Answer
$\text{The volume is}$
\begin{align}
& V = \frac{\pi}{15}
\end{align}
Work Step by Step
$\text{It is given that}$
\begin{align}
y = x^2 \ and \ y = x^3;
\end{align}
$\text{The function revolves around x = 1. Thus,}$
\begin{align}
y = x^2 \Rrightarrow x = \sqrt y \ and \ y = x^3 \Rrightarrow x = \sqrt[3] y
\end{align}
$\text{The intersections of two functions are}$
\begin{align}
\sqrt y = \sqrt[3] y \Rrightarrow y = 0 \ and y = 1 \Rrightarrow x = 0 \ and \ x = 1
\end{align}
$\text{The volume is}$
\begin{align}
& V =\pi \int_0^1 \left((\sqrt y - 1 )^2 - (\sqrt[3] y - 1)^2 \right)dy = \pi \int_0^1 \left(y-2y^{\frac{1}{2}} -y^{\frac{2}{3}} +2y^{\frac{1}{3}} \right)dy = \\
& = \pi \left(\frac{1}{2} - \frac{4}{3} -\frac{3}{5} + \frac{3}{2} \right) = \frac{\pi}{15}
\end{align}
