Use the disk method to find the volume of the solid of revolution bounded by the $x$-axis and the graphs of $f(x)=4 \sqrt{9 \cos x}$ $x=\frac{\pi}{6}$, and $x=\frac{\pi}{2}$ rotated about the $x$-axis. Enter your answer in terms of $\pi$.

Step 1 ：We are given the function \(f(x) = 4 \sqrt{9 \cos x}\) and we are asked to find the volume of the solid of revolution bounded by the x-axis and the graphs of \(f(x)\), \(x = \frac{\pi}{6}\), and \(x = \frac{\pi}{2}\) rotated about the x-axis.

Step 2 ：We use the disk method to find the volume. The disk method involves integrating the area of the cross-sections of the solid. In this case, the cross-sections are circles with radius \(f(x)\), so the area is \(\pi [f(x)]^2\). We need to integrate this from \(x = \frac{\pi}{6}\) to \(x = \frac{\pi}{2}\).

Step 3 ：First, we simplify the function \(f(x)\) to get \(f(x) = 12 \sqrt{\cos x}\).

Step 4 ：Then, we square \(f(x)\) and multiply by \(\pi\) to get the area of the cross-sections: \(\text{area} = 144 \pi \cos x\).

Step 5 ：Finally, we integrate the area from \(x = \frac{\pi}{6}\) to \(x = \frac{\pi}{2}\) to get the volume of the solid of revolution: \(\text{volume} = 72 \pi\).

Step 6 ：So, the volume of the solid of revolution is \(\boxed{72\pi}\).