Answer: \(\frac{\mathrm{d}S}{\mathrm{d}t}=\frac{4\pi}{3}\text{ cm}^2\text{ s}^{-1}\).

For a sphere,

\(V=\frac43\pi r^3.\)

When \(V=36\pi\),

\(\frac43\pi r^3=36\pi,\)

so \(r^3=27\), and therefore \(r=3\).

Differentiate \(V=\frac43\pi r^3\) with respect to \(r\):

\(\frac{\mathrm{d}V}{\mathrm{d}r}=4\pi r^2.\)

Using \(\frac{\mathrm{d}V}{\mathrm{d}t}=2\pi\),

\(2\pi=4\pi r^2\frac{\mathrm{d}r}{\mathrm{d}t}.\)

At \(r=3\),

\(\frac{\mathrm{d}r}{\mathrm{d}t}=\frac{2\pi}{36\pi}=\frac{1}{18}.\)

The surface area is \(S=4\pi r^2\), so

\(\frac{\mathrm{d}S}{\mathrm{d}r}=8\pi r.\)

At \(r=3\), \(\frac{\mathrm{d}S}{\mathrm{d}r}=24\pi\). Hence

\(\frac{\mathrm{d}S}{\mathrm{d}t}=\frac{\mathrm{d}S}{\mathrm{d}r}\frac{\mathrm{d}r}{\mathrm{d}t}=24\pi\cdot\frac{1}{18}=\frac{4\pi}{3}.\)