Why is calculus needed in this problem involving work and units of energy? Consider this problem:

Suppose it takes k units of energy to lift a cubic meter of water one meter. About how much energy E will it take to pump dry a circular hole one meter in diameter and 100 meters deep that is filled with water?

Here's how I tried to solve it:
Total volume of water = $\pi r^2h = \pi (\frac{1}{2})^2 100$
Energy required to move this volume of water by $1$ meter $=\pi(\frac{1}{2})^2100k$, and so energy required to move this move this volume of water by $100$ meters $= 100\pi(\frac{1}{2})^2100k = \frac{\pi k10^4}{4}$.
However, the answer given is: $\frac{\pi k10^4}{8}$. Where am I going wrong with my reasoning? Also, this problem was assigned in single variable calculus, under definite integrals. Why is calculus needed here at all?

Well, since everything's linear, and since the cross-section $A=\pi r^2$ is constant, consider a vertical element $dh$ near the top, which needn't move hardly at all, so $dE=0kAdh=0$. There's a "complementary" $dh$ element near the bottom that needs to move all $100m$, for which $dE=100kAdh$. Then the total for those two elements is $dE=(0+100)kAdh$.
Moving down a little from the top, say $z$ meters, and up a little from the bottom, you can do the same thing for two more $dh$ elements. And you can see that total in parentheses will just be $((0+z)+(100-z))=(100)$ same as before. And on and on for the whole thing.
So it's basically just the average, i.e., your "top"+"bottom" elements get exhausted at $50m$, whereby the answer's $50Ak$ rather than $100Ak$. And you're right -- the problem can be solved without calculus.
