To solve this problem, we need to find the volume of the solid of revolution formed by rotating the region $R$ around the x-axis. The region $R$ is bounded by the parabola $y = x^2$ and the straight line $y = 1$.

Step 1: Set up the integral

Since the solid is formed by rotating around the x-axis, we can use the disk method to calculate the volume. The volume $V$ of the solid of revolution can be found by the following integral:

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

In this case, $f(x)$ represents the distance from the x-axis to the curve $y = x^2$, and the region is bounded by $y = 1$. So, we'll subtract the area under the parabola from the area under the line to get the area of the cross-section at each point.

Step 2: Determine the bounds

The region $R$ is where $x^2 \leq y \leq 1$. So, solving for $x$, we have:

$x^2 \leq 1 \quad \Rightarrow \quad -1 \leq x \leq 1$

So the bounds of integration are $x = -1$ to $x = 1$.

Step 3: Set up the integral with the functions

The cross-sectional area at each point is given by the difference in radii between the line and the parabola:

$\text{Cross-sectional area} = \pi \left[(1)^2 - (x^2)^2 \right]$

So, the integral to find the volume is:

$V = \pi \int_{-1}^{1} \left[1 - (x^4)\right] \, dx$

Step 4: Evaluate the integral

Let's compute the integral:

$V = \pi \int_{-1}^{1} \left(1 - x^4\right) \, dx$

Break this down into two separate integrals:

$V = \pi \left[\int_{-1}^{1} 1 \, dx - \int_{-1}^{1} x^4 \, dx\right]$

First, calculate the integral of 1:

$\int_{-1}^{1} 1 \, dx = x \bigg|_{-1}^{1} = 1 - (-1) = 2$

Now, calculate the integral of $x^4$:

$\int_{-1}^{1} x^4 \, dx = \left[\frac{x^5}{5}\right]_{-1}^{1} = \frac{1}{5} - \left(-\frac{1}{5}\right) = \frac{2}{5}$

Putting it all together:

$V = \pi \left(2 - \frac{2}{5}\right) = \pi \cdot \frac{10 - 2}{5} = \frac{8\pi}{5}$

Thus, the volume of the solid of revolution is:

$\boxed{\frac{8\pi}{5}}$

Would you like further details or have any questions about the solution?

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Related Questions:

1. How does the disk method differ from the shell method in calculating volumes of revolution?
2. What if the region $R$ was rotated around the y-axis instead? How would the method change?
3. Can you solve the same problem using the shell method?
4. What would happen if the boundaries were different, say between $y = x^2$ and $y = x^4$?
5. How do you find the volume if the region is rotated around a line other than the axes?
6. What if the line was $y = 2$ instead of $y = 1$? How would the volume change?
7. How do you calculate the surface area of the solid of revolution for the same region?
8. What are the potential applications of calculating volumes of solids of revolution in real life?

Tip: Always double-check the bounds of integration when setting up an integral for volumes of revolution, as they play a crucial role in ensuring the accuracy of your solution.