Advanced Math/Volume Follow-up
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.
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"
Please explain this part of your solution in more detail. Thanks.
It's all chord math. The basic variables are
s = Arc Length
c = Chord Length
r = Radius
h = Height (outside radius to chord midpoint)
θ = Angle (in radians)
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
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