Calculus Problem: polar-coordinate integration of a slice/wedge of a right circular cylinder 
This image states the problem: As shown in the diagram, a solid right circular cylinder of radius 12 cm is sliced by a plane that passes through the center of the base circle.
Find the volume of the wedge-shaped piece that is created, given that its height is 16 cm (and its base has a radius of 12).1
I realized the easiest way to do this is to integrate with triangles from each side but I was curious to see if it would be possible to integrate this using polar coordinates.
I attempted to do this by integrating using the half-ellipses that are formed from the plane as cross sections and rotating that plane around the center of the base of the cylinder to create more ellipses. However, when I tried this I got $248$ instead of $\textbf{1536}$ (the correct answer). I'm confident in how I computed the integral--I think the mistake I made was in setting up the integration in the first place.
Here is my working:
Apologies if it's hard to follow
Could anyone help me set it up properly? Can you spot what I did wrong? Thanks!

First, place the center of the cylinder in $O(0, 0,0)$. Then, the equation of the plane is $z-\frac43 x=0$.
Here, a graph (point $P$ has coordinates $P(12, 0, 16)$ and represents the height of $\mathcal{V}$):

In cylindrical coordinates:
$$z-\frac43 \rho \cos(\theta)=0$$
So, the integral becomes:
$$\mathcal{V} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{12}\int_{0}^{\frac43 \rho \cos(\theta)}\rho dz d\rho d\theta$$
Solving:
$$\mathcal{V}=\frac43\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{0}^{12}\rho^2\cos(\theta)d\rho d\theta=\frac43\cdot\left(\int_{0}^{12}\rho^2 d\rho\right)\cdot\left(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos(\theta)d\theta\right)=\frac43\cdot\left[\frac13\rho^3\right]_{0}^{12}\cdot2=1536$$
