Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the line y = 3.
y = x
y = 2
x = 0

To find the volume of the solid generated by revolving the region bounded by the graphs of \(y = x\), \(y = 2\), and \(x = 0\) about the line \(y = 3\), we will use the **method of washers**.

### Step-by-Step Solution

1. **Visualize the Region**:
   - The line \(y = x\) is a straight line passing through the origin.
   - The line \(y = 2\) is a horizontal line.
   - The line \(x = 0\) is the y-axis.

   The bounded region lies between \(x = 0\) (the y-axis), \(y = x\) (from \(x = 0\) to \(x = 2\)), and \(y = 2\).

2. **Set Up the Integral for the Washer Method**:

   The volume \(V\) of the solid obtained by rotating a region around the line \(y = 3\) using the washer method is given by:
   \[
   V = \pi \int_{a}^{b} \left[ R_{\text{outer}}(x)^2 - R_{\text{inner}}(x)^2 \right] \, dx
   \]

   In this case, the outer radius \(R_{\text{outer}}(x)\) is the distance from \(y = 3\) to \(y = x\), and the inner radius \(R_{\text{inner}}(x)\) is the distance from \(y = 3\) to \(y = 2\).

   - \(R_{\text{outer}}(x) = 3 - x\) (distance from \(y = 3\) to \(y = x\))
   - \(R_{\text{inner}}(x) = 3 - 2 = 1\) (distance from \(y = 3\) to \(y = 2\))

3. **Determine the Limits of Integration**:

   The region is bounded horizontally from \(x = 0\) to \(x = 2\).

4. **Write the Integral**:

   Substituting into the formula for volume:
   \[
   V = \pi \int_{0}^{2} \left[ (3 - x)^2 - (1)^2 \right] \, dx
   \]

5. **Simplify and Compute the Integral**:

   Simplify the integrand:
   \[
   V = \pi \int_{0}^{2} \left[ (3 - x)^2 - 1 \right] \, dx
   \]

   Expand the square and simplify:
   \[
   (3 - x)^2 = 9 - 6x + x^2
   \]

   So the integral becomes:
   \[
   V = \pi \int_{0}^{2} (9 - 6x + x^2 - 1) \, dx = \pi \int_{0}^{2} (8 - 6x + x^2) \, dx
   \]

6. **Evaluate the Integral**:

   Compute the antiderivative:
   \[
   V = \pi \left[ 8x - 3x^2 + \frac{x^3}{3} \right]_{0}^{2}
   \]

   Plug in the limits of integration:
   \[
   V = \pi \left( \left(8(2) - 3(2)^2 + \frac{(2)^3}{3}\right) - \left(8(0) - 3(0)^2 + \frac{(0)^3}{3}\right) \right)
   \]

   Simplify:
   \[
   V = \pi \left(16 - 12 + \frac{8}{3}\right)
   \]

   Combine like terms:
   \[
   V = \pi \left(4 + \frac{8}{3}\right) = \pi \left(\frac{12}{3} + \frac{8}{3}\right) = \pi \left(\frac{20}{3}\right)
   \]

   So, the volume is:
   \[
   V = \frac{20\pi}{3}
   \]

### Final Answer
The volume of the solid generated by revolving the region about the line \(y = 3\) is \(\frac{20\pi}{3}\) cubic units.

---

Would you like more details on any part of this process, or do you have any questions?

Here are 5 related questions that might interest you:

1. How do you determine the outer and inner radii for different axis of rotation?
2. What is the difference between the washer method and the shell method?
3. How would the volume change if the region was revolved around the x-axis instead?
4. What if the region was revolved around a different horizontal line, like \(y = 1\)?
5. How do you find the volume of a solid generated by revolving a region around a vertical line?

**Tip:** When setting up integrals for volumes of revolution, always sketch the region and visualize the rotation to identify the correct radii and limits of integration.

Calculate the volume of the solid generated by revolving the region bounded by y = x, y = 2, and x = 0 about the line y = 3 using the washer method. Detailed step-by-step solution and integral setup provided.

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