1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Work integrals
  1. Integrals and work (Replies: 2)

  2. Integral, Work (Replies: 1)

  3. Work integral (Replies: 7)