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 r2=(s/2)2+(r−h)2 r2=0.25+(r−0.2)2 r=0.725 H=2⋅ r−h=2⋅ 0.725−0.2=54=1.25 A=r2⋅ arccos⁡((r−h)/r)=0.7252⋅ arccos⁡((0.725−0.2)/0.725)≐0.4 B=(r−h)⋅ 2⋅ r⋅ h−h2=(0.725−0.2)⋅ 2⋅ 0.725⋅ 0.2−0.22=2180=0.2625 S1=A−B=0.4−0.2625≐0.1375 S2=π⋅ r2=3.1416⋅ 0.7252≐1.6513 S=S2−S1=1.6513−0.1375≐1.5138 V=S⋅ a=1.5138⋅ 10≐15.1379≐15.138 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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