A hemispherical tank of radius 2 feet is positioned so that its base is circular. How much work is required to fill the tank with water through a hole in the base when the water source is at the base? (The weight-density of water is 62.4 pounds per cubic foot.)

784 ft·lb

Step-by-step explanation:

The amount of work required is equivalent to the work required to raise the total weight of water to the height of the centroid of the hemisphere.

The weight of the water is ...

Volume * density = (2/3) πr³ * 62.4 lb/ft³

= (2/3) π (2 ft) ³ (62.4 lb/ft³) ≈ 1045.5 lb

A reference page tells us that the centroid of a hemisphere is 3/8r above the circular base, so we're raising this weight of water by (3/8) (2 ft) = 3/4 ft.

The work done is ...

(3/4 ft) (1045.5 lb) ≈ 784 ft·lb