Answer
$\frac{3 \pi}{10}$ cubic feet
Work Step by Step
Using the assumption that the horizontal cross-sections form an annulus with an outer radius of $\sqrt{x}$ and an inner radius of $x^{2}$, the integer or the volume of the resulting solid is used.$$
\pi \int_{0}^{1}(\sqrt{x})^{2}-\left(x^{2}\right)^{2} d x
$$
So we've got
$$
\begin{array}{l}
\pi \int_{0}^{1} x-x^{4} d x \\
\pi\left[\frac{x^{2}}{2}-\frac{x^{5}}{5}\right]_{0}^{1} \\
\left[\frac{1}{2}-\frac{1}{5}-0\right]\pi \\
\left[\frac{5}{10}-\frac{2}{10}\right]\pi=\frac{3 \pi}{10} \text { cubic ft }
\end{array}
$$
