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The problem asks to find the volume of the solid formed by rotating the region bounded by the curves \( y = x^2 \) and \( y = 3x \) about the line \( y = -1 \).

### Steps to Solve:

1. **Find the intersection points** of the curves \( y = x^2 \) and \( y = 3x \) to determine the limits of integration. Set the equations equal to each other:
   \[
   x^2 = 3x
   \]
   \[
   x^2 - 3x = 0
   \]
   \[
   x(x - 3) = 0
   \]
   So, the curves intersect at \( x = 0 \) and \( x = 3 \).

2. **Set up the formula for the volume using the washer method**. Since we are rotating about the line \( y = -1 \), the radius of each washer is the distance from the curves to \( y = -1 \). For a given \( x \), the outer radius \( R(x) \) (from the curve \( y = 3x \)) is:
   \[
   R(x) = 3x - (-1) = 3x + 1
   \]
   And the inner radius \( r(x) \) (from the curve \( y = x^2 \)) is:
   \[
   r(x) = x^2 - (-1) = x^2 + 1
   \]

3. **Volume formula**: The volume of the solid is given by the integral of the area of the washers:
   \[
   V = \pi \int_{0}^{3} \left[ R(x)^2 - r(x)^2 \right] dx
   \]
   Substituting \( R(x) = 3x + 1 \) and \( r(x) = x^2 + 1 \), we get:
   \[
   V = \pi \int_{0}^{3} \left[ (3x + 1)^2 - (x^2 + 1)^2 \right] dx
   \]

4. **Expand both terms**:
   \[
   (3x + 1)^2 = 9x^2 + 6x + 1
   \]
   \[
   (x^2 + 1)^2 = x^4 + 2x^2 + 1
   \]

5. **Set up the integral**:
   \[
   V = \pi \int_{0}^{3} \left[ (9x^2 + 6x + 1) - (x^4 + 2x^2 + 1) \right] dx
   \]
   Simplifying the integrand:
   \[
   V = \pi \int_{0}^{3} \left( -x^4 + 7x^2 + 6x \right) dx
   \]

6. **Integrate term by term**:
   \[
   \int -x^4 dx = -\frac{x^5}{5}, \quad \int 7x^2 dx = \frac{7x^3}{3}, \quad \int 6x dx = 3x^2
   \]
   So, the volume integral becomes:
   \[
   V = \pi \left[ -\frac{x^5}{5} + \frac{7x^3}{3} + 3x^2 \right]_{0}^{3}
   \]

7. **Evaluate the integral at the limits**:
   At \( x = 3 \):
   \[
   -\frac{3^5}{5} + \frac{7(3^3)}{3} + 3(3^2) = -\frac{243}{5} + \frac{7(27)}{3} + 27
   \]
   Simplifying:
   \[
   -\frac{243}{5} + 63 + 27 = -\frac{243}{5} + 90 = \frac{-243 + 450}{5} = \frac{207}{5}
   \]
   At \( x = 0 \), all terms are zero.

Thus, the volume is:
\[
V = \pi \times \frac{207}{5} = \frac{207\pi}{5}
\]

The volume of the solid is \( \frac{207\pi}{5} \).

Would you like more details, or have any questions?

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### Related Questions:
1. What is the washer method used for in calculus?
2. How does changing the axis of rotation affect the volume of a solid?
3. What are the differences between the disk and washer methods?
4. Can you rotate the same region about the x-axis and find the volume?
5. How do you calculate the volume when rotating about vertical lines like \(x = 2\)?

### Tip:
Always check whether you need the outer and inner radii when using the washer method—this is key when the region has different top and bottom functions.

Find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and y = 3x about the line y = -1.

Discover how to calculate the volume of a solid by rotating a region bounded by y = x^2 and y = 3x about y = -1 using the washer method. This guide covers setting up the integral and evaluating it, including the calculation of the outer and inner radii.

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