# Math Help - Question about using integrals to find work

1. ## Question about using integrals to find work

My problem says that I need to figure out how much work has to be done to pump all of the water out over the side of a 5 foot tall circular pool, but the water is only 4 feet deep. I was wondering, would my integral be from 0 to 4 or 1 to 5?

Here is what I've set up:

Given: Weight of water = 62.4 lbs per cubic foot, diameter of pool = 24 ft

W = 62.4(144pi/25) integral (5y^3 - y^4)dy

Is that right? And would the integral be from 1 to 5 or 0 to 4?

2. Originally Posted by WahooMan
My problem says that I need to figure out how much work has to be done to pump all of the water out over the side of a 5 foot tall circular pool, but the water is only 4 feet deep. I was wondering, would my integral be from 0 to 4 or 1 to 5?

Here is what I've set up:

Given: Weight of water = 62.4 lbs per cubic foot, diameter of pool = 24 ft

W = 62.4(144pi/25) integral (5y^3 - y^4)dy

Is that right? And would the integral be from 1 to 5 or 0 to 4?
Take $h=0$ as the bottom of the pool with $h$ measured positive upward, to the rim of the pool is at $h=5$, and the top of the water is initially at $h=4$. Let $r$ be the radius of the pool. Let $\rho$ be the density of water and $g$ the acceleration due to gravity.

Consider a layer of thickness $\delta h$, at a height $h<4$ above the bottom of the pool. The work needed to pump this out is the change in potential energy when the layer is lifted form $h$ to $5$ which is:

$\delta W=(\pi r^2 \delta h)\times \rho \times g\times(5-h)$

So the total work is the sum of the work for all the layers of thickness $\delta h$ and proceeding to the limit we get:

$
W=\int_{h=0}^4 (\pi r^2 )\times \rho \times g\times(5-h)\ dh
$

CB

3. Thank you.

4. Slightly disappointed this thread isn't about using integrals to find employment.

5. Originally Posted by undefined