To find the volume of the solid obtained by rotating the region bounded by \( y = x^8 \), \( y = 1 \), and the line \( y = 5 \) about the line \( y = 5 \), we can use the method of washers.
First, we need to determine the points of intersection of the curves \( y = x^8 \) and \( y = 1 \):
\[ x^8 = 1 \]
\[ x = 1 \text{ or } x = -1 \]
The region of interest is bounded between \( x = -1 \) and \( x = 1 \), and it lies between \( y = x^8 \) and \( y = 1 \).
The radius of the washer at a general point \( x \) is the distance from \( y = 5 \) to \( y = x^8 \):
\[ \text{radius} = 5 - x^8 \]
The outer radius (distance from \( y = 5 \) to \( y = 1 \)) is:
\[ \text{outer radius} = 5 - 1 = 4 \]
The inner radius (distance from \( y = 5 \) to \( y = x^8 \)) is:
\[ \text{inner radius} = 5 - x^8 \]
Now, the volume \( V \) of the solid of revolution is given by the integral:
\[ V = \pi \int_{-1}^{1} \left[(4)^2 - (5 - x^8)^2 \right] \, dx \]
Calculate the integral:
\[ V = \pi \int_{-1}^{1} \left[ 16 - (25 - 10x^8 + x^{16}) \right] \, dx \]
\[ V = \pi \int_{-1}^{1} (x^{16} - 10x^8 + 9) \, dx \]
Integrate each term separately:
\[ \int x^{16} \, dx = \frac{x^{17}}{17}, \quad \int 10x^8 \, dx = \frac{10x^9}{9}, \quad \int 9 \, dx = 9x \]
Evaluate the definite integral from \( -1 \) to \( 1 \):
\[ V = \pi \left[ \frac{x^{17}}{17} - \frac{10x^9}{9} + 9x \right]_{-1}^{1} \]
\[ V = \pi \left( \left[ \frac{1}{17} - \frac{10}{9} + 9 \right] - \left[ \frac{1}{17} - \frac{10}{9} - 9 \right] \right) \]
\[ V = \pi \left( \frac{1}{17} - \frac{10}{9} + 9 - \frac{1}{17} + \frac{10}{9} + 9 \right) \]
\[ V = \pi \left( 18 \right) \]
\[ V = 18\pi \]
Therefore, the volume of the solid obtained by rotating the region bounded by \( y = x^8 \), \( y = 1 \), and \( y = 5 \) about the line \( y = 5 \) is \( \boxed{18\pi} \).
