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

1. ## "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.

2. 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)
$

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

its width is $\frac{7}{8}(8-y)
$

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.