \frac{1}{2}y \geqslant 0{\text{ for }}\left[ {0,4} \right] \to p\left( y \right) = 2 - \frac{1}{2}y{\text{ and }}q\left( y \right) = 0 \cr & {\text{We can represent the volume as:}} \cr & V = \int_0^4 {\pi \left[ {{{\left( {2 - \frac{1}{2}y} \right)}^2} - {{\left( 0 \right)}^2}} \right]} dy \cr & V = \int_0^4 {\pi {{\left( {2 - \frac{1}{2}y} \right)}^2}} dy \cr & V = - 2\pi \int_0^4 {{{\left( {2 - \frac{1}{2}y} \right)}^2}} \left( { - \frac{1}{2}} \right)dy \cr & {\text{Integrating we obtain}} \cr & V = - \frac{2}{3}\pi \left[ {{{\left( {2 - \frac{1}{2}y} \right)}^3}} \right]_0^4 \cr & V = - \frac{2}{3}\pi \left[ {{{\left( 0 \right)}^3} - {{\left( 2 \right)}^3}} \right]_0^4 \cr & V = \frac{{16}}{3}\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} $$