We can use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by the graph of $g(y) = \frac{8}{y}$ (on the right), the $y$-axis (on the left), and between $y = 4$ and $y = 6$, about the $x$-axis.

Step-by-Step Approach:

1. Formula for volume using cylindrical shells: The formula for the volume using cylindrical shells is:

   $V = 2\pi \int_{a}^{b} \text{radius} \times \text{height} \, dy$

   • The radius is the distance from the axis of rotation (the $x$-axis) to a point on the curve, which is $y$.
   • The height of each cylindrical shell is given by the function $g(y) = \frac{8}{y}$.

   So, the formula becomes:

   $V = 2\pi \int_{4}^{6} y \cdot \frac{8}{y} \, dy$

2. Simplifying the integral: Notice that $y \cdot \frac{8}{y} = 8$, so the integral simplifies to:

   $V = 2\pi \int_{4}^{6} 8 \, dy$

3. Evaluate the integral: Now, we integrate:

   $\int_{4}^{6} 8 \, dy = 8(y) \Big|_4^6 = 8(6 - 4) = 8 \times 2 = 16$

4. Final volume: Multiply by $2\pi$:

   $V = 2\pi \times 16 = 32\pi$

5. Approximating the result: Now, approximate $V$:

   $V \approx 32 \times 3.1416 = 100.53$

Final Answer:

The volume of the solid is approximately $\boxed{100.53}$ cubic units.

Would you like further details or clarification? Here are some related questions:

1. How does the method of cylindrical shells differ from the disk method?
2. What happens if the region is rotated about the y-axis instead?
3. How do we handle problems where the bounds are more complex?
4. Can we use cylindrical shells if the function was instead $g(y) = \frac{y}{8}$?
5. How would the solution change if the limits were from 3 to 5?

Tip: When solving problems using cylindrical shells, always check if the integral simplifies before performing complex operations—it can save time!