Spherical tank

The tank of a water tower is a sphere of radius 35ft. If the tank is filled to one quarter of full, what is the height of the water?

Result

h =  22.845 ft

Solution:

r=35 ft  V=43⋅ π⋅ r3=43⋅ 3.1416⋅ 353≐179594.38 ft3  V1=14⋅ V=14⋅ 179594.38≐44898.595 ft3  V2=π⋅ h23(3r−h)  V1=V2 π⋅ r33=π⋅ h23(3r−h) r3=h2(3r−h)  h1=−18.6231 h2=22.8446 h3=100.778  0<h<2r  h=h2=22.8446=22.845 ft  V2=π⋅ h23⋅ (3⋅ r−h)=3.1416⋅ 22.844623⋅ (3⋅ 35−22.8446)≐44898.5017 ft3r=35 \ \text{ft} \ \\ \ \\ V=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 35^3 \doteq 179594.38 \ \text{ft}^3 \ \\ \ \\ V_{1}=\dfrac{ 1 }{ 4 } \cdot \ V=\dfrac{ 1 }{ 4 } \cdot \ 179594.38 \doteq 44898.595 \ \text{ft}^3 \ \\ \ \\ V_{2}=\dfrac{ \pi \cdot \ h^2 }{ 3 } (3r-h) \ \\ \ \\ V_{1}=V_{2} \ \\ \dfrac{ \pi \cdot \ r^3 }{ 3 }=\dfrac{ \pi \cdot \ h^2 }{ 3 } (3r-h) \ \\ r^3=h^2(3r-h) \ \\ \ \\ h_{1}=-18.6231 \ \\ h_{2}=22.8446 \ \\ h_{3}=100.778 \ \\ \ \\ 0 < h < 2r \ \\ \ \\ h=h_{2}=22.8446=22.845 \ \text{ft} \ \\ \ \\ V_{2}=\dfrac{ \pi \cdot \ h^2 }{ 3 } \cdot \ (3 \cdot \ r-h)=\dfrac{ 3.1416 \cdot \ 22.8446^2 }{ 3 } \cdot \ (3 \cdot \ 35-22.8446) \doteq 44898.5017 \ \text{ft}^3

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