# The hemisphere

The hemisphere container is filled with water. What is the radius of the container when 10 liters of water pour from it when tilted 30 degrees?

Result

R =  19.079 cm

#### Solution:

$A = (30^\circ \rightarrow rad) = (30 \cdot \ \dfrac{ \pi }{ 180 } \ ) = 0.523598775598 \ \\ V_{ 1 } = 10 \ l = 10 \cdot \ 1000 \ cm^3 = 10000 \ cm^3 \ \\ \ \\ \cos A = r:R \ \\ \sin A = v:R \ \\ V_{ 2 } = \dfrac{ \pi v }{ 6 } \cdot \ (3r^2 +v^2) \ \\ \ \\ V = V_{ 1 }+V_{ 2 } = \dfrac{ 1 }{ 2 } \cdot \ \dfrac{ 4 }{ 3 } \pi R^3 = \dfrac{ 2 }{ 3 } \pi R^3 \ \\ \ \\ V_{ 2 } = \dfrac{ \pi R \sin A }{ 6 } \cdot \ (3(R \cos A)^2 +(R \sin A)^2) \ \\ \ \\ V_{ 2 } = \dfrac{ \pi R^3 \ \sin A }{ 6 } \cdot \ (3(\cos A)^2 +(\sin A)^2) \ \\ \ \\ \dfrac{ 2 }{ 3 } \pi R^3 = V_{ 1 } + \dfrac{ \pi R^3 \ \sin A }{ 6 } \cdot \ (3(\cos A)^2 +(\sin A)^2) \ \\ \ \\ k = \dfrac{ \pi \cdot \ \sin(A) }{ 6 } \cdot \ (3 \cdot \ (\cos(A))^2 +(\sin(A))^2) = \dfrac{ 3.1416 \cdot \ \sin(0.5236) }{ 6 } \cdot \ (3 \cdot \ (\cos(0.5236))^2 +(\sin(0.5236))^2) \doteq 0.6545 \ \\ \ \\ \dfrac{ 2 }{ 3 } \pi R^3 = V_{ 1 } + k \cdot \ R^3 \ \\ \ \\ R = \sqrt[3]{ \dfrac{ V_{ 1 } }{ \dfrac{ 2 }{ 3 } \cdot \ \pi - k } } = \sqrt[3]{ \dfrac{ 10000 }{ \dfrac{ 2 }{ 3 } \cdot \ 3.1416 - 0.6545 } } \doteq 19.079 \ cm \ \\ \ \\ \ \\ V = \dfrac{ 2 }{ 3 } \cdot \ \pi \cdot \ R^3 = \dfrac{ 2 }{ 3 } \cdot \ 3.1416 \cdot \ 19.079^3 \doteq 14545.4545 \ cm^3 \ \\ r = R \cdot \ \cos(A) = 19.079 \cdot \ \cos(0.5236) \doteq 16.5229 \ cm \ \\ v = R \cdot \ \sin(A) = 19.079 \cdot \ \sin(0.5236) \doteq 9.5395 \ cm \ \\ V_{ 2 } = \dfrac{ \pi \cdot \ v }{ 6 } \cdot \ (3 \cdot \ r^2 +v^2) = \dfrac{ 3.1416 \cdot \ 9.5395 }{ 6 } \cdot \ (3 \cdot \ 16.5229^2 +9.5395^2) = \dfrac{ 50000 }{ 11 } \doteq 4545.4545 \ cm^3 \ \\ \ \\ V_{ 8 } = V-V_{ 2 } = 14545.4545-4545.4545 = 10000 \ cm^3 \ \\ V_{ 8 } = V_{ 1 } \ \\ \ \\ \ \\ R = 19.079 \doteq 19.079 = 19.079 \ \text { cm }$

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