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**galactus** The area of the kth layer of water is $\displaystyle {\pi}(12)^{2}$ with height

$\displaystyle {\Delta}y$

So, we have $\displaystyle 144{\pi}{\Delta}y$ for the volume of the kth disc.

The force required to move the kth layer equals the weight of the layer,

which can be found by multiplying its volume by the weight density of

water, which is $\displaystyle 1000 \;\ kg/m^{3}$.

So, we have $\displaystyle 144000{\pi}\int_{0}^{4}\underbrace{(5-y)}_{\text{distance moved}}dy$