To find the volume of the solid formed by rotating the region bounded by the curves $y = x^3$, $x = 0$, and $y = 8$ around the x-axis, we will use the method of disks (also known as the disk method).

Step-by-step approach:

1. Identify the bounds:

   • The curve $y = x^3$ intersects the line $y = 8$ when $x^3 = 8$, which gives $x = 2$.
   • Therefore, the region is bounded by $x = 0$ on the left, $x = 2$ on the right, and the curve $y = x^3$ from below, with the line $y = 8$ from above.

2. Set up the volume integral using the disk method: The formula for the volume of a solid of revolution around the x-axis is:

   $V = \pi \int_{a}^{b} \left( R(x) \right)^2 \, dx$

   where $R(x)$ is the radius of the disk at a given $x$. In this case:

   • The radius is the distance from the x-axis to the curve, which is given by the difference between the line $y = 8$ and the curve $y = x^3$.
   • Thus, $R(x) = 8 - x^3$.

3. Write the volume integral: The bounds are from $x = 0$ to $x = 2$, so the volume integral becomes:

   $V = \pi \int_{0}^{2} \left( 8 - x^3 \right)^2 \, dx$

4. Expand the integrand:

   $(8 - x^3)^2 = 64 - 16x^3 + x^6$

   Therefore, the integral becomes:

   $V = \pi \int_{0}^{2} (64 - 16x^3 + x^6) \, dx$

5. Integrate each term:

   $\int_{0}^{2} 64 \, dx = 64x \Big|_0^2 = 64(2) - 64(0) = 128$

   $\int_{0}^{2} -16x^3 \, dx = -16 \left( \frac{x^4}{4} \right) \Big|_0^2 = -16 \times \frac{2^4}{4} = -16 \times 4 = -64$

   $\int_{0}^{2} x^6 \, dx = \frac{x^7}{7} \Big|_0^2 = \frac{2^7}{7} = \frac{128}{7}$

6. Combine the results:

   $V = \pi \left( 128 - 64 + \frac{128}{7} \right)$

   Simplifying:

   $V = \pi \left( 64 + \frac{128}{7} \right) = \pi \left( \frac{448}{7} + \frac{128}{7} \right) = \pi \left( \frac{576}{7} \right)$

7. Final answer:

   $V = \frac{576\pi}{7}$

Thus, the volume of the solid is $\frac{576\pi}{7}$ cubic units.

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Would you like further clarification or have any questions about the steps? Here are five related questions for deeper understanding:

1. How would the volume change if the region were rotated around the y-axis instead?
2. What would the result be if the upper boundary were $y = 4$ instead of $y = 8$?
3. How do you set up a similar problem for a different curve, such as $y = x^2$?
4. What is the physical interpretation of the disk method in real-world scenarios?
5. How does the method of washers differ from the method of disks in finding volumes?

Tip: When using the disk method, always check that the radius function reflects the distance from the axis of rotation to the curve.