Consider the region bounded to the right by the graph of $x=\frac{\sin (6 y)+10}{y}$, to the left by the $y$-axis, and above and below by $y=\frac{5 \pi}{3}$ and $y=\frac{\pi}{6}$. Using the shell method, what is the volume of the solid of revolution obtained by rotating this region about the $x$-axis? Enter an exact value in terms of $\pi$.

Step 1 ：Given the region bounded to the right by the graph of \(x=\frac{\sin (6 y)+10}{y}\), to the left by the y-axis, and above and below by \(y=\frac{5 \pi}{3}\) and \(y=\frac{\pi}{6}\). We are asked to find the volume of the solid of revolution obtained by rotating this region about the x-axis using the shell method.

Step 2 ：The shell method formula for the volume of a solid of revolution is \(V = 2\pi \int_a^b r(x)h(x)dx\), where \(r(x)\) is the radius of the cylindrical shell, \(h(x)\) is the height of the cylindrical shell, and \(a\) and \(b\) are the limits of integration.

Step 3 ：In this case, the radius of the cylindrical shell is \(y\), the height of the cylindrical shell is \(\frac{\sin (6 y)+10}{y}\), and the limits of integration are \(\frac{\pi}{6}\) and \(\frac{5 \pi}{3}\).

Step 4 ：Substituting these values into the shell method formula, we get \(V = 2\pi \int_{\frac{\pi}{6}}^{\frac{5 \pi}{3}} y \cdot \frac{\sin (6 y)+10}{y} dy\).

Step 5 ：Simplifying the integral, we get \(V = 2\pi (-\frac{1}{3} + 15\pi)\).

Step 6 ：Thus, the volume of the solid of revolution obtained by rotating the region bounded by the graph of \(x=\frac{\sin (6 y)+10}{y}\), the y-axis, and the lines \(y=\frac{5 \pi}{3}\) and \(y=\frac{\pi}{6}\) about the x-axis is \(\boxed{2\pi(-\frac{1}{3} + 15\pi)}\).