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To solve this problem, we need to calculate the volume of the solid obtained by rotating the region enclosed by the curves around the vertical axis \( x = -1 \).

### The region is bounded by:
- The hyperbola \( xy = 1 \), or \( y = \frac{1}{x} \),
- The horizontal line \( y = 0 \),
- The vertical lines \( x = 1 \) and \( x = 2 \).

This problem can be solved using the **method of cylindrical shells**, as we are rotating around a vertical line that is not an axis of symmetry. Here's a step-by-step outline of the solution:

### Step 1: Formula for the volume using cylindrical shells
The volume \( V \) of a solid of revolution generated by rotating a region around a vertical line using the method of cylindrical shells is given by the formula:
\[
V = 2\pi \int_{a}^{b} (r_{\text{shell}})(h_{\text{shell}}) \, dx
\]
Where:
- \( r_{\text{shell}} \) is the radius of the shell, which is the distance from the axis of rotation \( x = -1 \),
- \( h_{\text{shell}} \) is the height of the shell, which is the function \( y = \frac{1}{x} \),
- \( [a, b] \) is the interval for \( x \) (in this case, from 1 to 2).

### Step 2: Determine \( r_{\text{shell}} \) and \( h_{\text{shell}} \)
- The radius of the shell is the distance from \( x \) to \( x = -1 \), so:
  \[
  r_{\text{shell}} = x - (-1) = x + 1
  \]
- The height of the shell is given by the function \( y = \frac{1}{x} \).

### Step 3: Set up the integral
Now, we can set up the integral for the volume:
\[
V = 2\pi \int_{1}^{2} (x + 1) \cdot \frac{1}{x} \, dx
\]
Simplifying the integrand:
\[
V = 2\pi \int_{1}^{2} \left( 1 + \frac{1}{x} \right) \, dx
\]

### Step 4: Evaluate the integral
We now solve the integral:
\[
V = 2\pi \left[ x + \ln |x| \right]_{1}^{2}
\]
Substitute the limits of integration:
\[
V = 2\pi \left( \left[ 2 + \ln(2) \right] - \left[ 1 + \ln(1) \right] \right)
\]
Since \( \ln(1) = 0 \), this simplifies to:
\[
V = 2\pi \left( 2 + \ln(2) - 1 \right)
\]
\[
V = 2\pi (1 + \ln(2))
\]

### Final answer:
\[
V = 2\pi (1 + \ln(2)) \, \text{units}^3
\]

This is the volume of the solid obtained by rotating the given region around \( x = -1 \).

Would you like more details on any part of this solution?

### Related questions:
1. What would the volume be if the region were rotated around \( x = 0 \) instead of \( x = -1 \)?
2. How does the method of cylindrical shells differ from the disk or washer method in such problems?
3. What are the geometric implications of rotating this region around a horizontal line, such as \( y = 0 \)?
4. Could this problem be solved using parametric equations or polar coordinates?
5. How do changes in the boundary conditions (like the limits of integration) affect the volume calculation?

**Tip:** The method of cylindrical shells is particularly useful when rotating around vertical lines that are not the y-axis, as it avoids complex expressions for the radii of disks or washers.

Calcular el volumen del sólido obtenido al girar la región delimitada por las curvas xy = 1, y = 0, x = 1, x = 2, alrededor de x = -1

Learn how to calculate the volume of the solid obtained by rotating the region bounded by xy = 1, y = 0, and vertical lines x = 1, x = 2 around the axis x = -1 using the cylindrical shell method. A step-by-step calculus solution suitable for undergraduate students.

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