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I am in Calculus 2 and ran across this problem. I'm struggling with this subject in general, so I suspect I may be missing something fairly fundamental.

A circular swimming pool has a diameter of 14 m, the sides are 3 m high, and the depth of the water is 2.5 m. (The acceleration due to gravity is 9.8 m/s2 and the density of water is 1000 kg/m3.) How much work (in Joules) is required to: (a) pump all of the water over the side? (b) pump all of the water out of an outlet 2 m over the side?

The problem gives me: $$r: 7 m$$ $$h: 2.5 m$$ $$H: 3 m $$ $$Density = 1000 \frac{kg}{m^3}$$ $$Acceleration = 9.8 \frac{m}{s^2}$$

Just learned: $$Work = (Force)(distance)$$

So I need to solve for force: $$Force = (Mass)(Acceleration)$$

I don't have mass, so I'll solve for that: $$Mass = (Density)(Volume)$$

Solving for volume: $$Volume = h \pi r^2 = 2.5 \pi 49 = 384.8451001 m^3$$

Solving for mass: $$Mass = 384845.1001 kg$$

Solving for force: $$3771481.981 Newtons$$

So to solve for work: $$Work = \int_{0.5}^{3} (3771481.981x) dx$$ $$= 3771481.981 \int_{0.5}^{3} (x) dx$$

For part A) $$= 3771481.981\frac{x^2}{2}| 3 \leq x \leq 0.5$$

For part B) $$= 3771481.981\frac{x^2}{2}| 5 \leq x \leq 0.5$$

However this is wrong. Where am I going wrong?

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When you compute the work in your integral, the force should not be the force of gravity on the whole tank, but just on a slice of thickness $dh$, so the force should be $g\rho \pi r^2 dh$. That is because you pump that slice up a certain distance and other slices up a different distance. For part B, you can just lift all the water $2$ meters, then use the answer from part A. Otherwise, you should just integrate from $2.5$ to $5$ for part A as no water will only be lifted $0.5$ to $2.5$ meters.

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  • $\begingroup$ What is the intuition behind integrating from 2.5 to 5, instead of 0.5 to 5? $\endgroup$ – Storm Kiernan Jan 30 '16 at 22:24
  • $\begingroup$ Because the top layer of water is lifted $2.5$ meters and the bottom layer of water is lifted $5$ meters. $\endgroup$ – Ross Millikan Jan 30 '16 at 22:27
  • $\begingroup$ So, in general, the bounds of integration range from the distance to the top layer, to the distance to the bottom layer? $\endgroup$ – Storm Kiernan Jan 30 '16 at 22:34
  • $\begingroup$ That is right. You are thinking of raising each layer of water as required. The top layer needs the least lifting and the bottom layer the most. $\endgroup$ – Ross Millikan Jan 31 '16 at 0:19
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For part (a) we're assuming pumping puts all the water at the same level. So water at a lower level requires more work to pump out.

Imagine a cylinder of water at depth $x$ meter (measured from the top of the side of the pool) of height $dx$ meter in the pool. Mass of this cylinder = $49000\pi dx$ kg. Work needed to pump this cylinder out is $9.8\times49000\pi x dx$ J. So the total work required is

$$ \int_{0.5}^{3}9.8\times49000\pi x dx $$

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