Answer
$$V = \frac{1}{9}\pi $$
Work Step by Step
$$\eqalign{
& y = x - {x^4},{\text{ }}y = 0 \cr
& {\text{Using the washer method about the }}x{\text{ - axis}} \cr
& V = \int_a^b {\pi \left[ {f{{\left( x \right)}^2} - g{{\left( x \right)}^2}} \right]} dx \cr
& {\text{From the graph of the region shown below}} \cr
& x - x^4 \geqslant 0 {\text{ on the interval }}\left[ { 0,1} \right]{\text{ }} \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_0^1 {\pi \left[ {{{\left( {x - {x^4}} \right)}^2} - {{\left( 0 \right)}^2}} \right]} dx \cr
& V = \pi \int_0^1 {\left( {{x^2} - 2{x^5} + {x^8}} \right)} dx \cr
& {\text{Integrating}} \cr
& V = \pi \left[ {\frac{1}{3}{x^3} - \frac{{{x^6}}}{3} + \frac{{{x^9}}}{9}} \right]_0^1 \cr
& V = \pi \left[ {\frac{1}{3}{{\left( 1 \right)}^3} - \frac{{{{\left( 1 \right)}^6}}}{3} + \frac{{{{\left( 1 \right)}^9}}}{9}} \right] - \pi \left[ 0 \right] \cr
& {\text{Simplifying}} \cr
& V = \frac{1}{9}\pi \cr
& \cr
& {\text{Using the washer method about the }}y{\text{ - axis}} \cr
& V = \int_c^d {\pi \left[ {p{{\left( y \right)}^2} - q{{\left( y \right)}^2}} \right]} dy \cr
& y = x - {x^4} \cr
& {\text{It's complicated solve the function for }}x{\text{ for use the}} \cr
& {\text{shell method}}{\text{, so The washer method is easier to apply}} \cr} $$
