Cylinder Work Problem Calculus "Suppose it takes $k$ units of energy to lift a cubic meter of water one meter. About how much energy $E$ will take to remove all water from a cylinder with one meter in diameter and 100 meters deep? (Answer should depend on $k$.)"
It looks pretty straightforward but I encounter one problem.
So as I understood 
$E = k*\frac{V}{h}$ where V is the volume and h is the height.
substituting for volume we get
$$E = k\frac{\Pi}{4h}dh$$
after setting integral from 0 to 100 and solving I get 
$$ k\frac{\Pi}{4}(ln(100) - ln(0))$$
and obviously ln(0) is undefined.
Can you please point out what did I do wrong?
 A: You lost a factor $h$ because of the distance you must lift the volume.  If $A$ is the cross sectional area, a small unit of volume is $dV=Adh$.  If we measure $h$ with $0$ at the top of the tank and increasing downward, the small unit of volume at coordinate $h$ takes $dE=khdV=khAdh$ to remove it.  Now you can integrate over $h$ from $0$ to $100$.  You get something in $h^2$, which is reasonable because the average bit of water gets lifted $\frac h2$ and you have $h$ height of water.
A: the energy needed to lift one cubic meter of water for 1m is
$E=mgh=ke\Rightarrow e=\frac{mgh}{k}=\frac{\rho \cdot g}{k}\Rightarrow \rho \cdot g=e\cdot k$
Removing water from the reservoir is similar with bringing up cylindrical volumes of height dy located at depth y.
so the mass to lift is $dm=\rho\cdot \pi\cdot R^2\cdot dy$
the energy required to remove that mass is $dE = dm\cdot g\cdot y=$
the total energy spent is $E=\int_{0}^{h}dE=\rho\cdot\pi\cdot R^2\cdot g \cdot[\frac{y^2}{2}]_{0}^{h}=\rho\cdot\pi\cdot (0.5)^2\cdot g \cdot\frac{100^2}{2}=k\cdot e\cdot\pi \cdot \frac{10000}{8}=1250\cdot \pi\cdot k$ units of energy
