Question
A tank in the shape of a right circular cylinder is filled with water (62.5 lb/ft3). It has a height of 8 ft and a diameter of 10 ft. How much work is required to pump all the water to a spout that is 3 ft above the top of the tank?
Answers
Easy way: Consider the water to be concentrated at its center of gravity. How much work to lift that mass 4+3 feet?
pi*5^2*8*62.5 lb * 7ft = 274889 ft-lb
Calculus way: consider a thin slice of water of thickness dy. Its weight is
pi * 5^2 * dy * 62.5 = 4908.734 dy lb
Now lift the slice at distance y from the bottom of the tank. It must rise (8-y)+3 = 11-y ft.
So, the work done is the sum of all those works lifting the slices:
integral [0,8] 4908.734(11-y) dy = 274889 ft-lb
pi*5^2*8*62.5 lb * 7ft = 274889 ft-lb
Calculus way: consider a thin slice of water of thickness dy. Its weight is
pi * 5^2 * dy * 62.5 = 4908.734 dy lb
Now lift the slice at distance y from the bottom of the tank. It must rise (8-y)+3 = 11-y ft.
So, the work done is the sum of all those works lifting the slices:
integral [0,8] 4908.734(11-y) dy = 274889 ft-lb