Answer
$\text{The volume is}$
\begin{align}
& V = \frac{47\pi}{210}
\end{align}
Work Step by Step
$\text{It is given that}$
\begin{align}
y = x^2 \ and \ y = x^3;
\end{align}
$\text{The function revolves around y = -1.}$
$\text{The intersections of two functions are}$
\begin{align}
x^2 = x^3 \Rrightarrow x = 0 \ and x = 1 \Rrightarrow y= 0 \ and \ y = 1
\end{align}
$\text{The volume is}$
\begin{align}
& V =\pi \int_0^1 \left((x^2+1)^2-(x^3+1)^2 \right)dx =\\ & = \pi \int_0^1(x^4+2x^2-x^6-2x^3)dx = \\
& = \pi \times \left[\frac{x^5}{5} +\frac{2x^3}{3} - \frac{x^7}{7} -\frac{x^4}{2} \right]_0^1 = \pi \left(\frac{1}{5} +\frac{2}{3} - \frac{1}{7} -\frac{1}{2} \right) = \frac{47\pi}{210}
\end{align}
