Originally Posted by

**ebaines** This is easier than you think.

1. The weight of water in a layer dx thick is its mass times g, and the mass is the volume times density. So:$\displaystyle dw = \rho A g dx$, where A is the area of the cylinder. Here I'm using lower case w for weight.

2. The work to raise water of weight dw to by height h is dW = hdw. Here I'm using upper case W for work. So for a slice of water at height x, the distance it must be raised is 6-x, and the work required is dW=(6-x)dw.

3. The total amount of work to pump all the water to a height of 6 meters is $\displaystyle W = \int _0 ^3 (6-x)dw = \int _0 ^3 (6-x)\rho A g dx$