Answer
$V = \frac{{62}}{5}\pi $
Work Step by Step
$$\eqalign{
& y = {x^3},{\text{ }}y = 0,{\text{ }}x = 1,{\text{ }}x = 2 \cr
& {\text{Apply the shell method about the }}y{\text{ - axis}}{\text{.}} \cr
& V = \int_a^b {2\pi x} f\left( x \right)dx,{\text{ where }}0 \leqslant a \leqslant b \cr
& {\text{Therefore}}{\text{,}} \cr
& V = \int_1^2 {2\pi x\left( {{x^3}} \right)} dx \cr
& V = 2\pi \int_1^2 {{x^4}} dx \cr
& {\text{Integrating}} \cr
& V = 2\pi \left[ {\frac{{{x^5}}}{5}} \right]_1^2 \cr
& V = \frac{{2\pi }}{5}\left[ {{x^5}} \right]_1^2 \cr
& {\text{Evaluating}} \cr
& V = \frac{{2\pi }}{5}\left[ {{{\left( 2 \right)}^5} - {{\left( 1 \right)}^5}} \right] \cr
& V = \frac{{2\pi }}{5}\left( {32 - 1} \right) \cr
& V = \frac{{62}}{5}\pi \cr} $$
