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?


V =  15.138 m3


a=10 s=1 h=0.2 r2=(s/2)2+(rh)2 r2=0.25+(r0.2)2 r=0.725 H=2 rh=2 0.7250.2=54=1.25 A=r2 arccos((rh)/r)=0.7252 arccos((0.7250.2)/0.725)0.4 B=(rh) 2 r hh2=(0.7250.2) 2 0.725 0.20.22=2180=0.2625 S1=AB=0.40.26250.1375 S2=π r2=3.1416 0.72521.6513 S=S2S1=1.65130.13751.5138 V=S a=1.5138 1015.137915.138 m3a=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 \doteq 15.138 \ \text{m}^3

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