A spherical tank of radius 5 feet has its bottom half filled with water.

Question

How much work is done pumping the water to a point h=2 feet above the top of the tank? ρg = 62.4 lbs/ft³. Note: Define your coordinate system with its origin corresponding to the center of the sphere so that x²+y²=5².

a. What is h2?

b. What is h1?

c. If h(y) = H – y, then what is H?

d. What is d(vol) = A(x) dy?

e. Compute : W =∫ _h1^h2 ρg d(vol) h(y)

Answer 1

An alternate way of finding the work is to concentrate all the weight of the water at the center of mass of the inverted hemisphere. That will give the distance from the top of the water, which can be looked up in a table of center of mass for various shapes. In this case it is 3R/8. R=5 here. So the total distance to move this concentrated weight two feet above the sphere is 15/8+5+2 = 71/8 feet.

The force or weight of the water of the hemisphere is 62.4 × Volume = 62.4(4)πR³/6 = 31200π/6 = 5200π.

Work = Force × Distance = 5200π/(71/8) = 46150π ≈ 144984.5 ft-lbs

Which is the same answer Doug got.

Answer 2

desmos.com/calculator/lkp0zy3scn

Visit the above graph and use the slider on y_1 to visualize a slice of water being pumped to a point 2 feet above top of tank.