Chapter 6 - Section 6.2 - Volume - 6.2 Exercises - Page 456: 19

$V = \frac{5}{{14}}\pi $

$$\eqalign{ & y = {x^3},{\text{ }}y = \sqrt x ;{\text{ about the }}x{\text{ - axis}} \cr & {\text{Graph of the region shown below }}\left( {{\text{Using Geogebra}}} \right) \cr & \cr & {\text{For this exercise the inner radius is:}} \cr & {r_i} = {x^3} - 0 \cr & {r_i} = {x^3} \cr & {\text{And the outer radius is given by:}} \cr & {r_o} = \sqrt x - 0 \cr & {r_o} = \sqrt x \cr & {\text{The cross - sectional area is}} \cr & A\left( x \right) = \pi {\left( {{r_o}} \right)^2} - \pi {\left( {{r_i}} \right)^2} \cr & A\left( x \right) = \pi {\left( {\sqrt x } \right)^2} - \pi {\left( {{x^3}} \right)^2} \cr & A\left( x \right) = \pi \left( x \right) - \pi \left( {{x^6}} \right) \cr & A\left( x \right) = \pi \left( {x - {x^6}} \right) \cr & {\text{The volume is given by}} \cr & V = \int_a^b {A\left( x \right)} dx \cr & V = \int_0^1 {\pi \left( {x - {x^6}} \right)} dx \cr & V = \pi \int_0^1 {\left( {x - {x^6}} \right)} dx \cr & {\text{Integrating}} \cr & V = \pi \left[ {\frac{{{x^2}}}{2} - \frac{{{x^7}}}{7}} \right]_0^1 \cr & V = \pi \left[ {\frac{{{{\left( 1 \right)}^2}}}{2} - \frac{{{{\left( 1 \right)}^7}}}{7}} \right] - \pi \left[ {\frac{{{{\left( 0 \right)}^2}}}{2} - \frac{{{{\left( 0 \right)}^7}}}{7}} \right] \cr & V = \pi \left( {\frac{5}{{14}}} \right) \cr & V = \frac{5}{{14}}\pi \cr} $$