Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.2 Volumes By Slicing; Disks And Washers - Exercises Set 5.2 - Page 363: 39

$\text{The volume is}$ \begin{align} V = 40000\pi ft^3 \end{align}

$\text{It is given that}$ \begin{align} r = f(x) = \frac{x^2}{4} ft; 0 \leq x \leq 20 \end{align} $\text{Thus, the volume is}$ \begin{align} V = \pi \int_0^{20} \frac{x^4}{16} \ dx = \pi \left[\frac{x^5}{80} \right]_0^{20} = 40000\pi ft^3 \end{align}