Let $\mathrm{R}$ be the region bounded by the following curves. Find the volume of the solid generated when $\mathrm{R}$ is revolved about the $x$-axis. Recall that $\cos ^{2} x=\frac{1}{2}(1+\cos 2 x)$. \[ y=\cos 2 x, y=0, x=0 \] The volume of the region revolved about the $x$-axis is cubic units. (Type an exact answer.)

Step 1 ：Given the curve \(y = \cos 2x\), the x-axis, and the y-axis, we need to find the volume of the solid generated by revolving the given region around the x-axis.

Step 2 ：Using the disk method, the area of a cross-sectional disk is given by \(A(x) = \pi y^2\). Substituting the given equation for y, we get \(A(x) = \pi (\cos 2x)^2\).

Step 3 ：We know that \(\cos^2 x = \frac{1}{2}(1 + \cos 2x)\), so we can simplify the area function to \(A(x) = \pi \frac{1}{2}(1 + \cos 4x)\).

Step 4 ：Integrate the area function along the x-axis from \(x = 0\) to \(x = \frac{\pi}{4}\) to find the volume: \(V = \int_{0}^{\frac{\pi}{4}} A(x) dx\).

Step 5 ：Simplify the integral to get the volume: \(V = \frac{\pi}{4} + \frac{\pi}{16}\).

Step 6 ：\(\boxed{V = \frac{5\pi}{16}}\) cubic units.