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QUESTION: A cylindrical oil tank is laying on it's side on a level rack. A stick is placed through a hole in the upper side, and indicates 1 foot of oil in the tank.

Dimensions of tank: diameter- 10 ft. length- 20 ft.

Calculate number of gallons of oil in the tank.

Thanks.

ANSWER: Look at the tank in cross-section, which is a circle of radius r = 5 ft.
The oil is the segment with height h = 1 ft.

θ = 2·arccos((r - h)/r) = 2·arccos(4/5) ≅ 1.29 radians.

Area of segment = r²(θ-sinθ)/2 ≅ 4.09 ft²
Volume of oil in tank = 20×4.09 = 81.75 ft³
81.75 ft³ × 7.48 gallons/ft³ = 611.5 gallons

[an error occurred while processing this directive]---------- FOLLOW-UP ----------

QUESTION: "θ = 2·arccos((r - h)/r) = 2·arccos(4/5) ≅ 1.29 radians.

Area of segment = r²(θ-sinθ)/2 ≅ 4.09 ft²"

It's all chord math. The basic variables are

s = Arc Length
c = Chord Length
h = Height (outside radius to chord midpoint)
d = Apothem (distance from chord to center of circle)
K = Area of segment (chord to outside)
Using any two of these variables, you can derive the others.

You were given r = 5 and h = 1, so d = r - h = 4.

To determine θ, look at the right triangle:
You can see that cos(θ/2) = d/r, so θ = 2·arccos(d/r).

area of the sector = (πr²)·θ/(2π) = r²θ/2

Note that sin(θ/2) = c/(2r).
By the double-angle formula, sin(θ/2) = sinθ/(2cos(θ/2)) = r·sinθ/(2d).
area of the right triangle = (½)(c/2)d
= (½)(r²sinθ/(2d))d
= r²sinθ/4

area of the oil segment = (area of sector) - 2(area of right triangle)
= r²θ/2 - 2(r²sinθ/4)
= r²θ/2 - r²·sinθ/2
= r²(θ - sinθ)/2

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