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Rectangular tank is base of 4 feet by 5 feet and a height of 4 feet that is full of water. The water weighs 62.4 pounds per cubic foot.

This is what I got. $$Volume = (5*4*4) $$

$${\Delta}F = 62.4 * 80 $$

$${\Delta}W = 4992(4-y)$$

$$\int_{0}^{4} 4992(4-y) dy = 0 $$ $$\int_{2}^{4} 4992(4-y) dy = -9984 $$

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  • $\begingroup$ My book doesn't go over this problem at all. It showed an example of emptying a spherical tank. It didn't explain how to solve other problems. $\endgroup$ – user275564 Mar 11 '15 at 21:48
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You should just change your $\Delta F$.

$$\Delta F = 62.4 \cdot 5\cdot 4 \cdot \Delta y$$

$$\Delta W = 62.4 \cdot 5\cdot 4 \cdot (4-y) \Delta y $$

Then you can set up your definite integral accordingly.

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  • $\begingroup$ Thanks so much! The answer is correct but I am still confused. What did I do wrong or how did you come up with your answer? $\endgroup$ – user275564 Mar 11 '15 at 22:09
  • $\begingroup$ Your $\Delta F$ is the force on the whole tank. When you replace the whole height $4$ by $\Delta y$, it becomes the force on the slice of the volume. The idea of integral is to add up the work done on each slice of volume. $\endgroup$ – KittyL Mar 11 '15 at 22:23
  • $\begingroup$ Oh, I get it now. My book talked about none of this. Thanks so much :) $\endgroup$ – user275564 Mar 11 '15 at 23:45

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