If volume of a sphere increases by 72.8%, what is change of its surface area?

Recall that

• volume $V= \frac43 \pi r^3$
• surface area $S= 4 \pi r^2$

then assume

• $V+\Delta V=\frac43 \pi (r+\Delta r)^3 =1.728 V=1.728\cdot \frac43 \pi r^3$

and find $\Delta r$.

Use $V= \frac43 \pi r^3$ and $S= 4 \pi r^2$ to establish

$$S=(36\pi)^{1/3}V^{2/3}$$

With the new volume $V'=1.728V=1.2^3V$, the new surface is

$$S'=(36\pi)^{1/3}(1.2^3V)^{2/3}=(1.2)^2S=1.44S$$ Thus, the surface increases 44%, exactly.