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Homework Help: Work integrals

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A cylindrical gasoline tank with radius 4ft and length 15 ft is buried under a service station. The top of the tank is 10ft underground. Find the amount of work needed to pump all the gasoline in the tank to a nozzle 3 ft above the ground. Gasoline weighs 42lb/ft^3.

    3. The attempt at a solution

    So the way I have started is by setting a cylinder with the specified radius centered at (0,0). With that I know the top is at (0,4), bottom is at (0,-4) and the place it needs to reach is y = 17 (13ft above the top of the cylinder).

    I divide the cylinder into strips of thickness delta y. So I know I need to find the weight of each of those strips and multiply it by the distance it must travel (17-y).

    I have the density so I really now just need the volume of those strips. Volume = length * width * thickness. I have the thickness (delta y), the length (15), but I need the width.

    I can't figure out how to find the width of the strips. I know that at y = 0 the width is 8 and at y=4 and y = -4 the width is 0.

    Something like
    [tex] width = 2\sqrt{16-y^2}[/tex]
    would tell me the width with respect to y, but that would only be true from y=-4 to y=4, and it needs to stop after that.

    Any advice would be appreciated thanks.
  2. jcsd
  3. Oct 12, 2008 #2
    Wait tell me if this would work:

    [tex]width = 2\sqrt{16-y^2}[/tex]
    [tex]thickness = \Delta(y)[/tex]
    [tex]length = 15[/tex]

    [tex]weight = 2\sqrt{16-y^2} * \Delta(y) * 15 * 42[/tex]

    and the height = 17 - y
    so the integral would be:

    [tex]\int_{-4}^{4} 1260\sqrt{16-y^2}(y-17) dy[/tex]
  4. Oct 12, 2008 #3


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    Science Advisor

    One important thing you did not specify: is the axis of the cylinder vertical or horizontal? You seem to be assuming it is horizontal. If so then, yes, [itex]width= \sqrt{16- y^2}[/itex]. You will only be integrating from y= -4 to 4 so the fact that it "has to stop at 4" is irrelevant. Looks to me like your final integral is correct.
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