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To solve this problem, we need to find the volume of the solid generated by revolving the region enclosed by the curves \( y = 2x - 1 \) and \( y = x^2 - 1 \) around the line \( y = -2 \).

### Step 1: Finding the intersection points
The region enclosed by the curves is between the points where \( y = 2x - 1 \) and \( y = x^2 - 1 \) intersect. To find these points, set the equations equal to each other:
\[
2x - 1 = x^2 - 1
\]
Simplify:
\[
2x = x^2 \implies x(x - 2) = 0
\]
So, the intersection points are at \( x = 0 \) and \( x = 2 \).

### Step 2: Volume using the washer method
Since the region is revolved around \( y = -2 \), we use the washer method. The formula for the volume of a solid generated by revolving a region around a horizontal line is:

\[
V = \pi \int_{a}^{b} \left[ (R_{\text{outer}}(x))^2 - (R_{\text{inner}}(x))^2 \right] dx
\]
Where:
- \( R_{\text{outer}}(x) \) is the distance from the curve farther from \( y = -2 \) to the axis of revolution.
- \( R_{\text{inner}}(x) \) is the distance from the curve closer to \( y = -2 \) to the axis of revolution.

In this case:
- The outer curve is \( y = 2x - 1 \), so \( R_{\text{outer}}(x) = (2x - 1) - (-2) = 2x + 1 \).
- The inner curve is \( y = x^2 - 1 \), so \( R_{\text{inner}}(x) = (x^2 - 1) - (-2) = x^2 + 1 \).

### Step 3: Setting up the integral
Now, we set up the integral from \( x = 0 \) to \( x = 2 \):
\[
V = \pi \int_{0}^{2} \left[ (2x + 1)^2 - (x^2 + 1)^2 \right] dx
\]
We can expand both terms:
\[
(2x + 1)^2 = 4x^2 + 4x + 1
\]
\[
(x^2 + 1)^2 = x^4 + 2x^2 + 1
\]
Thus, the integrand becomes:
\[
(4x^2 + 4x + 1) - (x^4 + 2x^2 + 1) = -x^4 + 2x^2 + 4x
\]

### Step 4: Compute the integral
Now, compute the integral:
\[
V = \pi \int_{0}^{2} (-x^4 + 2x^2 + 4x) dx
\]
\[
V = \pi \left[ \frac{-x^5}{5} + \frac{2x^3}{3} + 2x^2 \right]_{0}^{2}
\]
Evaluate at \( x = 2 \):
\[
V = \pi \left[ \frac{-32}{5} + \frac{16}{3} + 8 \right]
\]
Simplifying the expression:
\[
V = \pi \left[ \frac{-32}{5} + \frac{16}{3} + \frac{24}{3} \right] = \pi \left[ \frac{-32}{5} + \frac{40}{3} \right]
\]
We can find a common denominator (15) to simplify further:
\[
V = \pi \left[ \frac{-96}{15} + \frac{200}{15} \right] = \pi \cdot \frac{104}{15}
\]

### Final Answer:
\[
V = \frac{104\pi}{15}
\]

Let me know if you need any further clarification or details!

### Related Questions:
1. How would the volume change if the region was revolved around the x-axis instead?
2. What if the region was revolved around the line \( y = -3 \) instead of \( y = -2 \)?
3. Can you use the shell method to solve this problem, and how would it differ?
4. What would happen if the curves were translated upward by 2 units?
5. How would the limits of integration change if the intersection points were different?

### Tip:
Always sketch the curves and the axis of revolution for problems involving solids of revolution, as it helps visualize the outer and inner radii.

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.

Learn how to calculate the volume of the solid obtained by revolving the region enclosed by y = 2x - 1 and y = x^2 - 1 around y = -2 using the washer method. Includes step-by-step integration and explanation of concepts, suitable for calculus students.

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