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Revolving a tank around the y axis: Work needed to pump water

  1. Jun 20, 2010 #1
    1. A tank is formed by revolving y = 3x^2, x = [0,2] around the y axis is filled to the 4 feet point with water (w = 62.4 lb/ft^3). Find the work necessary to pump the water out of the tank over the top.

    I got the integral being from a = 0 and b = 4 and integrating [(12y-y^2)/3]dy with w and pi as constants outside of the integral.

    And after the integration I got:

    224/9 pi (62.4)

    And when I multiplied the 62.4 out it came out to be 1553.066 pi

    Would this be correct? Just seems bit of a weird number?

    [I asked this question earlier...and received help on it on an earlier thread, but I just wanted to see if the numbers I got are correct]
     
  2. jcsd
  3. Jun 20, 2010 #2

    HallsofIvy

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    At each y, the figure rotated around the y-axis is a disk or radius [itex]x= (y/3)^{1/2}[/itex] and so has area [itex]\pi x^2= (\pi/3)y[/itex] and a very thin disk of thickness dy would have volume [itex](\pi/3)y dy[/itex].

    The weight of that disk is [itex](\pi w/3)y dy[/itex] and must be lifted the remaining 12- y feet so the work done in lifting just that disk is (force times distance) [itex](\pi w/3) y(12- y)dy[/itex].

    Integrate that from y= 0 to y= 4- that appears to be exactly what you have done!:wink:
     
  4. Jun 20, 2010 #3
    Wow. Ok.

    Thanks! The number just seems really weird! Haha.

    Thats all!

    Thanks though! :]
     
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