To solve the problem of finding the volume of the solid generated by revolving the plane region bounded by the curve $y = x$, the x-axis, and a given interval around the x-axis using the shell method, follow these steps:

Shell Method Setup

The shell method involves integrating cylindrical shells along the axis perpendicular to the axis of revolution (here, the x-axis). For the shell method, the formula for volume is:

$V = 2\pi \int_a^b \text{radius} \times \text{height} \, dx$

In this case:

• The radius of the shell is the distance from the x-axis, which is $y = x$.
• The height of the shell is $x$ because the function $y = x$ defines the curve.

Since we are revolving around the x-axis, we need to set up the integral using the y-values. Therefore, we solve for $x$ in terms of $y$, and our shell method integral becomes:

$V = 2\pi \int_a^b y \cdot x \, dy$

We need limits of integration $a$ and $b$, which correspond to the region you are rotating. Let's assume we are revolving the region from $y = 0$ to $y = 1$ for simplicity.

Step-by-Step Integration

1. Convert the function: Since $y = x$, the radius and height of the shells are $y$ and $x = y$, respectively.

2. Set up the integral: $V = 2\pi \int_0^1 y \cdot y \, dy = 2\pi \int_0^1 y^2 \, dy$

3. Integrate: $V = 2\pi \left[ \frac{y^3}{3} \right]_0^1 = 2\pi \left( \frac{1^3}{3} - \frac{0^3}{3} \right) = 2\pi \cdot \frac{1}{3}$

4. Evaluate: $V = \frac{2\pi}{3}$

Final Answer

The volume of the solid generated by revolving the region bounded by $y = x$ and the x-axis around the x-axis is $\frac{2\pi}{3}$.

Would you like more details, or have any further questions?

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Here are five related questions for further exploration:

1. How would the result change if the region was rotated around the y-axis instead?
2. What is the volume if the region $y = x$ is rotated around the line $y = 2$?
3. How does the disk method compare to the shell method for this problem?
4. Can you set up and solve the same volume problem using polar coordinates?
5. What if the curve was $y = x^2$ instead of $y = x$? How would the volume change?

Tip: Always check the axis of revolution carefully, as it can change the method you use and the form of the integral!