Answer
$\text{The volume is}$
\begin{align}
V = \frac{9\pi}{14}
\end{align}
Work Step by Step
$\text{It is given that}$
\begin{align}
y = x^3; \ x = 0 \ \ and \ \ y = 1
\end{align}
$\text{The given region is revolved about y = 1. Meaning that the cross }$
$\text{section generates a cylindircal surface of height $\sqrt[3] y$ and radius (1-y).}$
$\text{Thus, the volume is}$
\begin{align}
V = 2\pi \int_0^1 (1-y)\sqrt[3] y \ dy = 2\pi \left[\frac{3y^{\frac{4}{3}}}{4} - \frac{3y^{\frac{7}{3}}}{7}\right]_0^1 = \frac{9\pi}{14}
\end{align}
