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Math Help - Lost on Question 18...What am I doing wrong???

  1. #1
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    Question Lost on Question 18...What am I doing wrong???

    Example: The conical tank from y = 2x is filled to within 2 ft of the top with the olive oil weighing 57 lb/ft^3. How much work does it take to pump the oil to the rim of the tank.

    Delta V = Pi (radius)^2 (thickness) = Pi(1/2y)^2 delta y = pi/4 y^2delta y ft^3
    F(y) = 57 delta V = 57 Pi/4 y^2 delta y lb.
    Delta W = 57pi/4 (10-y) y^2
    W = Integral from 0 to 8 [ 57pi/4 (10-y)y^2dy ~ 30,561 ft-lb

    Now the problem: Supposed that, instead of being full , the tank in Example 5 is only half full. How much work does it take to pump the remaining oil to a level 4 feet above the top of the tank? Answer: 60,042 ft-lb.

    ok my process

    For problem 18
    Since it's 1/2 the volume capacity...I took half of the volume of Example 3 and the integral changes and kept everything else the same....
    What's wrong?
    V = 1/2[(pi)(1/2y)^2(delta y)] = 1/2 (Pi/4y^2delta y) ft^3
    F(y) = (57 lb/ft^3) 1/2[pi/4 y^2 delta y] = 57/2 [pi/4 y^2 delta y]
    W = 57/2 [integral from 0 to 6 pi/4 y^2 (10-y) dy = 57pi/8 [integral from 0 to 6 y^2(10-y))
    = 57pi/8 [integral from 0 to 6 10y^2-y^3] = 57pi/8[10y^3/3-y^4/4] from 0 to 6
    = 57pi/8 [10(6)^3/3 - (6)^4/4] = 8,864 ft lb
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  2. #2
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    Quote Originally Posted by Nimmy View Post
    Example: The conical tank from y = 2x is filled to within 2 ft of the top with the olive oil weighing 57 lb/ft^3. How much work does it take to pump the oil to the rim of the tank.

    Delta V = Pi (radius)^2 (thickness) = Pi(1/2y)^2 delta y = pi/4 y^2delta y ft^3
    F(y) = 57 delta V = 57 Pi/4 y^2 delta y lb.
    Delta W = 57pi/4 (10-y) y^2
    W = Integral from 0 to 8 [ 57pi/4 (10-y)y^2dy ~ 30,561 ft-lb

    Now the problem: Supposed that, instead of being full , the tank in Example 5 is only half full. How much work does it take to pump the remaining oil to a level 4 feet above the top of the tank? Answer: 60,042 ft-lb.
    Where did you get this? Doesn't it seem strange that pumping half the oil would take nearly twice the work?

    ok my process

    For problem 18
    Since it's 1/2 the volume capacity...I took half of the volume of Example 3 and the integral changes and kept everything else the same....
    What's wrong?
    V = 1/2[(pi)(1/2y)^2(delta y)] = 1/2 (Pi/4y^2delta y) ft^3
    F(y) = (57 lb/ft^3) 1/2[pi/4 y^2 delta y] = 57/2 [pi/4 y^2 delta y]
    W = 57/2 [integral from 0 to 6 pi/4 y^2 (10-y) dy = 57pi/8 [integral from 0 to 6 y^2(10-y))
    = 57pi/8 [integral from 0 to 6 10y^2-y^3] = 57pi/8[10y^3/3-y^4/4] from 0 to 6
    = 57pi/8 [10(6)^3/3 - (6)^4/4] = 8,864 ft lb
    Last edited by mr fantastic; August 22nd 2009 at 06:43 PM. Reason: Fixed quote tag for second quote.
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  3. #3
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    Well my prof...told me that....

    For #18: You're right that the volume is half of what it was before, but
    the 1/2 doesn't play into the V equation (which is the one you use for the
    volume of a "slab."). Instead, use the fact that we have half the initial
    volume to find the new height of the oil in the cone. This will make your
    work formula a little different.

    I got the answer from the professor.


    Quote Originally Posted by HallsofIvy View Post
    Where did you get this? Doesn't it seem strange that pumping half the oil would take nearly twice the work?
    ok my process

    For problem 18
    Since it's 1/2 the volume capacity...I took half of the volume of Example 3 and the integral changes and kept everything else the same....
    What's wrong?
    V = 1/2[(pi)(1/2y)^2(delta y)] = 1/2 (Pi/4y^2delta y) ft^3
    F(y) = (57 lb/ft^3) 1/2[pi/4 y^2 delta y] = 57/2 [pi/4 y^2 delta y]
    W = 57/2 [integral from 0 to 6 pi/4 y^2 (10-y) dy = 57pi/8 [integral from 0 to 6 y^2(10-y))
    = 57pi/8 [integral from 0 to 6 10y^2-y^3] = 57pi/8[10y^3/3-y^4/4] from 0 to 6
    = 57pi/8 [10(6)^3/3 - (6)^4/4] = 8,864 ft lb[/QUOTE][/QUOTE]
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