Calculus 2: Work Done by Pumping a Water Tank

Question

A water tank in the shape of a hemispherical bowl of radius 4 m is filled with water to a depth of 2 m. How much work is required to pump all the water over the top of the tank? (The density of the water is 1000 kg/m3. Assume g = 9.8 m/s2.)

The answer is 352800π but I do not understand how the book came to this answer. Please I would like an explanation

Answer 1

There are a couple ways to set up a coordinate system to model this problem. Here is how I did it. Letting (0,0) be the point on which the tank touches the ground, the center of the sphere would be located at (0,4). A cross section of the sphere through its diameter would be a circle with equation x² + (y-4)² = 16. This is used to establish a relationship between x and y, where x² = 16 - (y-4)^2. 

 

The volume of a typical disk of water y units from the ground could be represented by πx² = π(16-(y-4)²) = 

π(8y-y²). Each disk of water must get lifted 4-y m. 

 

Without specifying the units we have: 1000(9.8)π ∫₀² (4-y)(8y-y²) dy. (going from 0 to 2 because the water is at a depth of 2). Using the definition of Work to set up the integral. You should be able to take it from there.