# 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:

$A=30 ^\circ \rightarrow\ \text{rad}=30 ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ =30 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ =0.5236=π/6 \ \\ V_{1}=10 \ l \rightarrow cm^3=10 \cdot \ 1000 \ cm^3=10000 \ cm^3 \ \\ \ \\ \cos A=r:R \ \\ \sin A=v:R \ \\ V_{2}=\dfrac{ \pi v }{ 6 } \cdot \ (3r^2 +v^2) \ \\ \ \\ V=V_{1}+V_{2}=\dfrac{ 1 }{ 2 } \cdot \ \dfrac{ 4 }{ 3 } \pi R^3=\dfrac{ 2 }{ 3 } \pi R^3 \ \\ \ \\ V_{2}=\dfrac{ \pi R \sin A }{ 6 } \cdot \ (3(R \cos A)^2 +(R \sin A)^2) \ \\ \ \\ V_{2}=\dfrac{ \pi R^3 \ \sin A }{ 6 } \cdot \ (3(\cos A)^2 +(\sin A)^2) \ \\ \ \\ \dfrac{ 2 }{ 3 } \pi R^3=V_{1} + \dfrac{ \pi R^3 \ \sin A }{ 6 } \cdot \ (3(\cos A)^2 +(\sin A)^2) \ \\ \ \\ k=\dfrac{ \pi \cdot \ \sin(A) }{ 6 } \cdot \ (3 \cdot \ (\cos(A))^2 +(\sin(A))^2)=\dfrac{ 3.1416 \cdot \ \sin(0.5236) }{ 6 } \cdot \ (3 \cdot \ (\cos(0.5236))^2 +(\sin(0.5236))^2) \doteq 0.6545 \ \\ \ \\ \dfrac{ 2 }{ 3 } \pi R^3=V_{1} + k \cdot \ R^3 \ \\ \ \\ R=\sqrt[3]{ \dfrac{ V_{1} }{ \dfrac{ 2 }{ 3 } \cdot \ \pi - k } }=\sqrt[3]{ \dfrac{ 10000 }{ \dfrac{ 2 }{ 3 } \cdot \ 3.1416 - 0.6545 } } \doteq 19.079 \ \text{cm} \ \\ \ \\ \ \\ V=\dfrac{ 2 }{ 3 } \cdot \ \pi \cdot \ R^3=\dfrac{ 2 }{ 3 } \cdot \ 3.1416 \cdot \ 19.079^3 \doteq 14545.4545 \ \text{cm}^3 \ \\ r=R \cdot \ \cos(A)=19.079 \cdot \ \cos(0.5236) \doteq 16.5229 \ \text{cm} \ \\ v=R \cdot \ \sin(A)=19.079 \cdot \ \sin(0.5236) \doteq 9.5395 \ \text{cm} \ \\ V_{2}=\dfrac{ \pi \cdot \ v }{ 6 } \cdot \ (3 \cdot \ r^2 +v^2)=\dfrac{ 3.1416 \cdot \ 9.5395 }{ 6 } \cdot \ (3 \cdot \ 16.5229^2 +9.5395^2) \doteq \dfrac{ 50000 }{ 11 } \doteq 4545.4545 \ \text{cm}^3 \ \\ \ \\ V_{8}=V-V_{2}=14545.4545-4545.4545=10000 \ \text{cm}^3 \ \\ V_{8}=V_{1} \ \\ \ \\ \ \\ R=19.079 \doteq 19.079=19.079 \ \text{cm}$

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