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Math Help - Integrals and work again

  1. #1
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    Integrals and work again

    Water in a cylinder of height 10 ft and radius 4 ft is to be pumped out. Find the work required if

    (a) The tank is full of water and the water is to be pumped over the top of the tank.

    (b) The tank is full of water and the water must be pumped to a height 5 ft above the top of the tank.

    (c) The depth of water in the tank is 8 ft and the water must be pumped over the top of the tank.


    I just want to know here.. how I would have to change my setup of the problem to account for each condition.. I get confused on what my limits of integration are..(i think that's what changes here.)

    Thanks in advance.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jacobpm64 View Post
    Water in a cylinder of height 10 ft and radius 4 ft is to be pumped out. Find the work required if

    (a) The tank is full of water and the water is to be pumped over the top of the tank.

    (b) The tank is full of water and the water must be pumped to a height 5 ft above the top of the tank.
    a) We are going to pump all the water from the tank and lift it to a specified height. So let's take this height as our reference level for potential energy. We are going to lift all this water up to the reference height, so the work done will simply be equal to the negative of the potential energy stored in the water.

    So how to find the potential energy of the water? Imagine the cylinder as a series of disks, each with a height dh. Each disk will have a volume of:
    dV = (pi)r^2 * dh
    Thus each disk will contain a mass of water equal to:
    dm = (rho) * dV (where rho is the density of water.)

    dm = (rho)*(pi)r^2 * dh

    Thus the potenial energy of the disk will be:
    d(PE) = dm * g * h, where h is the height of the disk.

    d(PE) = g*(rho)*(pi)*r^2 * hdh

    Now integrate over the height coordinate of the cylinder:
    Int(d(PE)) = PE = Int(g*(rho)*(pi)*r^2 * hdh, h, 0, -H) where H is the height of the cylinder. (Recall that we require + to be upward, so we are integrating the cylinder from 0 to -H: we are integrating "upside down.")

    PE = -(1/2)g*(rho)*(pi)*r^2*H^2

    Thus W = -PE = -(1/2)g*(rho)*(pi)*r^2*H^2

    b) This is the same idea as a) except that now we are pumping the water up 5 ft more. So add a Mg(5 ft) to the answer in a) where M is the total mass of the water in the cylinder.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jacobpm64 View Post
    Water in a cylinder of height 10 ft and radius 4 ft is to be pumped out. Find the work required if
    (c) The depth of water in the tank is 8 ft and the water must be pumped over the top of the tank.
    I'll leave this as mostly an exercise. What's changed here from a) is that now we start integrating the height at 10 ft - 8 ft = 2 ft and integrate down to 10 ft, instead of integrating from 0 ft to 10 ft.

    -Dan
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