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Work removing water from trough

  1. Feb 13, 2009 #1
    1. The problem statement, all variables and given/known data

    A trough is 2 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^4 from x=-1 to x=1. The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot.


    2. Relevant equations



    3. The attempt at a solution

    I found the volume of the trough by integrating the inverse of the function defining the cross section (y^1/4) from 0 to 1, and multiplying it by 4 (2 to take into account both parts of the cross section, and 2 to account for the length of the trough). The volume came out as 3.2 cubic feet. I then found the total weight of the water by multiplying that by 62, and I obtained a weight of 198.4 pounds.

    In order to find the amount of work, I believed that all I had to do is find the average height at which the water is raised out of the trough, and multiply the entire weight of the water by that value.

    Using the normal formula for finding the average of the function would not work (because the cross section is a filled in area rather than just a line), so I find the average point, I integrated the inverse of x^4 from 0 to 1 (this defines a filled in half of the cross section), and I obtained a value of 0.8 from that integral. I then cut that value in half to 0.4, and integrated the inverse function again from 0 to some number n. The result was an expression for n that equaled half of the integral from 0 to 1. Solving for n gave me an average height of 0.57435. Because that height is the average height of the water and not the height that the water is moved, the average distance that the water is moved is then equal to 1-0.57435=0.42565

    Then I multiplied that value by the total weight, and obtained:
    198.4*0.42565=84.449

    Which was apparently wrong.
     
  2. jcsd
  3. Feb 13, 2009 #2

    HallsofIvy

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    Imagine a thin "layer" of water at a fixed distance, say d, below the top of the trough. That whole layer is lifted a distance d and so the work done to lift that particular layer is the weight of the layer times d. The layer is a rectangular solid with length 2, width 2x and depth "dy". The height, y, above the bottom of the trough is, since the trough is 1 foot deep, 1- d so that d= 1- y. The volume of that layer is 2(2x)(1-y)dy. We also know that y= x4 so 2x= 2y1/4. The volume is 4y1/4(1-y)dy If the density of water is 62 pound/ft3 the weight of that layer is 4(62)y1/4(1-y)dy= 248(y1/4- y5/4)dy. Now "add" that up for all possible different values of y: that would be a Riemann sum which converts to the integral
    [tex]248\int_0^1 (y^{1/4}- y^{5/4})dy[/tex]
     
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