To solve this problem, let's go through the steps to find the volume of the solid of revolution formed by revolving the region $R$ around the $y$-axis. The region $R$ is bounded by $f(x) = \frac{1}{x}$, the $x$-axis, and the interval $[1,3]$.

Solution Outline:

1. Identify the function and limits:

   • We are given $f(x) = \frac{1}{x}$ from $x = 1$ to $x = 3$.

2. Volume of Revolution around the $y$-axis:

   • The formula for the volume of a solid of revolution about the $y$-axis is: $V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx$
   • Here, $f(x) = \frac{1}{x}$, so the volume integral becomes: $V = 2\pi \int_{1}^{3} x \cdot \frac{1}{x} \, dx$

3. Simplify the Integral:

   • Simplifying $x \cdot \frac{1}{x}$ gives $1$, so we have: $V = 2\pi \int_{1}^{3} 1 \, dx$
   • This integral now becomes: $V = 2\pi \left[ x \right]_{1}^{3}$

4. Evaluate the Integral:

   • Substitute the limits: $V = 2\pi (3 - 1) = 2\pi \cdot 2 = 4\pi$

Answer:

The volume of the solid of revolution is $\boxed{4\pi}$, which corresponds to option (a).

Would you like further details on any of these steps, or do you have other questions?

Here are some follow-up questions to consider:

1. How would the solution change if we were revolving around the $x$-axis instead?
2. What is the significance of the function $f(x) = \frac{1}{x}$ in calculus problems involving areas and volumes?
3. How would this calculation differ if we used the shell method around a different axis?
4. Could we compute the same volume using a different method, like the disk/washer method?
5. How does changing the interval affect the resulting volume of revolution?

Tip: When dealing with solids of revolution, carefully choose between the disk/washer method and the shell method, depending on the axis of rotation and the function's form.