# 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:

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

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