The surface area \( S \) of a sphere is \( S = 4\pi r^2 \), where \( r \) is the radius. The volume \( V \) of the sphere is \( V = \frac{4}{3} \pi r^3 \). We are given \( \frac{dS}{dt} = 2 \, \text{cm}^2/\text{sec} \) and \( r = 6 \, \text{cm} \). We need to find \( \frac{dV}{dt} \).
Step 1: Find \( \frac{dr}{dt} \). Differentiate \( S = 4\pi r^2 \) with respect to \( t \) to get \( \frac{dS}{dt} = 8\pi r \frac{dr}{dt} \). Rearrange to \( \frac{dr}{dt} = \frac{\frac{dS}{dt}}{8\pi r} \). Substitute the given values: \( \frac{dr}{dt} = \frac{2}{8\pi \cdot 6} = \frac{1}{24\pi} \, \text{cm/sec} \).
Step 2: Find \( \frac{dV}{dt} \). Differentiate \( V = \frac{4}{3}\pi r^3 \) with respect to \( t \) to get \( \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} \). Substitute \( r = 6 \) and \( \frac{dr}{dt} = \frac{1}{24\pi} \): \( \frac{dV}{dt} = 4\pi (6)^2 \cdot \frac{1}{24\pi} \).
Simplify: \( \frac{dV}{dt} = 4\pi \cdot 36 \cdot \frac{1}{24\pi} = \frac{144}{24} = 6 \, \text{cm}^3/\text{sec} \).
Final Answer:
\[\boxed{6 \, \text{cm}^3/\text{sec}}\]