Work: finding the work done in filling a tank

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For finding the work done in filling a tank. If we are given radius (i.e. radius r = 5) and height (h = 15), should it be done like:

. . .S (pi) r^2 * y

...or:

. . .S (pi) r^2 (height-y)dy

I don't know if you understand what I'm trying to say. Pretty much, I know that in the integral we are multiplying the area (pir^2) * (something). I just don't know if the height is just "y" or if it is (y- height). Thank you.
 
work = INTEGRAL(W A L T)

where ...
W = Weight density of the liquid (a constant)
A = the cross-sectional Area of a representative horizontal "slice" of liquid, in terms of y (this can be a constant also if the cross-sectional area remains unchanged, as in a right circular cylinder)
L = vertical distance that a representative horizontal slice is to be Lifted, also in terms of y
T = horizontal slice Thickness ... dy.

the limits of integration will be the upper and lower y-values where the liquid resides at the start of the problem.
 
"vertical distance that a representative horizontal slice is to be Lifted, also in terms of y"

I'm still confused on this part. Does this mean that I look to see where its being pumped from? SO if its from ground, then its just y? Thanks.
 
Maybe I can explain a little.

The kth layer has a finite thickness. The upper and lower surfaces are different distances from the bottom(the origin).

If the layer is thin, the difference in the distances is small, so we can say the entire layer is a single distance from the origin. So, the work needed to pump the kth layer over the top is, in your case, (15-y).

Here's a sloppy drawing to either help or confuse more.

workfq5.gif
 
ah!! i get it! the picture definitely helped a lot, thank you galactus.
 
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