Answer
The volume is increasing at $25600\pi \frac{mm^{3}}{s}$.
Work Step by Step
The radius of the sphere is $40 mm$.
$V(r) = \frac{4}{3} \pi r^{3}$
$\frac{dV}{dt} = \frac{4}{3}\pi \times \frac{d(r^{3})}{dt}$
Use the chain rule $(\frac{dr}{dt})$:
$\frac{dV}{dt} = \frac{4}{3}\pi \times \frac{d(r^{3})}{dt} \times \frac{dr}{dt}$
$\frac{dV}{dt} = \frac{4}{3}\pi \times 3r^{2} \times \frac{dr}{dt}$
$\frac{dV}{dt} = 4\pi r^{2} \times \frac{dr}{dt}$
$\frac{dr}{dt} = 4\frac{mm}{s}$
$\frac{dV}{dt} = 4\pi (40)^{2} \times (4)$
$\frac{dV}{dt} = 4\pi (1600) \times (4)$
$\frac{dV}{dt} = 4\pi (6400)$
$\frac{dV}{dt} = 25600\pi \frac{mm^{3}}{s}$
