Q:

Calculate volume of a hemispherical barrel

A hemispherical barrel of 3 meters of radius is fill for 2 meters of water. I need to calculate the content volume in $m^3$.
I tried this solution: change to polar coordinates, the radius is between 0 and 3, the barrel is hemispherical so $\theta$ is between 0 and $\pi$, and, I supposed that z is the height, so it is between 0 and 2. So:  
$\int_0^\pi d\theta\int_0^3\rho d\rho\int_0^2 dz = 9\pi$  
but it doesn't work. I know the result that is $\frac{28}3\pi$.
Can you help me and explain where I'm wrong.

A:

Note that $\rho$ depends on $z$ as follows:

So $\rho=\sqrt{(3)^2-(3-z)^2}$.

