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Math Help - Work Using the Integral? Help!

  1. #1
    Member zangestu888's Avatar
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    Work Using the Integral? Help!

    Question:
    ?
    A tank in the form of a hemispherical bowl of radius 4 m is full of water. Find the work required to
    pump all of the water to a point 2 m above the tank.??

    i dont know how to start this, please i need a starting point anyone much appreciated!
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  2. #2
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    Quote Originally Posted by zangestu888 View Post
    Question:
    ?
    A tank in the form of a hemispherical bowl of radius 4 m is full of water. Find the work required to
    pump all of the water to a point 2 m above the tank.??

    i dont know how to start this, please i need a starting point anyone much appreciated!
    work = \int WALT

    W = weight density of the fluid
    A = cross-sectional area as a function of y
    L = lift distance of a representative horizontal cross section as a function of y
    T = cross sectional thickness ... dy

    sketch the graph of the circle x^2 + y^2 = 4^2 ... use the lower half to represent the edge of the bowl.

    radius of a representative horizontal "slice" of liquid is r = x, so cross-sectional area ... A = \pi x^2 = \pi(4^2 - y^2)

    lift distance of a representative 'slice" ... L = 2-y

    weight density of water is 9800 \, N/m^3

    all the "slices" of water lie between y = -4 and y = 0

    work in Joules is given by ...

    Work = \int_{-4}^0 9800 \pi(4^2 - y^2) (2 - y) \, dy
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  3. #3
    Member zangestu888's Avatar
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    I dont understand how you get the lift distance?, and also the weight density of the water, and the limits of intergration? how? thank you
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  4. #4
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    Quote Originally Posted by zangestu888 View Post
    I dont understand how you get the lift distance?, and also the weight density of the water, and the limits of intergration? how? thank you
    a representative horizontal slice of water has position y ... this representative slice is to be lifted to the position y = 2 ... therefore, the distance of lift is 2-y.

    water has a mass density of 1000 kg/m^3
    the weight of one kg of water is 9.8 N
    so ... weight density of water is 9800 N/m^3

    as stated in my initial response, all the horizontal slices of water reside between the positions y = -4 and y = 0 ... the integral sums the work required to lift all these slices between y = -4 and y = 0.
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