Eight

Eight small Christmas balls with a radius of 1 cm have the same volume as one large Christmas ball. What has a bigger surface: eight small balls, or one big ball?

Correct result:

x =  8

Solution:

n=8 r=1 cm  V1=n⋅ 43⋅ π⋅ r3=8⋅ 43⋅ 3.1416⋅ 13≐33.5103 cm3  V2=V1=33.5103≐33.5103 cm3  V2=43⋅ π⋅ r23  r2=3⋅ V2/4/π3=3⋅ 33.5103/4/3.14163=2 cm  S1=n⋅ 4π⋅ r2=8⋅ 4⋅ 3.1416⋅ 12≐100.531 cm2 S2=4π⋅ r22=4⋅ 3.1416⋅ 22≐50.2655 cm2  S1>S2  x=8n=8 \ \\ r=1 \ \text{cm} \ \\ \ \\ V_{1}=n \cdot \ \dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r^3=8 \cdot \ \dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 1^3 \doteq 33.5103 \ \text{cm}^3 \ \\ \ \\ V_{2}=V_{1}=33.5103 \doteq 33.5103 \ \text{cm}^3 \ \\ \ \\ V_{2}=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r_{2}^3 \ \\ \ \\ r_{2}=\sqrt[3]{ 3 \cdot \ V_{2}/4/\pi}=\sqrt[3]{ 3 \cdot \ 33.5103/4/3.1416}=2 \ \text{cm} \ \\ \ \\ S_{1}=n \cdot \ 4 \pi \cdot \ r^2=8 \cdot \ 4 \cdot \ 3.1416 \cdot \ 1^2 \doteq 100.531 \ \text{cm}^2 \ \\ S_{2}=4 \pi \cdot \ r_{2}^2=4 \cdot \ 3.1416 \cdot \ 2^2 \doteq 50.2655 \ \text{cm}^2 \ \\ \ \\ S_{1}>S_{2} \ \\ \ \\ x=8

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