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We are tasked with finding the volume of a solid obtained by rotating a region about the line \( y = 1 \) using the method of disks or washers. The given boundaries are:

- \( x = 0 \)
- \( y = 1 \)
- \( x = y^5 \)

### Step-by-Step Process:

1. **Identify the boundaries**:
   - The region is bounded by \( x = 0 \), \( y = 1 \), and \( x = y^5 \).

2. **Set up the integral** using the washer method:
   - The formula for the volume of a solid generated by rotating about a horizontal line (like \( y = 1 \)) using the washer method is:
     \[
     V = \pi \int_a^b \left( R_{\text{outer}}(y)^2 - R_{\text{inner}}(y)^2 \right) \, dy
     \]
     where \( R_{\text{outer}}(y) \) is the distance from the line \( y = 1 \) to the outer curve, and \( R_{\text{inner}}(y) \) is the distance to the inner curve.

3. **Determine the radii**:
   - For this problem:
     - The outer radius (the distance from \( y = 1 \) to the x-axis) is constant at \( 1 \).
     - The inner radius is the distance from the line \( y = 1 \) to the curve \( x = y^5 \). This distance is \( 1 - y \), where \( y \) is the value on the curve.

4. **Write the integral**:
   The volume is computed as:
   \[
   V = \pi \int_0^1 \left( 1^2 - (1 - y)^2 \right) \, dy
   \]
   Simplifying the expression inside the integral:
   \[
   1^2 - (1 - y)^2 = 1 - (1 - 2y + y^2) = 1 - 1 + 2y - y^2 = 2y - y^2
   \]
   So the integral becomes:
   \[
   V = \pi \int_0^1 (2y - y^2) \, dy
   \]

5. **Integrate**:
   Now, integrate term by term:
   \[
   \int_0^1 (2y - y^2) \, dy = \left[ y^2 - \frac{y^3}{3} \right]_0^1
   \]
   Evaluating at the limits \( 0 \) and \( 1 \):
   \[
   \left( 1^2 - \frac{1^3}{3} \right) - \left( 0^2 - \frac{0^3}{3} \right) = \left( 1 - \frac{1}{3} \right) = \frac{2}{3}
   \]

6. **Multiply by \( \pi \)**:
   \[
   V = \pi \times \frac{2}{3} = \frac{2\pi}{3}
   \]

Thus, the volume of the solid is:
\[
V = \frac{2\pi}{3} \, \text{cubic units}
\]

---

Would you like more details or have any questions? Here are some related questions for practice:

1. What would the volume be if the region were rotated about the x-axis?
2. How would the integral change if the curve was \( x = y^3 \) instead of \( x = y^5 \)?
3. What is the geometric significance of the disk method versus the washer method?
4. How would you calculate the surface area of the solid generated in this problem?
5. What would the limits of integration be if the region extended to \( y = 2 \)?

**Tip:** In problems involving rotation about a horizontal line, always carefully consider the radii as distances from the line of rotation!

The volume of the solid obtained by rotating the region enclosed by x = 0, y = 1, x = y^5 about the line y = 1 can be computed using the method of disks or washers via an integral with limits of integration a = 0 and b = 1.

Explore the step-by-step process to calculate the volume of a solid obtained by rotating a region enclosed by x = 0, y = 1, and x = y^5 about the line y = 1 using the method of disks or washers. The integral runs from 0 to 1 and involves concepts like the Fundamental Theorem of Calculus.

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