Solution:
\(Vol =A_{base} h\)
The area of the base if constant, while the height is increase as water pours in.
\(\frac{dVol}{dt} = \frac{\pi 8^2}{4} \frac{dh}{dt}\)
\(24 = 16 \pi \frac{dh}{dt}\)
\(\frac{dh}{dt} = \frac{24}{16\pi} = 0.48 \frac{\text{ft}}{\text{min}}\)
The surface is rising at approximately 0.48 ft per min.