Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=x^6, y =1 about y=5. 

We are tasked with finding the volume of the solid obtained by rotating the region bounded by \( y = x^6 \) and \( y = 1 \) about the axis \( y = 5 \). To solve this, we will use the **method of cylindrical shells**.

### Step-by-step solution:

1. **Sketch the region**:
   The region is bounded by the curve \( y = x^6 \), the horizontal line \( y = 1 \), and we are rotating around \( y = 5 \), which is a horizontal axis above the region.

2. **Set up the cylindrical shells**:
   The formula for the volume using cylindrical shells is:

   \[
   V = 2\pi \int_a^b (r_{\text{shell}})(h_{\text{shell}}) \, dx
   \]

   Here:
   - \( r_{\text{shell}} \) is the distance from the shell to the axis of rotation \( y = 5 \), which is \( r_{\text{shell}} = 5 - y \).
   - \( h_{\text{shell}} \) is the height of the shell, which is determined by the curve equation \( y = x^6 \).

3. **Express the radius and height**:
   - The radius from \( y = 5 \) to a point on the curve is \( r_{\text{shell}} = 5 - y = 5 - x^6 \).
   - The height of the shell is the difference between the upper boundary \( y = 1 \) and the curve \( y = x^6 \). Therefore, \( h_{\text{shell}} = 1 - x^6 \).

4. **Determine the limits of integration**:
   The curves intersect at \( y = 1 \) and \( y = x^6 \). Solving \( x^6 = 1 \), we find that the limits of integration are from \( x = 0 \) to \( x = 1 \).

5. **Set up the integral**:
   The volume integral becomes:

   \[
   V = 2\pi \int_0^1 (5 - x^6)(1 - x^6) \, dx
   \]

6. **Expand the integrand**:
   \[
   (5 - x^6)(1 - x^6) = 5(1 - x^6) - x^6(1 - x^6)
   \]
   Simplifying:
   \[
   = 5 - 5x^6 - x^6 + x^{12}
   \]
   \[
   = 5 - 6x^6 + x^{12}
   \]

7. **Evaluate the integral**:
   Now, integrate term by term:
   \[
   V = 2\pi \int_0^1 \left( 5 - 6x^6 + x^{12} \right) \, dx
   \]
   \[
   = 2\pi \left[ 5x - \frac{6x^7}{7} + \frac{x^{13}}{13} \right]_0^1
   \]
   Substituting the limits:
   \[
   V = 2\pi \left( 5 - \frac{6}{7} + \frac{1}{13} \right)
   \]
   \[
   V = 2\pi \left( 5 - \frac{6}{7} + \frac{1}{13} \right)
   \]

   To simplify the expression inside the parentheses, find a common denominator:
   \[
   V = 2\pi \left( \frac{455}{91} - \frac{78}{91} + \frac{7}{91} \right)
   \]
   \[
   V = 2\pi \left( \frac{455 - 78 + 7}{91} \right)
   \]
   \[
   V = 2\pi \left( \frac{384}{91} \right)
   \]

   Therefore, the volume is:
   \[
   V = \frac{768\pi}{91}
   \]

### Final Answer:
The volume of the solid is \( \frac{768\pi}{91} \) cubic units.

Would you like further details or have any questions?

Here are five related questions to explore:

1. How would the volume change if the axis of rotation was \( y = 3 \) instead of \( y = 5 \)?
2. Can you find the volume if we rotate the same region about the x-axis?
3. What is the volume of the solid if we rotate the region around the line \( x = 1 \)?
4. How does the method of disks or washers compare to the method of cylindrical shells for this problem?
5. What if the region is rotated around the line \( y = -1 \)? How would the integral change?

**Tip**: The method of cylindrical shells is often simpler when rotating around a horizontal or vertical line that is not an axis.

Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis. y=x^6, y =1 about y=5.

Learn how to calculate the volume of the solid obtained by rotating the region bounded by y=x^6 and y=1 about y=5 using the cylindrical shell method. This detailed solution covers calculus concepts like integration and solid volumes, suitable for Grades 11-12 or Calculus I students.

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