Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.2 Volumes By Slicing; Disks And Washers - Exercises Set 5.2 - Page 362: 6

$$\,\,V = \frac{2}{{35}}\pi $$

$$\eqalign{ & {\text{From the graph we have }} \cr & f\left( x \right) = {x^2}{\text{ and }}g\left( x \right) = {x^3}{\text{ with }}f\left( x \right) \geqslant g\left( x \right){\text{ on the interval }}\left[ {0,1} \right] \cr & {\text{The volume of the solid can be calculated using}} \cr & \,\,\,\,V = \int_a^b {\pi \left( {{{\left[ {f\left( x \right)} \right]}^2} - {{\left[ {g\left( x \right)} \right]}^2}} \right)dx} \cr & {\text{Thus}}{\text{,}} \cr & \,\,\,\,V = \int_0^1 {\pi \left( {{{\left[ {{x^2}} \right]}^2} - {{\left[ {{x^3}} \right]}^2}} \right)dx} \cr & \,\,\,\,V = \pi \int_0^1 {\left( {{x^4} - {x^6}} \right)dx} \cr & \,\,\,\,V = \pi \left[ {\frac{1}{5}{x^5} - \frac{1}{7}{x^7}} \right]_0^1 \cr & {\text{Evaluating}} \cr & \,\,\,\,V = \pi \left[ {\frac{1}{5}{{\left( 1 \right)}^5} - \frac{1}{7}{{\left( 1 \right)}^7}} \right] - \pi \left[ {\frac{1}{5}{{\left( 0 \right)}^5} - \frac{1}{7}{{\left( 0 \right)}^7}} \right] \cr & \,\,\,\,V = \pi \left( {\frac{1}{5} - \frac{1}{7}} \right) \cr & \,\,\,\,V = \frac{2}{{35}}\pi \cr} $$