# Two balls

Two balls, one 8cm in radius and the other 6cm in radius, are placed in a cylindrical plastic container 10cm in radius. Find the volume of water necessary to cover them.

Result

V =  5322.626 cm3

#### Solution:

$r=8 \ \\ R=6 \ \\ x=2 \cdot \ 10-r-R=2 \cdot \ 10-8-6=6 \ \\ h=\sqrt{ (r+R)^2-x^2 }=\sqrt{ (8+6)^2-6^2 } \doteq 4 \ \sqrt{ 10 } \doteq 12.6491 \ \\ H=h + r + R=12.6491 + 8 + 6 \doteq 26.6491 \ \\ V_{1}=\pi \cdot \ 10^2 \cdot \ H=3.1416 \cdot \ 10^2 \cdot \ 26.6491 \doteq 8372.065 \ \\ V_{2}=4/3 \pi \cdot \ r^3=4/3 \cdot \ 3.1416 \cdot \ 8^3 \doteq 2144.6606 \ \\ V_{3}=4/3 \pi \cdot \ R^3=4/3 \cdot \ 3.1416 \cdot \ 6^3 \doteq 904.7787 \ \\ V=V_{1} - V_{2}-V_{3}=8372.065 - 2144.6606-904.7787 \doteq 5322.6258 \doteq 5322.626 \ \text{cm}^3$

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