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 \( \pi \).

To find the volume of the solid of revolution obtained by rotating the region bounded by the graph \( x=\frac{\sin (6 y)+10}{y} \), the \( y \)-axis, and the lines \( y=\frac{5 \pi}{3} \) and \( y=\frac{\pi}{6} \) about the \( x \)-axis using the shell method, we can use the formula: \[ V = 2\pi \int_{a}^{b} (radius)(height) \, dy. \] In this scenario, \( a = \frac{\pi}{6} \) and \( b = \frac{5\pi}{3} \). The radius of the shell at a given \( y \) is \( y \) (the distance from the \( x \)-axis to the shell), and the height of the shell is given by the function \( x=\frac{\sin(6y) + 10}{y} \). Thus, the volume is calculated as follows: \[ V = 2\pi \int_{\frac{\pi}{6}}^{\frac{5\pi}{3}} y \cdot \frac{\sin(6y) + 10}{y} \, dy. \] This simplifies to: \[ V = 2\pi \int_{\frac{\pi}{6}}^{\frac{5\pi}{3}} (\sin(6y) + 10) \, dy. \] Now, we can split the integral: \[ V = 2\pi \left( \int_{\frac{\pi}{6}}^{\frac{5\pi}{3}} \sin(6y) \, dy + \int_{\frac{\pi}{6}}^{\frac{5\pi}{3}} 10 \, dy \right). \] Now we compute each integral. 1. For the first integral: \[ \int \sin(6y) \, dy = -\frac{1}{6} \cos(6y). \] Evaluating this from \( y = \frac{\pi}{6} \) to \( y = \frac{5\pi}{3} \): \[ -\frac{1}{6} \left[ \cos\left(6 \cdot \frac{5\pi}{3}\right) - \cos\left(6 \cdot \frac{\pi}{6}\right) \right] = -\frac{1}{6} \left[ \cos(10\pi) - \cos(\pi) \right] \] \[ = -\frac{1}{6} \left[ 1 - (-1) \right] = -\frac{1}{6} (1 + 1) = -\frac{2}{6} = -\frac{1}{3}. \] So, \[ \int_{\frac{\pi}{6}}^{\frac{5\pi}{3}} \sin(6y) \, dy = -\frac{1}{3}. \] 2. For the second integral: \[ \int 10 \, dy = 10y. \] Evaluating from \( y = \frac{\pi}{6} \) to \( y = \frac{5\pi}{3} \): \[ 10 \left[ \frac{5\pi}{3} - \frac{\pi}{6} \right] = 10 \left[ \frac{10\pi}{6} - \frac{\pi}{6} \right] = 10 \left[ \frac{9\pi}{6} \right] = 15\pi. \] Putting the two results together: \[ V = 2\pi \left( -\frac{1}{3} + 15\pi \right) = 2\pi \left( 15\pi - \frac{1}{3} \right) = 30\pi^2 - \frac{2\pi}{3}. \] To express this as a single fraction: \[ V = \frac{90\pi^2}{3} - \frac{2\pi}{3} = \frac{90\pi^2 - 2\pi}{3}. \] Thus, the final answer for the volume \( V \) is: \[ \boxed{\frac{90\pi^2 - 2\pi}{3}}. \]