From a Single Variable Calc Textbook:

"An aquarium $2$ m long, $1$ m wide, and $1$ m deep is full of water. Find the work needed to pump half of the water out of the tank."

My attempt:

First, I multiplied 1000 (density of water) by 9.8 to help compute the weight of the water. Then, I figured I could use a rectangular cross-section to serve as my "disk." $(1-y)$ is the height, and I used $2y$ as the area.

$$\int_0^1(1-y)*2y*9800dy$$ $$19600\int_0^1(1-y)ydy$$ $$19600\int_0^1y-y^2dy$$ $$19600\left(\int_0^1ydy-\int_0^1y^2dy\right)$$ $$19600*(\frac{1}{2}-\frac{1}{3})$$ $$3266 J$$ $$1633 J$$

But the book says $2.45 * 10^3 J$

Thanks in advance for your help.


Presumably $y$ is the vertical axis with $0$ at the bottom of the tank. You should specify that. You are only to remove half the water, so your integral should run from $\frac 12$ to $1$. The $dy$ is a thin layer of water at the same starting altitude that you remove. The area of your layer is $2$, not $2y$, as it is a $1 \times 2$ rectangle. The work done to remove the thin layer is then $9800\cdot 2 \cdot (1-y) dy$ where $9800\cdot 2 dy$ is the weight of the water and $1-y$ is the height you lift it.

  • $\begingroup$ Okay, thanks. I just tested it out and I tried to evaluate $\int_0^{\frac{1}{2}}9800*2(1-y)dy$, but I got a different answer. I also set it up the way you said and got the right answer. What's the difference between the two? Presumably, it's more work because the water is coming from lower in the tank. $\endgroup$ – N. Bar Jul 7 '19 at 1:00
  • 1
    $\begingroup$ Yes. If you empty the top half the top layer of water takes no work at all to remove. The bottom part you remove needs to be lifted $\frac 12$ meter. When you integrate from $0$ to $\frac 12$ the first water is lifted $\frac 12$ meter and the last is lifted $1$ meter. The answers should differ by a factor $3$ because the average water is lifted $\frac 14$ meter in one case and $\frac 34$ meter in the other. $\endgroup$ – Ross Millikan Jul 7 '19 at 2:04
  • $\begingroup$ You must integrate from 0.5 to 1 in the above solution 9800 x 2(1-y)dy $\endgroup$ – pabodu Jul 8 '19 at 1:58

I am adding this non-calculus answer because it is a useful principle to get the answer to a question quickly and easily before you start. This gives you a cross-check on the correctness of your calculus.

  1. Consider the lump of water you are trying to remove.

  2. In terms of work, it is exactly equivalent to a point mass at its centre of gravity. If you want to convince yourself of this, take one drop from the bottom of the lump and raise it half way, then take one drop from the top and let it fall half way. Equal force, equal distance, so equal work.

  3. You now have a 1000kg point mass which needs to be raised $\frac 1 4$ metres.

  4. Enjoy!

I am not knocking the calculus, which I love. But knowing the answer before you start can be a big help in making sure you have got the calculus right.


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