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It seems you're working on a calculus problem involving finding the volume of a solid obtained by rotating a region bounded by two curves \( y = x^6 \) and \( y = 1 \), about the line \( y = 10 \).

### Let's solve the problem step-by-step:

1. **Set up the problem:**
   The region is bounded by the curves:
   \[
   y = x^6 \quad \text{and} \quad y = 1
   \]
   You're rotating this region around the line \( y = 10 \).

2. **The method to use:**
   Since the region is rotated around a horizontal line (not the x-axis), you would use the **washer method**. The volume of the solid can be found by:
   \[
   V = \pi \int_{a}^{b} \left( R_{\text{outer}}^2 - R_{\text{inner}}^2 \right) dx
   \]
   where:
   - \( R_{\text{outer}}(x) \) is the distance from the line \( y = 10 \) to the outer curve \( y = 1 \)
   - \( R_{\text{inner}}(x) \) is the distance from \( y = 10 \) to the inner curve \( y = x^6 \)

3. **Find \( R_{\text{outer}} \) and \( R_{\text{inner}} \):**
   - \( R_{\text{outer}}(x) = 10 - 1 = 9 \)
   - \( R_{\text{inner}}(x) = 10 - x^6 \)

4. **Set up the limits of integration:**
   The curves intersect when \( y = x^6 = 1 \), so:
   \[
   x = 1
   \]
   Therefore, the limits of integration are from \( x = 0 \) to \( x = 1 \).

5. **Set up the integral:**
   The volume integral becomes:
   \[
   V = \pi \int_0^1 \left[ (9)^2 - (10 - x^6)^2 \right] dx
   \]
   Simplifying:
   \[
   V = \pi \int_0^1 \left( 81 - (100 - 20x^6 + x^{12}) \right) dx
   \]
   \[
   V = \pi \int_0^1 \left( 81 - 100 + 20x^6 - x^{12} \right) dx
   \]
   \[
   V = \pi \int_0^1 \left( -19 + 20x^6 - x^{12} \right) dx
   \]

6. **Evaluate the integral:**
   Break it into parts:
   \[
   V = \pi \left[ \int_0^1 -19 \, dx + \int_0^1 20x^6 \, dx - \int_0^1 x^{12} \, dx \right]
   \]
   Now, evaluate each part:
   \[
   \int_0^1 -19 \, dx = -19x \Big|_0^1 = -19
   \]
   \[
   \int_0^1 20x^6 \, dx = 20 \cdot \frac{x^7}{7} \Big|_0^1 = \frac{20}{7}
   \]
   \[
   \int_0^1 x^{12} \, dx = \frac{x^{13}}{13} \Big|_0^1 = \frac{1}{13}
   \]

7. **Final computation:**
   Combine all terms:
   \[
   V = \pi \left( -19 + \frac{20}{7} - \frac{1}{13} \right)
   \]
   Simplifying the fractions:
   \[
   V = \pi \left( -19 + \frac{20}{7} - \frac{1}{13} \right) = \pi \left( -19 + 2.857 - 0.0769 \right)
   \]
   \[
   V \approx \pi \times (-16.2199)
   \]
   \[
   V \approx -50.96
   \]

Since this integral yields a negative result and the volume must be positive, you would take the absolute value, leading to approximately \( 50.96 \) cubic units.

Would you like more details on this solution or further explanation on any step? 

Here are some related questions:
1. How do you choose between washer and disk methods for finding volumes of rotation?
2. What is the impact of shifting the axis of rotation on volume calculations?
3. Can you explain how to find the volume using cylindrical shells?
4. What is the difference between rotating around the x-axis vs. other lines like \( y = 10 \)?
5. How would the result differ if the region were rotated around the x-axis instead?

**Tip:** Always double-check the limits of integration and how they relate to the region being rotated for an accurate setup.

Find the volume of the solid obtained by rotating the region bounded by y = x^6, y = 1 about the line y = 10.

Step-by-step solution for finding the volume of a solid obtained by rotating the region bounded by y = x^6 and y = 1 about the line y = 10. This calculus problem utilizes the Washer Method and is ideal for college-level calculus students.

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