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Math Help - Calculus 2: Excercise about "Work Needed"?

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    Calculus 2: Excercise about "Work Needed"?

    I'm not sure if this should be in this thread or in the Physics one, but it was given to me by my Calc2 professor. Thing is, I'm not sure if I'm in the right direction.

    "An aquarium measures 2m long, 1m wide and 1m deep and it is full of water. Calculate the work needed to empty it halfway. (Remember that density of water is given by 1000 \frac{kg}{m^3})"

    Here's what I have.

    Volume ==> V = 2m^3

    If we want half the volume, then V(\frac{1}{2}) = 2m^3(\frac{1}{2}), so the volume I want is 1m^3

    Would that make my integral go from 2 to 1? How would I go about the function I'm supposed to integrate?

    Thank you!
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    Quote Originally Posted by Gretchy View Post
    I'm not sure if this should be in this thread or in the Physics one, but it was given to me by my Calc2 professor. Thing is, I'm not sure if I'm in the right direction.

    "An aquarium measures 2m long, 1m wide and 1m deep and it is full of water. Calculate the work needed to empty it halfway. (Remember that density of water is given by 1000 \frac{kg}{m^3})"

    Here's what I have.

    Volume ==> V = 2m^3

    If we want half the volume, then V(\frac{1}{2}) = 2m^3(\frac{1}{2}), so the volume I want is 1m^3

    Would that make my integral go from 2 to 1? How would I go about the function I'm supposed to integrate?

    Thank you!
    take a representative horizontal "slice" of water ... length 2m , width 1m , thickness dy

    volume of the slice is 2 \, dy

    mass of the slice is 2000 \, dy

    weight of the slice is 2000g \, dy , where g = 9.8 \, m/s^2

    work to lift one slice to the top of the tank (force times distance) is 2000g \cdot (1-y) \, dy

    work (in Joules) to lift all the slices in the top half of the tank to the top of the tank ...

    \displaystyle W = \int_{\frac{1}{2}}^1 2000g \cdot (1-y) \, dy
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