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Math Help - Application of integration - Work

  1. #1
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    Application of integration - Work

    I was wondering if I could elicit some help with this problem. I feel as if I understand it but cannot get the right answer.




    It is number 21.
    These are the steps I took:
    • ΔF=(62.4ft/cm^3)(volume)
    • =(62.4)(1/3 π r2h)
    • Since this is a cone, the radius isn't constant. I imagined that the come was made by revolving the integral x=2/3 y about the y axis, which then gives the r at any given height, Δy
    • ΔF=62.4(1/3 π (2/3y)2Δy)
    • = 9.244π* y2Δy
    • Then
    • ΔW=ΔF*d
    • Let d= 6-y
    • ΔW=(9.244π* y2Δy)(6-y)
    • W=9.244π*int [0, 6] 6y2-y3
    • =9.244π(432-324)
    • =998.352π

    However the correct answer is 2995.2π...
    Thank you in advance!

    By the way I'm doing this on my phone so if the formatting is terrible, I'm sorry I'll try and revise it.
    Last edited by hayjude99; July 22nd 2012 at 09:55 AM.
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  2. #2
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    Re: Application of integration - Work

    work = the integral of WALT

    W = weight density of the liquid ... 62.4 \, lbs/ft^3

    A = cross-sectional area of a representative horizontal "slice" of liquid as a function of y ... \pi \left(\frac{2y}{3}\right)^2

    L = lift distance of a that representative slice as a function of y ... (6 - y)

    T = slice thickness ... dy

    the limits of integration are determined by where the horizontal slices reside w/r to y


    work in ft-lbs = \int_0^6 62.4\pi \left(\frac{2y}{3}\right)^2(6-y) \, dy
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    Re: Application of integration - Work

    Thanks, I had noticed I needed a factor of 3 to get to the right answer. My book doesn't seem to use the WALT method, do you know how this ties in to using the W=Fd method? I just don't understand why both methods don't work.

    Thanks again
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    Re: Application of integration - Work

    Quote Originally Posted by hayjude99 View Post
    Thanks, I had noticed I needed a factor of 3 to get to the right answer. My book doesn't seem to use the WALT method, do you know how this ties in to using the W=Fd method? I just don't understand why both methods don't work.
    both methods are the same ...

    the amount of work required to lift a single horizontal slice of water = (weight of that slice)(vertical displacement of that to the top of the cone)

    dW = \color{red}{F} \cdot \color{blue}{\Delta y}

    dW = \color{red}{mg} \cdot \color{blue}{\Delta y}

    dW = \color{red}{\rho \cdot \pi \left(\frac{2y}{3}\right)^2 \, dy \cdot g} \cdot \color{blue}{(6-y)}

    note that \rho \cdot g = weight density of water

    and \pi \left(\frac{2y}{3}\right)^2 \, dy is the volume of a representative slice
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    Re: Application of integration - Work

    Oh, I see what happened. I went into the problem think that the volume of the cone was 1/3 π r2h and then attempted to find the r in terms of y. But that radius in terms of y, when integrated , actually gives me the volume.
    Thank you
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