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Math Help - "A vertical cross section of a tank is shown..."

  1. #1
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    "A vertical cross section of a tank is shown..."

    "A vertical cross section of a tank is shown ( in attachment ). Assume that the tank is 16 feet long and full of water. ( delta = 62.4 pounds per cubic foot ) and that the water is to be pumped to a height 8 feet above the top of the tank. Find the work done in emptying the tank."

    So, work = force * distance. The force should be the weight of the tank, found by the volume ( l * w * h ) times delta. The distance should be ( 16 - y ).

    I'm at a complete loss here. Starting with the algebra involving that diagram to find volume. The 'help me solve this' doesn't elaborate on what it's doing.

    Next, I know the limits of integration are 0 to 8, but I do not understand why they are.
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  2. #2
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    let the 7' base be the x-axis and let the y-axis bisect the vertical cross section.

    the right side of the cross-section has equation x = \frac{7}{16}(8-y)<br />

    sketch in a horizontal "slab" of water in the tank ...

    its width is \frac{7}{8}(8-y)<br />

    its length is 16

    hence, the area of a representative slab is 16 \cdot \frac{7}{8}(8-y) = 14(8-y)

    the thickness of a representative "slab" is dy

    work lifting a representative "slab" = (weight-density)(area of slab)(distance the slab is lifted)(slab thickness)

    the work required to lift a single "slab" is ...

    dW = \delta \cdot 14(8-y) \cdot (16-y) \cdot dy

    all the horizontal "slabs" of liquid reside from y = 0 to y = 8 forming the limits of integration.
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