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The problem asks for the volume of the solid generated by revolving the region bounded by the curves \( y = \frac{1}{x} \), \( y = 0 \), \( x = 2 \), and \( x = 3 \) about the x-axis.

This is a typical problem where we use the **method of disks** to find the volume of a solid of revolution.

### Step-by-Step Solution:

1. **Formula for the Volume (Disk Method):**
   The volume \( V \) of a solid of revolution about the x-axis is given by:
   \[
   V = \pi \int_{a}^{b} [f(x)]^2 \, dx
   \]
   where \( f(x) \) is the function being revolved, and \( [a, b] \) is the interval over which the region is defined.

2. **Identifying the Boundaries and Function:**
   The function is \( f(x) = \frac{1}{x} \).
   The boundaries for the region are \( x = 2 \) and \( x = 3 \).

3. **Set up the Integral:**
   The volume will be calculated as:
   \[
   V = \pi \int_{2}^{3} \left( \frac{1}{x} \right)^2 \, dx
   \]
   Simplifying the integrand:
   \[
   V = \pi \int_{2}^{3} \frac{1}{x^2} \, dx
   \]

4. **Evaluate the Integral:**
   The integral of \( \frac{1}{x^2} \) is \( -\frac{1}{x} \), so:
   \[
   V = \pi \left[ -\frac{1}{x} \right]_{2}^{3}
   \]
   Substituting the limits of integration:
   \[
   V = \pi \left( -\frac{1}{3} + \frac{1}{2} \right)
   \]
   \[
   V = \pi \left( \frac{1}{2} - \frac{1}{3} \right)
   \]
   \[
   V = \pi \left( \frac{3}{6} - \frac{2}{6} \right) = \pi \times \frac{1}{6}
   \]
   So, the volume is:
   \[
   V = \frac{\pi}{6}
   \]

Thus, the correct answer is:
\[
\boxed{\frac{1}{6} \pi}
\]

Would you like a more detailed explanation of any step?

Here are five related questions:
1. How does the disk method differ from the shell method for solids of revolution?
2. What happens if we rotate the region around the y-axis instead of the x-axis?
3. Can the same approach be used for a curve like \( y = x^2 \)? How would the setup change?
4. How do we handle volumes for non-bounded regions in the plane?
5. What are other applications of integral calculus in physics or engineering?

**Tip:** Always check that the function \( f(x) \) and the limits of integration correspond to the correct axis of revolution to avoid mistakes.

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations y = 1/x, y = 0, x = 2, x = 3 about the x-axis.

Learn how to calculate the volume of the solid generated by revolving the region bounded by y = 1/x, y = 0, x = 2, and x = 3 around the x-axis. This step-by-step solution uses the Disk Method, focusing on key calculus concepts suitable for high school students in Grades 11-12.

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