The volume V of the cylinder is V = πr^2h. Taking the derivative of both sides with respect to time t, we get:
dV/dt = π(2rh.dr/dt + r^2dh/dt)
We are given that dh/dt = 2 mm/s and r = 6 mm, so we have:
dV/dt = π(2rh.dr/dt + r^2dh/dt) = π(2(6)(dr/dt)(2) + 6^2(2)) = 24π(dr/dt + 2)
We know that dV/dt = 0, since we are only interested in the instant when the volume is increasing at the rate of cubic millimeters per second. Solving for dr/dt, we get:
dr/dt = -2/(24π) = -1/(12π)
Now we can find the surface area S of the cylinder, not including the top and bottom of the cylinder, which is S = 2πrh. Substituting r = 6 mm and h = 2 mm, we get:
S = 2π(6)(2) = 24π square mm
Taking the derivative of S with respect to time t, we get:
dS/dt = 2πr(dh/dt) + 2πh(dr/dt)
Substituting r = 6 mm, h = 2 mm, dh/dt = 2 mm/s, and dr/dt = -1/(12π), we get:
dS/dt = 2π(6)(2) + 2π(2)(-1/(12π)) = 28π square mm per second
Therefore, the correct answer is a. The surface area is increasing by 28π square mm per second.