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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 $\displaystyle h=0$ as the bottom of the pool with $\displaystyle h$ measured positive upward, to the rim of the pool is at $\displaystyle h=5$, and the top of the water is initially at $\displaystyle h=4$. Let $\displaystyle r$ be the radius of the pool. Let $\displaystyle \rho$ be the density of water and $\displaystyle g$ the acceleration due to gravity.Quote:
Originally Posted by WahooMan
Consider a layer of thickness $\displaystyle \delta h$, at a height $\displaystyle h<4$ 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 $\displaystyle h$ to $\displaystyle 5$ which is:
$\displaystyle \delta W=(\pi r^2 \delta h)\times \rho \times g\times(5-h)$
So the total work is the sum of the work for all the layers of thickness $\displaystyle \delta h$ and proceeding to the limit we get:
$\displaystyle
W=\int_{h=0}^4 (\pi r^2 )\times \rho \times g\times(5-h)\ dh
$
CB
Thank you.
Slightly disappointed this thread isn't about using integrals to find employment. :(
Yep, employers just never give calculus skills sufficient credit!Quote:
Originally Posted by undefined
