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The problem requires finding the volume of the solid formed by revolving the region enclosed by the curves \( y = 2x - 1 \) and \( y = x^2 - 1 \) around the line \( y = -2 \).

### Steps to solve:
1. **Find the points of intersection** of the curves \( y = 2x - 1 \) and \( y = x^2 - 1 \).
   
   Set the two equations equal to each other:
   \[
   2x - 1 = x^2 - 1
   \]
   Simplifying:
   \[
   2x = x^2 \quad \Rightarrow \quad x^2 - 2x = 0 \quad \Rightarrow \quad x(x - 2) = 0
   \]
   So, the points of intersection are \( x = 0 \) and \( x = 2 \).

2. **Determine the volume using the washer method**: 
   The volume of the solid is found by integrating the difference between the squares of the outer radius and inner radius, based on the distance of each curve from the axis of rotation \( y = -2 \).

   - **Outer radius**: The curve \( y = 2x - 1 \), revolved around \( y = -2 \), has a radius:
     \[
     R(x) = (2x - 1) - (-2) = 2x - 1 + 2 = 2x + 1
     \]
   - **Inner radius**: The curve \( y = x^2 - 1 \), revolved around \( y = -2 \), has a radius:
     \[
     r(x) = (x^2 - 1) - (-2) = x^2 - 1 + 2 = x^2 + 1
     \]

3. **Set up the integral** for the volume:
   \[
   V = \pi \int_0^2 \left[ (2x + 1)^2 - (x^2 + 1)^2 \right] \, dx
   \]
   Expand the expressions:
   \[
   (2x + 1)^2 = 4x^2 + 4x + 1
   \]
   \[
   (x^2 + 1)^2 = x^4 + 2x^2 + 1
   \]
   Thus, the integral becomes:
   \[
   V = \pi \int_0^2 \left[ (4x^2 + 4x + 1) - (x^4 + 2x^2 + 1) \right] \, dx
   \]
   Simplify:
   \[
   V = \pi \int_0^2 \left[ -x^4 + 2x^2 + 4x \right] \, dx
   \]

4. **Compute the integral**:
   \[
   V = \pi \left[ -\frac{x^5}{5} + \frac{2x^3}{3} + 2x^2 \right]_0^2
   \]
   Evaluate at \( x = 2 \):
   \[
   V = \pi \left[ -\frac{32}{5} + \frac{16}{3} + 8 \right]
   \]
   Evaluate at \( x = 0 \) (all terms vanish):
   \[
   V = \pi \left( -\frac{32}{5} + \frac{16}{3} + 8 \right)
   \]

   Find a common denominator to simplify:
   \[
   V = \pi \left( -\frac{32}{5} + \frac{80}{15} + \frac{120}{15} \right) = \pi \left( -\frac{32}{5} + \frac{200}{15} \right)
   \]
   Simplifying further:
   \[
   V \approx \pi \left( -6.4 + 13.33 \right) = \pi \times 6.93
   \]
   Therefore:
   \[
   V \approx 21.77 \, \pi \, \text{units}^3
   \]

The volume is approximately \( 21.77\pi \) cubic units.

Would you like any further details on these steps?

### Related questions:
1. How is the washer method different from the disk method in finding volumes of solids of revolution?
2. What is the general formula for the volume of a solid revolved around the x-axis?
3. How do you determine which curve gives the outer radius and which gives the inner radius?
4. Can the same integral be applied if the region was revolved around the y-axis?
5. What changes in the volume formula if the axis of rotation is shifted further down or up?

#### Tip: 
When integrating using the washer method, always make sure to clearly identify the outer and inner radii before setting up the integral.

Find the volume of the solid that results when the region enclosed by y = 2x - 1 and y = x^2 - 1 is revolved around y = -2.

Calculate the volume of a solid formed by revolving the region enclosed by y = 2x - 1 and y = x^2 - 1 around y = -2. This solution involves the washer method and step-by-step integration to determine the exact volume. Ideal for high school students studying calculus.

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