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:

$r_{1}=1 \ \\ r_{2}=r_{1} - r_{1}/3=1 - 1/3 \doteq \dfrac{ 2 }{ 3 } \doteq 0.6667 \ \\ \ \\ V_{1}=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r_{1}^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 1^3 \doteq 4.1888 \ \\ V_{2}=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r_{2}^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 0.6667^3 \doteq 1.2411 \ \\ \ \\ p_{1}=100 \cdot \ \dfrac{ V_{1}-V_{2} }{ V_{1} }=100 \cdot \ \dfrac{ 4.1888-1.2411 }{ 4.1888 }=\dfrac{ 1900 }{ 27 }=70.37 \%$
$S_{1}=4 \pi \cdot \ r_{1}^2=4 \cdot \ 3.1416 \cdot \ 1^2 \doteq 12.5664 \ \\ S_{2}=4 \pi \cdot \ r_{2}^2=4 \cdot \ 3.1416 \cdot \ 0.6667^2 \doteq 5.5851 \ \\ \ \\ p_{2}=100 \cdot \ \dfrac{ S_{1}-S_{2} }{ S_{1} }=100 \cdot \ \dfrac{ 12.5664-5.5851 }{ 12.5664 }=\dfrac{ 500 }{ 9 }=55.556 \%$

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