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April 14th, 2017, 12:18 PM   #1
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Calculating the volume of a circular pool

A swimming pool is circular with a 40 meter diameter. The depth is constant along east-west
lines and increases linearly from 2 meters at the south end to 7 meters at the north end.
Find the volume of the pool.

Answer: 1800pi

This is what I came up with (but it's incorrect). Please anyone tell me what I did wrong...


$\displaystyle \int_{-20}^{20}\int_{-\sqrt{400 - x^2}}^{\sqrt{400 - x^2}} -1/8 (y+20) - 2 dydx$

Last edited by skipjack; April 14th, 2017 at 03:33 PM.
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April 14th, 2017, 01:23 PM   #2
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You say the swimming pool is "circular", but that only applies to the surface of the pool. Are we to assume that the sides of the pool are vertical and that the bottom of the pool flat?

If so, then modeling the surface of the pool as a disk with center at (0, 0), and radius 20 m, then, we can write the equation of the surface of the pool at $\displaystyle x^2+ y^2= 400$ so we can take x going from -20 to 20 and, for each x, y from $\displaystyle -\sqrt{40- x^2}$ to $\displaystyle \sqrt{400- x^2}$. That you have.

Now the depth "increases linearly from 2 meters at the south end to 7 meters at the north" so setting z= ay+ b, we have z= a(-20)+ b= 2 and z= a(20)+ b= 7. Subtracting the first of those equation from the second eliminates b and give 40a= 5 so that a= 5/40= 1/8. Putting that into the second equation, 20(1/8)+ b= b+ 5/2= 7 so b= 7- 5/2= 9/2. z= y/8+ 9/2. That is the same as z= (1/8)(y+ 20)+ 2 which is what you would have if you removed those "-" signs.
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Last edited by skipjack; April 14th, 2017 at 03:36 PM.
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April 14th, 2017, 02:35 PM   #3
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Yes. It is a cylinder pool. The bottom of the pool is flat but decline (or incline) at a degree.

Last edited by skipjack; April 14th, 2017 at 03:37 PM.
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April 14th, 2017, 04:00 PM   #4
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Volume = average depth × surface area = 4.5m × $\pi$(20m)² = 1800$\pi$ m³.
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