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Math Help - Pumping a cylindrical tank

  1. #1
    kuahji
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    Pumping a cylindrical tank

    A storage tank is a right circular cylinder 20ft long and 8 ft in diameter with its axis horizontal. If the tank is half full of olive oil weighing 57 lb/ft^3, find the work done in emptying it through a pipe that runs from the bottom of the tank to an outlet that is 6 ft above the top of the tank.
    My work
    I set the bottom of the tank at the point (0,0) & then drew a circle with a radius of 4.
    dV=2x(20)dy
    then solved the the circle's equation for x, x= \sqrt{(16-y^2)}
    dV=40 \sqrt{(16-y^2)}
    F(y)=57(40) \sqrt{(16-y^2)}
    10-y should be the distance the work must do
    W=2280 \int(10-y) \sqrt{(16-y^2)}
    Then I distributed the (10-y)
    W=22800 \int \sqrt{(16-y^2)}- 2280 \inty \sqrt{(16-y^2)}
    For part two, I set u=16-y^2 & got
    W=22800 \int \sqrt{(16-y^2)}+ 1140 \inty \sqrt{(16-y^2)}
    This is the part where I get lost, I did it a bit differently from the solutions manual, but at this point the manual shows inserting 4 \pi as follows
    22800(4 \pi)+ 1140 \inty \sqrt{(16-y^2)} (evaluated from 0 to -4), I tried from (0 to 4) in my solution.
    I don't understand why the solutions manual is doing this particular step, it says its the area of a semicircle, but can anyone explain why I use it in this problem & where the integral disappears to?
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by kuahji View Post
    A storage tank is a right circular cylinder 20ft long and 8 ft in diameter with its axis horizontal. If the tank is half full of olive oil weighing 57 lb/ft^3, find the work done in emptying it through a pipe that runs from the bottom of the tank to an outlet that is 6 ft above the top of the tank.
    ...

    This is the part where I get lost, I did it a bit differently from the solutions manual, but at this point the manual shows inserting 4 \pi as follows
    22800(4 \pi)+ 1140 \inty \sqrt{(16-y^2)} (evaluated from 0 to -4), I tried from (0 to 4) in my solution.
    I don't understand why the solutions manual is doing this particular step, it says its the area of a semicircle, but can anyone explain why I use it in this problem & where the integral disappears to?

    in fact, it was the area of a quarter circle:
    note that the "function" was g(y) = \sqrt{16 - y^2}, which is actually a semicircle on the right side of the y-axis.. but since it was only integrated from -4 to 0, then the integral was the "area" of the region of the circle on the fourth quadrant only.. and since the whole circle is of radius 4, then A_{circle} = \pi r^2 = 16\pi and you only need the quarter of it and therefore, \frac{A_{circle}}{4} = 4\pi
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