A hemispherical water tank with radius 6 meters is filled to a depth of h meters. The volume of the tank is given to be V = (1/3)pih(108-h^2), where 0 < h < 6. If water is being bumped into the tank at a rate of 3 cubic meters per minute, find the rate of change of the depth of the water when h = 2 meters.

My Work:

dr/dt = 3

I need dv/dt when h is 2.

I then replaced h by r/3 and proceeded to substituted

r/3 for h in the given formula before taking the implicit derivative.

I simplified and found the new volume equation to be V = 12pi - (pi/81)*r^3.

I differentiated my new volume formula implicitly, substituted r = 6 or the given radius of the water tank and simplified.

My Answer is dv/dt = 32pi meters per minute.

Book's answer is (3/32pi) meters per minute.

How do I find the right answer? What are the steps for this question?