Calculating Work Done on Emptying a Half-Full Cylindrical Reservoir

[demonelite123]

A cylindrical reservoir of diameter 4 feet and height 6 feet is half-full of water weighing w pounds per cubic foot. Find the work done in emptying the water over the top.

i put the cylinder on the x-y axes, so y represents height and x represents radius of the cylinder. so the volume of a slice of water ΔV = πx^2 (Δy), and since x = 2, ΔV = 4π(Δy).

a slice of the water at height y is lifted (6 - y) feet so therefore the work,
W = w*4π(Δy)*(6 - y)

so the integral looks like 4πw ∫ (6-y) dy.

the only problem i have is how to decide the limits of integration. since you are trying to empty the water over the top, the water at the very bottom should travel to the top right? so the limits should be from 0 to 6 i think. but my book says the limits are from 0 to 3 because the reservoir is only half full. can somebody please explain this to me?

 

[bm0p700f]

What is the height of the top layer of water and what is height of the bottom layer?

Also why is everything in feet. Most of the world now uses SI units?

 

[kuruman]

Perhaps it will be easier to see what is going on if you did it in two steps. I know that water is incompressible, but suppose you were able to scrunch all the water molecules and bring them to the initial level of 3.0 feet. Step 1: How much work against gravity is needed to do that? Step 2: Now that you have all the water at the same height, how much extra work is needed to push it the additional 3 feet over the top? Add the two for the answer.