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Let's break down the problem step-by-step to set up the integral without integrating, as requested:

### Problem:
You are asked to find the integral that computes the **volume** of the solid generated by rotating the region **R**, bounded by the curves:

- \( x = y^2 - 2y \) 
- \( y = x \)

about the **x-axis**.

### Steps to solve:

1. **Sketch the curves and determine the region**:
   - First, let's analyze the given curves.
     - The first curve is a quadratic in terms of \( y \): \( x = y^2 - 2y \), which can be rewritten as \( x = (y - 1)^2 - 1 \), representing a parabola shifted right and downward.
     - The second curve is linear: \( y = x \), which is a straight line passing through the origin with a slope of 1.
   - These two curves intersect at points that bound the region \( R \).

2. **Find the intersection points**:
   To find the intersection points, set the two equations equal to each other:
   \[
   x = y^2 - 2y \quad \text{and} \quad x = y
   \]
   So, equate:
   \[
   y^2 - 2y = y
   \]
   Simplify the equation:
   \[
   y^2 - 3y = 0
   \]
   Factor it:
   \[
   y(y - 3) = 0
   \]
   Thus, \( y = 0 \) and \( y = 3 \).

   Now, plug these \( y \)-values back into \( x = y \) to find the corresponding \( x \)-values:
   - For \( y = 0 \), \( x = 0 \).
   - For \( y = 3 \), \( x = 3 \).

   So, the region is bounded by the points \( (0, 0) \) and \( (3, 3) \).

3. **Draw a typical approximating rectangle**:
   - Since we are rotating about the **x-axis**, the approximation method uses a vertical strip along the \( y \)-axis. The width of the strip is \( dy \), and the height extends from the curve \( y = x \) to the parabola \( x = y^2 - 2y \).

4. **Sketch a typical disk/washer**:
   - The region is rotated about the \( x \)-axis, so the volume can be computed using the **disk/washer method**.
   - The formula for the volume of a washer is:
     \[
     V = \pi \int \left[\text{outer radius}^2 - \text{inner radius}^2\right] \, dy
     \]
   - Here, the outer radius is the distance from the x-axis to the curve \( x = y^2 - 2y \), and the inner radius is the distance to the curve \( y = x \).

5. **Set up the integral**:
   - The outer radius for the washer is given by the curve \( x = y^2 - 2y \).
   - The inner radius for the washer is given by the curve \( x = y \).
   - The limits of integration are from \( y = 0 \) to \( y = 3 \).
   - The volume integral is:
     \[
     V = \pi \int_0^3 \left[(y^2 - 2y)^2 - y^2\right] \, dy
     \]

### Final integral setup:
\[
V = \pi \int_0^3 \left[(y^2 - 2y)^2 - y^2\right] \, dy
\]

This is the integral that computes the volume of the solid formed by rotating the region \( R \) around the \( x \)-axis.

---

Do you have any questions or need further clarification? Here are five follow-up questions related to this problem:

1. How would you approach solving the integral we just set up?
2. Could this problem be solved using the **cylindrical shell method** instead of the washer method?
3. What happens if we rotate the region about the **y-axis** instead of the x-axis?
4. How do you handle finding volumes when curves are rotated about lines other than the axes?
5. What is the geometric interpretation of the difference between the outer radius and inner radius?

**Tip**: Always double-check the intersection points when setting up integrals, as they define the boundaries for your region of integration.

Let R be the region bounded by the curves x = y^2 - 2y and y = x. Set up an integral to compute the volume of the solid generated by rotating R about the x-axis. Do not integrate.

Set up the integral to find the volume of the solid formed by rotating the region bounded by the curves x = y^2 - 2y and y = x around the x-axis. Learn about the washer method and curve intersection for undergraduate calculus students.

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