The surface of a spherical balloon is increasing at the rate of . Find the rate of change of volume…

Surface area of a sphere: $S = 4\pi r^2$. $\frac{dS}{dt} = 8\pi r \frac{dr}{dt} = 4 \implies \frac{dr}{dt} = \frac{1}{2\pi r}$. Volume of a sphere: $V = \frac{4}{3}\pi r^3$. $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 4\pi r^2 \left(\frac{1}{2\pi r}\right) = 2r$. When $r = 12\text{ cm}$: $\frac{dV}{dt} = 2(12) = 24\text{ cm}^3/\text{sec}$.