Answer
$$\frac{3}{5}$$
Work Step by Step
Solve to x
$$
y^{1 / 3}=x
$$
The volume is $\int_{a}^{b} A(y) d y$
The cross sections are squares with base $y^{1 / 3}=x$
Thus, $y^{2 / 3}=\left(y^{1 / 3}\right)^{2}=A(y)$
The integral part of the volume is
$\int_{0}^{1} y^{2 / 3} d y$
$$
\begin{array}{c}
{\left[\frac{3}{5} y^{5 / 3}\right]_{0}^{1}} \\
\frac{3}{5}=\frac{3}{5}(1)^{5 / 3}-\frac{3}{5}(0)^{5 / 3}
\end{array}
$$
