Answer
$\text{The volume is}$
\begin{align}
V = \frac{1}{30}
\end{align}
Work Step by Step
$\text{It is given that}$
\begin{align}
y = x^2 \ and \ y = x
\end{align}
$\text{The intersections of the functions are}$
\begin{align}
x^2 = x \Rrightarrow x = 0 \ and \ x = 1 \Rrightarrow y =0 \ and \ y = 1
\end{align}
$\text{To find the volume, we have to find the area of the cross section.}$
$\text{It is given that the cross section is squre, therefore:}$
\begin{align}
A(x) = (x-x^2)^2 = x^2-2x^3+x^4
\end{align}
$\text{Thus, the volume is}$
\begin{align}
V = \int_0^1 (x^2-2x^3+x^4) \ dx = \left[\frac{x^3}{3} - \frac{x^4}{2} + \frac{x^5}{5} \right]_0^1 = \left(\frac{1}{3} - \frac{1}{2} + \frac{1}{5} \right) = \frac{1}{30}
\end{align}
