The radius of the sphere we reduce by 1/3 of the original radius. How much percent does the volume and surface of the sphere change?

Correct result:

p1 =  70.37 %
p2 =  55.556 %

#### Solution:

$r_{1}=1 \ \\ r_{2}=r_{1} - r_{1}/3=1 - 1/3 \doteq \dfrac{ 2 }{ 3 } \doteq 0.6667 \ \\ \ \\ V_{1}=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r_{1}^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 1^3 \doteq 4.1888 \ \\ V_{2}=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r_{2}^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 0.6667^3 \doteq 1.2411 \ \\ \ \\ p_{1}=100 \cdot \ \dfrac{ V_{1}-V_{2} }{ V_{1} }=100 \cdot \ \dfrac{ 4.1888-1.2411 }{ 4.1888 }=\dfrac{ 1900 }{ 27 }=70.37 \%$
$S_{1}=4 \pi \cdot \ r_{1}^2=4 \cdot \ 3.1416 \cdot \ 1^2 \doteq 12.5664 \ \\ S_{2}=4 \pi \cdot \ r_{2}^2=4 \cdot \ 3.1416 \cdot \ 0.6667^2 \doteq 5.5851 \ \\ \ \\ p_{2}=100 \cdot \ \dfrac{ S_{1}-S_{2} }{ S_{1} }=100 \cdot \ \dfrac{ 12.5664-5.5851 }{ 12.5664 }=\dfrac{ 500 }{ 9 }=55.556 \%$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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