# How much work is done in pumping water out over the top edge in order to empty (a) half the tank (b) all of the tank.

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. $$Volume = (5*4*4)$$

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

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

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

• 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. – user275564 Mar 11 '15 at 21:48

You should just change your $\Delta F$.
$$\Delta F = 62.4 \cdot 5\cdot 4 \cdot \Delta y$$
$$\Delta W = 62.4 \cdot 5\cdot 4 \cdot (4-y) \Delta y$$
• 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. – KittyL Mar 11 '15 at 22:23