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To solve this problem using the shell method, we will set up an integral for the volume of the solid formed by revolving the region bounded by the curves \( y = x^2 \), \( y = 3 - 2x \), and the line \( x = 0 \) about the \( y \)-axis.

### Steps to Solution:

1. **Identify the Bounds of Integration:**
   The intersection points of \( y = x^2 \) and \( y = 3 - 2x \) will give the limits for \( x \). Setting the equations equal:
   \[
   x^2 = 3 - 2x
   \]
   Rearranging this, we get:
   \[
   x^2 + 2x - 3 = 0
   \]
   Factoring:
   \[
   (x + 3)(x - 1) = 0
   \]
   So, \( x = -3 \) and \( x = 1 \). Since \( x \geq 0 \), we’ll only use \( x = 0 \) and \( x = 1 \) as our limits.

2. **Set Up the Shell Method Integral:**
   For the shell method, the volume \( V \) is given by:
   \[
   V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dx
   \]
   Here:
   - **Radius:** The distance from the \( y \)-axis is \( x \).
   - **Height:** The difference between the two curves, \( (3 - 2x) - x^2 \).

   So, we set up the integral:
   \[
   V = 2\pi \int_{0}^{1} x \left( (3 - 2x) - x^2 \right) \, dx
   \]

3. **Simplify the Integrand:**
   Expanding the integrand:
   \[
   V = 2\pi \int_{0}^{1} x (3 - 2x - x^2) \, dx
   \]
   Distribute \( x \):
   \[
   V = 2\pi \int_{0}^{1} (3x - 2x^2 - x^3) \, dx
   \]

4. **Integrate with Respect to \( x \):**
   Now, integrate each term:
   \[
   V = 2\pi \left[ \frac{3x^2}{2} - \frac{2x^3}{3} - \frac{x^4}{4} \right]_{0}^{1}
   \]
   Evaluating at \( x = 1 \) and \( x = 0 \):
   \[
   V = 2\pi \left( \frac{3(1)^2}{2} - \frac{2(1)^3}{3} - \frac{(1)^4}{4} \right)
   \]
   \[
   = 2\pi \left( \frac{3}{2} - \frac{2}{3} - \frac{1}{4} \right)
   \]

5. **Combine Terms:**
   To combine these, find a common denominator. The common denominator for \( \frac{3}{2} \), \( \frac{2}{3} \), and \( \frac{1}{4} \) is 12:
   \[
   \frac{3}{2} = \frac{18}{12}, \quad \frac{2}{3} = \frac{8}{12}, \quad \frac{1}{4} = \frac{3}{12}
   \]
   So,
   \[
   V = 2\pi \left( \frac{18 - 8 - 3}{12} \right)
   \]
   \[
   = 2\pi \cdot \frac{7}{12} = \frac{7\pi}{6}
   \]

### Final Answer:
The volume of the solid is:
\[
V = \frac{7\pi}{6}
\]

Would you like further details or have any questions on specific steps?

### Related Questions:
1. How would you set up the integral if the region was rotated about the \( x \)-axis instead?
2. What are the differences between the shell and disk/washer methods for volume calculations?
3. How would you determine the volume if the region was bounded by different functions?
4. Can this integral be solved using numerical methods if it were too complex to integrate by hand?
5. How would the solution change if the region was rotated about a line other than the \( y \)-axis?

### Tip:
When using the shell method, always check which axis the rotation is around, as this determines the radius (distance to the axis) and the height of the shell.

Use the shell method to find the volume of the solid generated by revolving the regions bounded by the curves and lines about the y-axis.

y = x^2, y = 3 - 2x, x = 0, for x >= 0

Calculate the volume of a solid generated by revolving the region bounded by y = x^2, y = 3 - 2x, and x = 0 around the y-axis using the shell method. This solution involves integral calculus and applies the shell method formula.

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