Find the work required to pump all the water

[nameVoid]

the ends of an 8 foot long water trough are equilateral triangles having sides of length 2 feet. If the trough is full of water, find the work required to pump all the water over the top.

 

[nameVoid]

possible trapazoid section? area?

 

[HallsofIvy]

I presume the trough has the vertex of the triangle down! Don't just try to remember formulas- think about what work means. The work done to lift a "piece of water" out of the trough is the weight of the piece times the height it must be lifted: the distance from the top base of the triangle to the piece. We can handle all "pieces of water" that have that same distance at once by thinking of a "layer of water" at a given distance from the top base. That will be a rectanglar layer having length 8 feet, depth dx, and width w, depending on what x is. That is, its volume will be 8w(x)dx and so its weight will be \gamma (8w(x)dx) where \gamma is the density of water (look it up) in pounds per cubic foot. That layer has to be lifted x feet to the top of the trough so the work done is 8\gamma w(x)x dx. Integrate that from 0 to the height of the triangle.

Now, determining w(x). Draw a picture: an equilateral triangle with vertex at the bottom, base at the height and a horizontal line x units below the base. The length of that line is w(x). "Similar triangles" is a good way to find its length. Another way would be to find the equations of the two lines, both through (0, 0) and one through (1, \sqrt{3}), the other through (-1, \sqrt{3}). Be careful what you call "x" and what "w". Do you see where I got the \sqrt{3}?