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⋅ 10−r−R=2⋅ 10−8−6=6 h=(r+R)2−x2=(8+6)2−62≐4 10≐12.6491 H=h+r+R=12.6491+8+6≐26.6491 V1=π⋅ 102⋅ H=3.1416⋅ 102⋅ 26.6491≐8372.065 V2=4/3π⋅ r3=4/3⋅ 3.1416⋅ 83≐2144.6606 V3=4/3π⋅ R3=4/3⋅ 3.1416⋅ 63≐904.7787 V=V1−V2−V3=8372.065−2144.6606−904.7787≐5322.6258≐5322.626 cm3r=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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