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An oil container in the form of a right circular cylinder with a horizontal axis is filled to a

An oil container in the form of a right circular cylinder with a horizontal axis is filled to a
Bu haber 16 Temmuz 2022 - 11:17 'de eklendi ve kez görüntülendi.

The tank is on its side, so the above calculation won’t work.

Draw a side view of the end of the tank. Since the radius of the tank is 5, the oil surface is 2 feet chord from the center of the circle. This means that the width of the oil surface is 2√(5^2-2^2) = 2√21 ft and forms an angle θ where sin(θ/2) = √21/5
So the area of ​​the oil is 1/2 r^2 (θ – sinθ)
Since sin(θ/2) = √21/5, sinθ = 2(√21/5)(2/5) = 4√21/25
so the area is 1/2 * 25 (0.823 – 4√21/25) = 1.122 ft^2
That makes the oil volume 1.122 * 12 = 13.464 ft^3

The cylinder is on its side, so the question is a bit trickier.

Make a sketch of the circular cross-section
Draw the waterline at a depth of 3 feet, connect this chord to the ends of the
Diameter.
You should see 2 congruent right triangles, each with
Hypotenuse of 5 and a leg of 2 . (5-3 = 2)
Find base:
x^2 + 2^2 = 5^2
x = √21
So the length of the chord (the waterline) = 2√21
(Never used this, switched the strategy to using angles.)

Center angle: half center angle:
cosθ = 2/5, θ = 66.42°
i.e. central angle = 132.84°

area of sector :

sector/(π(5^2)) = 132.84/360
Sector = 28.982 ft^2

Area of ​​the triangle on the chord:
= 1/2 (5)(5)sin 132.84
= 9.165 feet^2

area of segment = 28.982 – 9.165 = 19.8168

Water volume = surface area * length
= 19.8168*12 = 237.8 feet^3

Check my calculations.

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