# Horizontal Cylindrical Segment

How much fuel is in the tank of horizontal cylindrical segment with a length 10m, width of level 1 meter and level is 0.2 meters below the upper side of the tank?

Result

V =  15.138 m3

#### Solution:

$a = 10 \ \\ s = 1 \ \\ h = 0.2 \ \\ r^2 = (s/2)^2+(r-h)^2 \ \\ r^2 = 0.25 + (r-0.2)^2 \ \\ r = 0.725 \ \\ H = 2 \cdot \ r-h = 2 \cdot \ 0.725-0.2 = \dfrac{ 5 }{ 4 } = 1.25 \ \\ A = r^2 \cdot \ \arccos((r-h)/r) = 0.725^2 \cdot \ \arccos((0.725-0.2)/0.725) \doteq 0.4 \ \\ B = (r-h) \cdot \ \sqrt{ 2 \cdot \ r \cdot \ h-h^2 } = (0.725-0.2) \cdot \ \sqrt{ 2 \cdot \ 0.725 \cdot \ 0.2-0.2^2 } = \dfrac{ 21 }{ 80 } = 0.2625 \ \\ S_{ 1 } = A - B = 0.4 - 0.2625 \doteq 0.1375 \ \\ S_{ 2 } = \pi \cdot \ r^2 = 3.1416 \cdot \ 0.725^2 \doteq 1.6513 \ \\ S = S_{ 2 }-S_{ 1 } = 1.6513-0.1375 \doteq 1.5138 \ \\ V = S \cdot \ a = 1.5138 \cdot \ 10 \doteq 15.1379 = 15.138 \ m^3$

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