Sphere radius

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 =  70.3704 %
p₂ =  55.5556 %

Solution:

$$\begin{gathered} r_1=1 \\ {r}_2=r_1-r_1\mathrm{/}3=1-1\mathrm{/}3={\displaystyle \frac{2}{3}}\doteq 0.6667 \\ {V}_1={\displaystyle \frac{4}{3}}\cdot \pi \cdot r_1^3={\displaystyle \frac{4}{3}}\cdot 3.1416\cdot 1^3\doteq 4.1888 \\ {V}_2={\displaystyle \frac{4}{3}}\cdot \pi \cdot r_2^3={\displaystyle \frac{4}{3}}\cdot 3.1416\cdot 0.6667^3\doteq 1.2411 \\ {p}_1=100\cdot {\displaystyle \frac{V_1-V_2}{V_1}}=100\cdot {\displaystyle \frac{4.1888-1.2411}{4.1888}}={\displaystyle \frac{1900}{27}}=70.3704\mathrm{\%} \end{gathered}$$

$$\begin{gathered} S_1=4\pi \cdot r_1^2=4\cdot 3.1416\cdot 1^2\doteq 12.5664 \\ {S}_2=4\pi \cdot r_2^2=4\cdot 3.1416\cdot 0.6667^2\doteq 5.5851 \\ {p}_2=100\cdot {\displaystyle \frac{S_1-S_2}{S_1}}=100\cdot {\displaystyle \frac{12.5664-5.5851}{12.5664}}={\displaystyle \frac{500}{9}}=55.5556\mathrm{\%} \end{gathered}$$

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