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The given problem asks to find the volume of the solid generated when the region bounded by the curves:

\[
y = x - x^4 \quad \text{and} \quad y = 0
\]

is revolved about the x-axis.

### Step 1: Identify the boundaries
The curve \( y = x - x^4 \) intersects the x-axis (where \( y = 0 \)) at \( x = 0 \) and \( x = 1 \). This can be verified by solving \( x - x^4 = 0 \):

\[
x(1 - x^3) = 0 \quad \Rightarrow \quad x = 0 \quad \text{or} \quad x = 1
\]

Thus, the region is bounded between \( x = 0 \) and \( x = 1 \).

### Step 2: Use the Disk Method
Since we are revolving the region about the x-axis, we can use the disk method to calculate the volume. The volume \( V \) of the solid is given by:

\[
V = \pi \int_{a}^{b} [f(x)]^2 \, dx
\]

where \( f(x) = x - x^4 \), and the limits of integration are from \( x = 0 \) to \( x = 1 \).

### Step 3: Set up the integral
The integral for the volume becomes:

\[
V = \pi \int_{0}^{1} (x - x^4)^2 \, dx
\]

### Step 4: Expand the integrand
First, expand \( (x - x^4)^2 \):

\[
(x - x^4)^2 = x^2 - 2x^5 + x^8
\]

Now the integral becomes:

\[
V = \pi \int_{0}^{1} (x^2 - 2x^5 + x^8) \, dx
\]

### Step 5: Integrate term by term
Now, integrate each term separately:

\[
\int x^2 \, dx = \frac{x^3}{3}, \quad \int x^5 \, dx = \frac{x^6}{6}, \quad \int x^8 \, dx = \frac{x^9}{9}
\]

So the volume integral becomes:

\[
V = \pi \left[ \frac{x^3}{3} - \frac{2x^6}{6} + \frac{x^9}{9} \right]_{0}^{1}
\]

### Step 6: Evaluate the definite integral
At \( x = 1 \):

\[
\frac{1^3}{3} - \frac{2 \times 1^6}{6} + \frac{1^9}{9} = \frac{1}{3} - \frac{2}{6} + \frac{1}{9} = \frac{1}{3} - \frac{1}{3} + \frac{1}{9} = \frac{1}{9}
\]

At \( x = 0 \), all terms are 0.

Thus, the volume is:

\[
V = \pi \times \frac{1}{9} = \frac{\pi}{9}
\]

### Final Answer:
The volume of the solid is:

\[
V = \frac{\pi}{9}
\]

---

Would you like more details or explanations on any of the steps?

Here are five related questions you might find useful:

1. How does the disk method differ from the shell method for finding volumes?
2. What is the significance of expanding \( (x - x^4)^2 \) in the integral?
3. Can this volume be computed using cylindrical shells? If yes, how?
4. How would the volume calculation change if the region was revolved around the y-axis?
5. What are some real-world applications of computing volumes by revolution?

**Tip:** For curves that are symmetric around the axis of revolution, you can often simplify the integral by considering only one half of the region.

The region bounded by the curves y = x - x^4 and y = 0 is revolved about the x-axis. Find the volume of the resulting solid.

Learn how to calculate the volume of a solid generated by revolving the region bounded by the curves y = x - x^4 and y = 0 about the x-axis using the Disk Method. Includes detailed steps and key calculus concepts, suitable for advanced high school or college-level students.

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