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We are tasked with finding the volume of the solid obtained by rotating the region bounded by the curves \( y^2 = x \) and \( x = 3y \) about the line \( y = -1 \).

### Step-by-Step Solution:

#### 1. **Identify the curves:**
   - The first curve is \( y^2 = x \), or equivalently \( x = y^2 \).
   - The second curve is \( x = 3y \).

   These curves will intersect at certain points, and we need to determine the bounds of integration.

#### 2. **Find the points of intersection:**
   To find where the curves intersect, set \( y^2 = 3y \) (since both equations equal \( x \)):
   \[
   y^2 = 3y \implies y^2 - 3y = 0 \implies y(y - 3) = 0.
   \]
   Therefore, \( y = 0 \) and \( y = 3 \).

   So, the region is bounded between \( y = 0 \) and \( y = 3 \).

#### 3. **Set up the volume integral:**
   We will use the **method of cylindrical shells** to calculate the volume of the solid, since we are rotating around \( y = -1 \), which is parallel to the axis of the curves but not through the curves themselves.

   The formula for the volume using cylindrical shells is:
   \[
   V = 2\pi \int_a^b ( \text{radius} )( \text{height} ) \, dy.
   \]

   - **Radius:** The distance from the curve to the line \( y = -1 \) is \( y - (-1) = y + 1 \).
   - **Height:** The height of each shell is the horizontal distance between the curves, which is \( x = 3y \) minus \( x = y^2 \), so the height is \( 3y - y^2 \).

   Therefore, the volume integral is:
   \[
   V = 2\pi \int_0^3 (y + 1)(3y - y^2) \, dy.
   \]

#### 4. **Expand the integrand:**
   Expand \( (y + 1)(3y - y^2) \):
   \[
   (y + 1)(3y - y^2) = y(3y - y^2) + 1(3y - y^2) = 3y^2 - y^3 + 3y - y^2 = 2y^2 - y^3 + 3y.
   \]

   So, the volume integral becomes:
   \[
   V = 2\pi \int_0^3 (2y^2 - y^3 + 3y) \, dy.
   \]

#### 5. **Integrate:**
   Now, integrate each term:
   \[
   \int_0^3 2y^2 \, dy = \frac{2y^3}{3} \Big|_0^3 = \frac{2(27)}{3} = 18,
   \]
   \[
   \int_0^3 -y^3 \, dy = -\frac{y^4}{4} \Big|_0^3 = -\frac{81}{4},
   \]
   \[
   \int_0^3 3y \, dy = \frac{3y^2}{2} \Big|_0^3 = \frac{3(9)}{2} = \frac{27}{2}.
   \]

#### 6. **Compute the total volume:**
   Combine all the results:
   \[
   V = 2\pi \left( 18 - \frac{81}{4} + \frac{27}{2} \right).
   \]
   To simplify, find a common denominator:
   \[
   V = 2\pi \left( 18 - \frac{81}{4} + \frac{54}{4} \right) = 2\pi \left( 18 - \frac{27}{4} \right) = 2\pi \left( \frac{72}{4} - \frac{27}{4} \right) = 2\pi \left( \frac{45}{4} \right).
   \]
   Therefore:
   \[
   V = \frac{90\pi}{4} = 22.5\pi.
   \]

### Final Answer:
The volume of the solid is \( \boxed{22.5\pi} \) cubic units.

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Would you like any further details or explanations on this solution?

Here are 5 related questions for practice:
1. How would the volume change if the region were rotated around the line \( y = 1 \)?
2. How would the problem be solved if the region were rotated about the \( x \)-axis?
3. What is the volume if the curves were \( y^2 = x \) and \( x = 4y \) instead?
4. Can this problem be solved using the disk or washer method? If yes, how?
5. What happens to the volume if the curves do not intersect?

**Tip:** When solving rotation problems, identifying the right method (shell, disk, or washer) is key to simplifying the integration process.

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

This solution details the process of finding the volume of a solid obtained by rotating the region between y^2 = x and x = 3y around y = -1. Using the method of cylindrical shells, the step-by-step integration process is explained. Ideal for undergraduate calculus students.

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