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!

I'm going to deviate slightly from your integral setup but hopefully the point is clear.
Put the base of the cylinder in the $(x,y)$ plane. Align the diameter of the wedge - or using the far more interesting name, the ungula - with the plane $y=0$. Each semielliptical cross section is made in a plane $z=tx$, where $0\le t \le \frac43$. Each semiminor axis will be the same as the base radius $12$. For a given value of $t$, the semimajor axis of the cross section has length $\sqrt{144+144t^2}$ (i.e. the distance from the origin to the point $\left(12,0,12t\right)$). Hence the faces of a cross section have an average area of about $72\pi \sqrt{1+t^2}$.
Now, you may be tempted to conclude the volume must be
$$\int_0^{\frac43} 72\pi \sqrt{1+t^2} \, dt$$
but this would be wrong, and for the same reason your integral is wrong. In fact, this gives the volume of a cone of height $\frac43$ with elliptical cross sections possessing the same semiaxes.
You effectively ended up treating each cross section as though they have the same thickness. But the solid we care about isn't made by stacking elliptical half-disks like a warped roll of coins, it's more like putting a misshapen orange back together from separate wedges. (One such wedge is shown in red.) The non-uniform thickness of the wedge needs to be accounted for.

To approximate the volume of a given wedge, we can compare it to the volume of a sliver of an oblate-spheroidal shell - take the arc of an ellipse with semimajor/-minor axes $a$ and $b$, respectively, and revolve it by a small angle $\theta$ about its minor axis. This volume $V_{\rm sliver}$ is proportional to the spheroid's total volume when the arc is revolved by $2\pi$, such that
$$\frac{V_{\rm sliver}}\theta = \frac{\frac{4\pi}3 a^2b}{2\pi} \implies V_{\rm sliver} = \frac23 a^2 b \theta$$
A small increase in the angle of $\Delta \theta$ causes $V_{\rm sliver}$ to increase to
$$V'_{\rm sliver} = \frac23 a^2 b (\theta + \Delta \theta)$$
so that the overall change in sliver volume amounts to
$$\Delta V = V'_{\rm sliver}-V_{\rm sliver} = \frac23 a^2 b (\theta + \Delta \theta) - \frac23 a^2 b \theta = \frac23 a^2 b \Delta \theta$$
Now divide both sides by $\Delta\theta$ to get the ratio of change in volume to change in angle. Letting $\Delta\theta\to0$ (note the convergence to non-constant), we have
$$\lim_{\Delta\theta\to0} \frac{\Delta V}{\Delta \theta} = \frac{dV}{d\theta} = \frac23 a^2 b$$
Then the volume of an ungula would be obtained by the definite integral (omitting the domain)
$$V = \int dV = \int \frac23 a^2 b \, d\theta$$

For a given sliver that makes up our ungula, for which $a=12\sqrt{1+t^2}$ and $b=12$, we have the relation $\cos(\theta)=\frac{12}a$. It follows that
$$\theta = \cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right) = \tan^{-1}(t) \implies \frac{d\theta}{dt} = \frac{1}{1+t^2}$$
and hence, using the chain rule, the ungula's volume is
$$\begin{align*}
V &= \int\limits_{t\in\left[0,\frac43\right]} \frac23 a^2 b \, d\theta \\[1ex]
&= \int_0^{\frac43} \frac23 \left(12\sqrt{1+t^2}\right)^2 \cdot 12 \, \frac{d\theta}{dt} \, dt \\[1ex]
&= 1152 \int_0^{\frac43} (1+t^2) \, \frac{dt}{1+t^2} \\[1ex]
&= 1152 \cdot \frac43 = \boxed{1536}
\end{align*}$$
It turns out the integral you got isn't too far off. Let $t=\tan(\theta)$. As we have $t\in\left[0,\frac43\right]$, we get $\theta\in\left[0,\tan^{-1}\left(\frac43\right)\right]$. Then
$$V = 1152 \int_0^{\tan^{-1}\left(\frac43\right)} \sec^{\color{red}{2}}(\theta) \, d\theta$$
