Work to empty trough

Question

 trough is 4 meters long, 1 meters wide, and 1 meters deep.

The vertical cross-section of the trough parallel to an end is shaped like an isosceles triangle (with height 1 meters, and base, on top, of length 1 meters). 

The trough is full of water which has density 1000 kg/m3. 

Find the amount of work required to empty the trough by pumping the water over the top. 

Work =  Joules

Answer 1

Set up a coordinate system with an origin at the deepest triangle vertex and a vertical y axis pointing up.

At height y above the vertex, the horizontal width of the triangle will be w = y, since at height 1, the width is 1.

Consider a horizontal slab of water at height y above the vertex:

width = y

length = 4

thickness = dy

mass = density . width . length . thickness = 1000 * y * 4 * dy = 4000ydy

This slab has to be raised to the rim of the trough by a vertical distance 1 - y

The necessary work: dW = mass . g. (1 - y) = 4000ydy . g . (1-y), where g = 9.8 m/s^2 is the gravitational acceleration.

The total work is the sum of the work for lifting all the slabs to the rim i.e. the integral:

W = integral of 4000 g y (1-y) dy for y = 0 to 1.