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