We are tasked with finding the volume of the solid generated by revolving the region bounded by the curves $y = 6x - x^2$ and $y = 0$ about the line $x = 8$. To achieve this, we will use the shell method.

Step-by-Step Solution:

1. Identify the region: The curves that bound the region are:

   • $y = 6x - x^2$, a parabola opening downwards.
   • $y = 0$, the x-axis.

   We need to find the intersection points of these curves, i.e., solve: $6x - x^2 = 0$ Factoring: $x(6 - x) = 0$ So, $x = 0$ and $x = 6$.

2. Set up the shell method formula: The volume using the shell method is given by: $V = 2\pi \int_{a}^{b} ( \text{radius} ) \times ( \text{height} ) \, dx$ where:

   • The radius is the horizontal distance from the curve to the line of rotation $x = 8$. This distance is $8 - x$.
   • The height of the shell is given by the value of $y$, which is $6x - x^2$.

   Therefore, the volume integral becomes: $V = 2\pi \int_{0}^{6} (8 - x)(6x - x^2) \, dx$

3. Expand the integrand: Let's expand the expression inside the integral: $(8 - x)(6x - x^2) = 8(6x - x^2) - x(6x - x^2)$ Simplifying both terms: $= 48x - 8x^2 - 6x^2 + x^3$ $= 48x - 14x^2 + x^3$

4. Evaluate the integral: Now, we integrate term by term: $V = 2\pi \int_{0}^{6} (48x - 14x^2 + x^3) \, dx$

   • The integral of $48x$ is $24x^2$,
   • The integral of $-14x^2$ is $-\frac{14}{3}x^3$,
   • The integral of $x^3$ is $\frac{x^4}{4}$.

   So, the volume is: $V = 2\pi \left[ 24x^2 - \frac{14}{3}x^3 + \frac{x^4}{4} \right]_{0}^{6}$

5. Substitute the limits: First, plug in $x = 6$:

   = 24(36) - \frac{14}{3}(216) + \frac{1296}{4} = 864 - 1008 + 324 = 180$$ When $$x = 0$$, all terms vanish. Therefore, the volume is: $$V = 2\pi \times 180 = 360\pi$$

Final Answer:

The volume of the solid is $\boxed{360\pi}$ cubic units.

Would you like further clarification on any part of the solution? Here are some related questions you can explore:

1. How does the shell method compare to the disk/washer method in this type of problem?
2. What if the line of rotation were $x = -8$? How would the setup change?
3. How do you find the volume when the region is rotated about the y-axis using the shell method?
4. What if the parabola had a different equation, say $y = 5x - x^2$?
5. How does the choice of the axis of rotation affect the resulting solid's shape?

Tip: When using the shell method, the radius is always the distance from the axis of rotation to the representative shell. Keep this in mind when setting up the problem!