Question

A water heater that has the shape of a right cylindrical tank with a radius of 1 foot and a height of 4 feet is being drained. how fast is the water draining out of the tank in cubic feet​ / minute if the water level is dropping at 6​ inches/min?

Answer 1

 For any prism-shaped geometry, the volume (V) is assumed by the product of cross-sectional area (A) and height (h). 

 V = Ah 

Distinguishing with respect to time gives the relationship between the rates. 
dV/dt = A*dh/dt

 in the meantime the area is not altering 

dV/dt = π*(1 ft)^2*(-0.5 ft/min) 

dV/dt = -π/2 ft^3/min ≈ -1.571 ft^3/min 

Water is draining from the tank at the rate of π/2 ft^3/min.