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Work done against gravity on a right circular cone

  1. Jun 15, 2011 #1
    1. The problem statement, all variables and given/known data
    A right circular cone has vertex down and is 10 feet tall with base radius 5 feet. The cone is filled with a fluid having varying density. The density varies linearly with distance to the top. Here "varies linearly" means the quantities are related by an equation of at most degree 1. At the top of the cone, the density is 80 lbs/ft^3, and at the bottom the density is 120 lbs/ft^3. How much work in ft-lbs is needed to pump out all the fluid to the top of the cone.

    2. Relevant equations

    3. The attempt at a solution

    Well, I figured out that the equation for the density would be 120 - 4y (taking 120 - 80, then dividing that answer by the height gives 4). The total work will be found by integrating from 0 to 10. From physics, I know that work is equal to force times distance. In order to get the force you need to know the mass (multiplied by gravity). The mass is found by multiplying the volume of a layer of the solid by the density.

    I'm having trouble finding the area at any height of the cone. I know that the radius changes as the height increases eventually getting to 5 feet.

    m = pV = (120-4y)(π(r(y)^2))
    F = 9.8(120-4y)(π(r(y)^2))
    W = ∫[0,10] 9.8(120-4y)(π(r(y)^2))y dy
    Last edited: Jun 15, 2011
  2. jcsd
  3. Jun 15, 2011 #2


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    Look at the slanted side of the cone; say it's in the xy plane. It goes from (0,0) to (5,10). What is the equation of that line? The radius you want is the x on that line in terms of y.
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