Related rates of a cone I am in an intro calculus class and I have a problem which I am unsure about. It reads:

Water is being poured into a conical reservoir at a rate of pi cubic feet per second. The reservoir has a radius of 6 fees across the top and a height of 12 feet. At what rate is the depth of the water increasing when the depth is 6 feet?

Here's what I came up with for the problem:
h = height of cone
r = radius of cone 
V = volume of cone
$\frac{r}{h}$ = $\frac{1}{2}$
V = $ \frac{\pi r^2h}{3}$
$\frac{dV}{dt}$ = $\frac {2\pi r \frac{dr}{dt}h}{3} + $$\frac{1}{3} \pi r^2 
\frac {dh}{dt}$
(Using product rule)
Because r = $\frac{h}{2},
$$\frac{dv}{dt}$ = $\frac{2\pi h}{2}\frac{dh}{dt}\frac{h}{6} + \frac{\pi (\frac{h}{2})^2}{3}
\frac{dh}{dt}$
I solved for $\frac{dh}{dt}$
$\frac{dh}{dt}$ = $\frac{12\frac{dv}{dt}}{3\pi h^2}$
I plugged in $\pi$ for $\frac{dv}{dt}$ and 6 for h and got
1/9 or 0.111
Can anyone tell me if this is correct? Thanks if advance!

Since the dimension of the cone is such that $r = 6$ft and $h=12$ft, we have $\dfrac{r}{h} = \dfrac{6}{12}$, which means $r = \dfrac{h}{2}$. We must substitute this into the equation of the volume of a cone before differentiating. It follows that
$V=\pi r^2 \dfrac{h}{3}$ 
$V=\pi \bigg(\dfrac{h}{2}\bigg)^2 \dfrac{h}{3}$
$V=\pi \dfrac{h^2}{4} \dfrac{h}{3}$
$V= \pi \dfrac{h^3}{12}$
$\dfrac{dV}{dt} = \dfrac{\pi}{12} 3h^2 \dfrac{dh}{dt}$
$\pi = \dfrac{\pi}{12} 3(6)^2 \dfrac{dh}{dt}$
$\pi = 9\pi \dfrac{dh}{dt}$
$\dfrac{dh}{dt} = \dfrac{1}{9}$ 
The height of the cone is increasing at a rate of $\dfrac{1}{9}$ feet per second . 
