work

[ProBasket]

You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved perfectly into the water. However, now that you want to swim, you must remove all of the Kool-Aid contaminated water. The swimming pool is round, with a 14 foot radius. It is 12 feet tall and has 1 feet of water in it.
How much work is required to remove all of the water by pumping it over the side? Use the physical definition of work, and the fact that the weight of the Kool-Aid contaminated water is 65.7lbs/ft^3


first of all, i figure the bounds of integration to be from 1 to 12.

volume = pi(14)^2

=196pi*ft = 65.7lbs/ft^3

so...
196pi(65.7)

my function:
distance would just be equal to x right?
\int_{1}^{12} 196*pi(65.7) * x dx

[learningphysics]

[QUOTE=ProBasket]You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved perfectly into the water. However, now that you want to swim, you must remove all of the Kool-Aid contaminated water. The swimming pool is round, with a 14 foot radius. It is 12 feet tall and has 1 feet of water in it.
How much work is required to remove all of the water by pumping it over the side? Use the physical definition of work, and the fact that the weight of the Kool-Aid contaminated water is 65.7lbs/ft^3


first of all, i figure the bounds of integration to be from 1 to 12.
[/tex]

No (unless I'm misunderstanding). You want to integrate over the height of water that needs to be pumped. There is only 1 foot of water. Draw a picture and label an axis. You can let the bottom of the pool be 0 and the top of the water be at 1. So you'd integrate from 0 to 1. The weight of an amount dx is 196pi(65.7)dx. What is the distance that this amount needs to be pushed through?

[Nenad]

[QUOTE=ProBasket]You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved perfectly into the water. However, now that you want to swim, you must remove all of the Kool-Aid contaminated water. The swimming pool is round, with a 14 foot radius. It is 12 feet tall and has 1 feet of water in it.
How much work is required to remove all of the water by pumping it over the side? Use the physical definition of work, and the fact that the weight of the Kool-Aid contaminated water is 65.7lbs/ft^3


first of all, i figure the bounds of integration to be from 1 to 12.
[/tex]

No (unless I'm misunderstanding). You want to integrate over the height of water that needs to be pumped. There is only 1 foot of water. Draw a picture and label an axis. You can let the bottom of the pool be 0 and the top of the water be at 1. So you'd integrate from 0 to 1. The weight of an amount dx is 196pi(65.7)dx. What is the distance that this amount needs to be pushed through?

The water needs to be pumped out of the pool, so the bounds of integration would be from 1 to 12. Youre not pumping the water from nside the pool to a height of 1m. The pools depth is 12m.

Regards,

Nenad

[learningphysics]

The water needs to be pumped out of the pool, so the bounds of integration would be from 1 to 12. Youre not pumping the water from nside the pool to a height of 1m. The pools depth is 12m.

Regards,

Nenad

But there is only 1 foot of water in the pool. You are integrating over the water that needs to be pumped, not the distance that needs to be travelled.

Anyway here's the integral I figured:
\int_{0}^{1} 196*pi(65.7) * (12 - x) dx

[Nenad]

I see. My mistake. you're right, that integrals is correct. I mixed up the distance traveled with the integrals bounds.

Regards,

Nenad