The radius of a sphere is increased by $10 \%$. Prove that the volume will be increased by $33.1 \%$ approximately.

(N/A) The volume of a sphere $V = \frac{4}{3} \pi r^{3}$.
Let the initial radius be $r$. If the radius is increased by $10 \%$,the new radius $r'$ becomes $r + 0.1r = 1.1r$.
The new volume $V'$ is given by $V' = \frac{4}{3} \pi (r')^{3} = \frac{4}{3} \pi (1.1r)^{3}$.
Calculating $(1.1)^{3} = 1.331$,so $V' = 1.331 \times (\frac{4}{3} \pi r^{3}) = 1.331V$.
The increase in volume is $\Delta V = V' - V = 1.331V - V = 0.331V$.
The percentage increase in volume is $\frac{\Delta V}{V} \times 100 = \frac{0.331V}{V} \times 100 = 33.1 \%$.
Thus,the volume increases by $33.1 \%$.