Calc 2 problem, Calculating Work Done by Pumping Water from Pool

[dli42395]

A circular swimming pool has a diameter of 12 meters, the sides are 3 meters high, and the depth of the water is 2.5 meters. The acceleration due to gravity is 9.8 m/s^2 and the density of water is 1000 kg/m^2

How much work is required to:
(a) pump all of the water over the side?
(b) pump all of the water out of an outlet 1 m over the side?

a) I tried integrating from 2.5 to 3: (1000)(9.8)(3)(36pi), and that wasn't correct

 

[HallsofIvy]

"Tried integrating" what from 3.5 to 4? Surely not that expression you give- that has no variable, it must be the result of your integration.

Imagine a "layer" of water x m above the bottom of the pool with thickness dx. The pool has cross sectional area $\displaystyle 36\pi$ so that "layer" has volume $\displaystyle 36\pi dx$. Since "the acceleration due to gravity is 9.8 m/s^2 and the density of water is 1000 kg/m^2", (that should be "kg/m^3") water has weight density 9800 Newtons per cubic meter and the weight of that "layer" of water is 352800\pi dx Newtons. Since the pool has height 3 m, we must lift that "layer" a distance 3- x meters which will require $\displaystyle 352800\pi (3- x)dx$ Joules of work. To find the work done in lifting all of the water, integrate that from x= 0 to x= 2.5.

For (b), the distance the water is lifted is 4- x since the water must be lifted an additional meter to the outlet.

 

[dli42395]

Thank you! That makes sense now! And sorry about the typos!

 

[Ishuda]

Thank you! That makes sense now! And sorry about the typos!

Just remember that if you were pumping it into another pool of the same dimensions and orientation such that the effect was just a horizontal translation of a 'patch of water', the total work done would be zero even if you were completely drained from pumping all that water.

 

[HallsofIvy]

The moral being- get a job working with your mind, not your muscles!