The depth is constant along east-west lines and increases linearly from 1 ft at the south end to 6 ft at the north end. Find the volume of water in the pool.

The method of the previous answer is fine but he has forgotten to square the 20 so final result should be almost 4400 ft^3.

If the depth increase linearly from one side to another, that means you can use the mean as a constant depth for all the pool.

Depth = (1 + 6) / 2 = 7 / 2 = 3.5 ft

Radius r = 40 / 2 = 20 ft

Volume = Area * depth

Volume = pi*r^2 * 3.5

Volume = 3.1416*20 * 3.5

Volume = 219.9 ft^3

EDIT :

Thank you mathsman. You are right.

Inattention error :

3.1416 * 20^2 * 3.5 = 4398.24 ft^3

Found a “pool calculator”

http://www.swimmingpool.com/maintenance/testing-yo…

My problem has 30 ft diameter. 4 ft at one end and 9 ft at the other. I put the dimensions into the site above and it gave me the volume in gallons. (34377 gal) converted to cubic feet (4595.54 ft^3). Question wanted my answer rounded to the nearest whole number. I put 4596 and it was wrong. Then I tried 4595 and it was right. Pretty nifty. Try it out yourself.

diameter = 10ft depth: 5 to 10 ft linearly average depth = (5 + 10)/2 = 15/2 = 7.5ft Volume = area * average depth V = pi * radius^2 * 7.5 V = pi * (10/2)^2 * 7.5 V = pi * 5^2 * 7.5 V = pi * 25 * 7.5 V = 589.05ft^3

(πd^2(h_1+h_2))/8

where d is the diameter and h_1/h_2 are the two depths (6ft and 1ft)

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