Calculus - Fluid Leaving Cone Vessel 
Suppose that fluid flows out of the bottom of a cone-shaped vessel at the rate of 3 cu ft/min. If the height of the cone is 3 times the radius, how fast is the height of the fluid decreasing when the fluid is 6 inches deep in the middle?

I'm unsure whether my approach is correct and whether my answer is right.
My approach to solving this was the following:
Since we're trying to find how fast the height of the cone decreases, we need to find $\frac{dh}{dt}$. To do this, we can use the chain rule.
$$\frac{dh}{dt} = \frac{dh}{dr} \frac{dr}{dt}$$
Where $r$ is the radius and $t$ is the time.
Since the height is 3 times the radius, $\frac{dh}{dr} = 3$. The derivative of the radius with respect to time can be solved by finding:
$$\frac{dV}{dt} = \frac{dV}{dr} \frac{dr}{dt}$$
$$\frac{dV}{dt} = 3$$ and $$\frac{dV}{dr} = 3\pi r^2$$
$$\therefore \frac{dr}{dt} = \frac{1}{\pi r^2}$$
Plugging in, $$\frac{dh}{dt} = 3 \biggl(\frac{1}{\pi r^2}\biggl) = \frac{3}{\pi r^2}$$
So, if the height is 6 inches (0.5 feet) deep in the middle, we know the radius is $\frac{1}{6}$ feet or 2 inches. Therefore, the height decreases at a rate of $\frac{3}{\pi * 2^2} = \frac{3}{4 \pi}$ in / min. 
Just to clarify, I would like to know whether this is the right answer and, if not, then whether my approach was correct or not. The book I got this problem out of is called Calculus by Morris Kline.
Thanks for the help.
 A: Well, let's see . . . 
If I recall correctly, the volume $V$ of a cone of height $h$ and radius $r$ is given by
$V = \dfrac{\pi r^2 h}{3}; \tag 1$
and here we have
$h = 3r, \tag 2$
that is,
$r = \dfrac{h}{3}; \tag 3$
then, in terms of $h$,
$V = \dfrac{\pi h (h/3)^2}{3} = \dfrac{\pi h^3}{27}; \tag 4$
now if the volume of the cone of fluid changes in such a way as to maintain the proportionality (2), (3) 'twixt $r$ and $h$, we have, by the chain rule,
$\dfrac{dV}{dt} = \dfrac{\pi h^2}{9} \dfrac{dh}{dt}, \tag 5$
and we merely need to plug in the numbers, taking care that the units are consistent:  we have
$V = 3 \; \text{cu. ft./min.}; \; h = 6 \; \text{in.}; \tag 6$
now let's see, 
$1 \; \text{cu. ft.} = (12)^3 \; \text{cu. in.} = 1728 \; \text{cu. in.}; \tag 7$
$3 \; \text{cu. ft.} = 3 \times (12)^3 \; \text{cu. in.} = 3 \times 1728 \; \text{cu. in.} = 5184 \; \text{cu. in.}; \tag 8$
from (5),
$\dfrac{dh}{dt} = \dfrac{9}{\pi h^2} \dfrac{dV}{dt}, \tag 9$
and plugging in the numbers
$\dfrac{dh}{dt} = \dfrac{9}{\pi(6 \; \text{in.})^2} (5184 \; \text{cu. in./min.}) \cong \dfrac{1296}{\pi} \; \text{in./min.} \cong412.74 \; \text{in./min.}; \tag{10}$
that's a pretty rapid decrease of depth; indeed, at that rate $h = 0$ in about
$\Delta t \cong \dfrac{6}{412.74} \; \text{min.} \cong .015 \; \text{min.} \cong .90 \; \text{sec.} \tag{11}$
The basic approach to such problems, using the chain rule and geometrical information to relate the rates of different variables, is correct and indeed often done.  
It's the arithmetic that scares me.  I've tried to lay it out carefully; if I made a mistake, please leave a comment.
