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.