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?
 A: You don't have to lift every cup of water out by the whole $100$m.  If a given horizontal cross-section of water is a depth $d$ meters, you need to move that cross-section only $d$ meters, not $100$m.  So to find the whole amount of work done you need to add up the work done for each thin cross-section, which depends on the depth; and you can do that with an appropriate integral.
A: 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.
A: It's usually an overstatement to say that something needs Calculus, so even though you have some good answers, I'm going to solve the problem without using Calculus at all.  Your solution was that the total volume of water is $ 25 \pi $ cubic metres, and this is raised $ 100 $ metres, so the total energy required is the product of these numbers and $ k $.  As Jair Taylor's answer pointed out, this is flawed because not all of the water is raised the full height of $ 100 $ metres.  But as John Forkosh's answer pointed out, because the cross sections are all the same, everything is linear.  So we can just say that the average height the water is raised is $ 50 $ metres.  Multiply $ 25 \pi $, $ 50 $, and $ k $, and you get the correct answer, $ 1250 \pi k $ units of energy.  (If the cross sections had different areas, say because the hole were shaped like a cone or a paraboloid, which are typical textbook problems, then it would be harder to avoid Caclulus.  Indeed, for the paraboloid, I wouldn't know how to do it without Calculus.)
