Answer
$$V = \frac{6}{7}\pi $$
Work Step by Step
$$\eqalign{
& y = {x^3},{\text{ }}y = 1,{\text{ and }}x = 0 \cr
& y = {x^3} \to x = \root 3 \of y \cr
& {\text{Use the Shell method about the }}x{\text{ - axis}} \cr
& V = \int_c^d {2\pi y\left[ {p\left( y \right) - q\left( y \right)} \right]} dy \cr
& {\text{From the graph we can see that}} \cr
& \root 3 \of y > 0{\text{ on the interval }}\left[ {0,1} \right] \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_0^1 {2\pi y\left( {\root 3 \of y - 0} \right)} dy \cr
& V = 2\pi \int_0^1 {y\root 3 \of y } dy \cr
& V = 2\pi \int_0^1 {{y^{4/3}}} dy \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {\frac{{{y^{7/3}}}}{{7/3}}} \right]_0^1 \cr
& V = \frac{6}{7}\pi \left[ {{y^{7/3}}} \right]_0^1 \cr
& {\text{Simplifying}} \cr
& V = \frac{6}{7}\pi \left( {1 - 0} \right) \cr
& V = \frac{6}{7}\pi \cr} $$
