The radius of a sphere is increasing at a rate of $3 \mathrm{~mm} / \mathrm{s}$. How fast is the volume increasing (in $\mathrm{mm}^{3} / \mathrm{s}$ ) when the diameter is $40 \mathrm{~mm}$ ? (Round your answer to two decimal places.) \[ \mathrm{mm}^{3} / \mathrm{s} \]

Step 1 ：Given the rate of change of the radius (dr/dt) is 3 mm/s and the diameter is 40 mm, so the radius is 20 mm.

Step 2 ：Using the formula dV/dt = (4 * pi * r^2) * dr/dt, we can plug in the values: dV/dt = (4 * pi * (20)^2) * 3.

Step 3 ：Calculate the final answer: \(dV/dt = 15079.64 \, \mathrm{mm}^{3} / \mathrm{s}\)

Step 4 ：\(\boxed{15079.64}\) is the rate at which the volume is increasing when the diameter is 40 mm.