Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving the plane region about the x-axis.
y = 2 − x, The line starts at (2, 0) and ends at (6, −4).

To find the volume of the solid generated by revolving the plane region around the x-axis using the shell method, we need to express the volume integral in terms of cylindrical shells. The shell method involves integrating with respect to \(y\), and for a region revolved around the x-axis, the formula for the volume of a solid is:

\[
V = 2 \pi \int_{a}^{b} y \, (r(y)) \, (h(y)) \, dy
\]

where:
- \(y\) is the height from the axis of rotation (in this case, the x-axis).
- \(r(y)\) is the radius of the shell, which is simply the distance from the axis of rotation.
- \(h(y)\) is the height of the shell, which in this case is the length of the line segment in terms of \(y\).

Given the line \(y = 2 - x\), which starts at (2, 0) and ends at (6, -4), let's rewrite the function in terms of \(x\):

### Step 1: Rewrite the line equation in terms of \(y\)

Starting with:

\[
y = 2 - x \implies x = 2 - y
\]

We will integrate with respect to \(y\), and the line ranges from \(y = 0\) to \(y = -4\).

### Step 2: Set up the volume integral using the shell method

The radius of the shell \(r(y)\) is the distance from the x-axis, which is just \(|y|\). The height of the shell \(h(y)\) is the horizontal distance from \(x = 2\) to \(x = 6\), which in terms of \(y\), simplifies to:

\[
h(y) = x_{\text{right}} - x_{\text{left}} = (6 - (2 - y))
\]

So, we can express:

\[
h(y) = 4 + y
\]

### Step 3: Set up the definite integral

The volume \(V\) can now be represented as:

\[
V = 2 \pi \int_{-4}^{0} |y| \cdot (4 + y) \, dy
\]

Since \(y\) is negative in this interval, \(|y| = -y\). Therefore, the integral simplifies to:

\[
V = 2 \pi \int_{-4}^{0} (-y)(4 + y) \, dy
\]

Simplify further:

\[
V = 2 \pi \int_{-4}^{0} (-4y - y^2) \, dy
\]

### Step 4: Evaluate the integral

Now compute the integral:

\[
V = 2 \pi \left[ \int_{-4}^{0} (-4y - y^2) \, dy \right]
\]

Let's integrate term-by-term:

\[
\int (-4y - y^2) \, dy = \left( -2y^2 - \frac{y^3}{3} \right) + C
\]

Evaluating from \(y = -4\) to \(y = 0\):

\[
V = 2 \pi \left[ \left( -2(0)^2 - \frac{(0)^3}{3} \right) - \left( -2(-4)^2 - \frac{(-4)^3}{3} \right) \right]
\]

Simplify the expression:

\[
V = 2 \pi \left[ 0 - \left( -2(16) - \frac{-64}{3} \right) \right]
\]

\[
V = 2 \pi \left[ 0 - \left( -32 + \frac{64}{3} \right) \right]
\]

Combine like terms:

\[
V = 2 \pi \left[ - \left( -32 + \frac{64}{3} \right) \right] = 2 \pi \left[ \frac{-96 + 64}{3} \right]
\]

\[
V = 2 \pi \left( \frac{-32}{3} \right)
\]

\[
V = \frac{64 \pi}{3}
\]

### Conclusion

The volume of the solid generated by revolving the region about the x-axis is:

\[
V = \frac{64 \pi}{3}
\]

Would you like more details on any part of the process or have any specific questions? 

### Related Questions:
1. How do you use the disk method to find the volume for the same region?
2. Can we use the washer method for this problem, and how would it change?
3. What if the region were revolved around the y-axis instead of the x-axis?
4. How does the shell method differ from the disk method conceptually?
5. Can you apply the shell method to a different line, like \(y = x - 3\)?

**Tip:** Always check the limits and boundaries when using integral methods, as mistakes here can lead to incorrect solutions.

Use the shell method to write and evaluate the definite integral that represents the volume of the solid generated by revolving the plane region about the x-axis. y = 2 − x, The line starts at (2, 0) and ends at (6, −4).

This problem involves calculating the volume of a solid generated by revolving the region under the line y = 2 - x around the x-axis using the shell method. The solution includes setting up the integral, applying the shell method formula, and evaluating the definite integral. Suitable for students in Grades 11-12.

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