To solve the problem using the shell method, we first analyze the given functions and the region they enclose. The three given functions are:

1. $x + y = 8$
2. $y = x$
3. $y = 0$ (the x-axis)

Step 1: Rearrange the Equations

For the shell method, we need to express $x$ in terms of $y$ because we will be revolving around the x-axis. The equations need to be rearranged as follows:

1. $x + y = 8$ → $x = 8 - y$
2. $y = x$ → $x = y$

Step 2: Set up the Integral

The shell method involves integrating cylindrical shells parallel to the axis of rotation. The formula for the volume of a solid generated by revolving a region about the x-axis using the shell method is:

$V = 2\pi \int_{a}^{b} y \cdot \left( \text{height of shell} \right) \, dy$

The height of each shell is the horizontal distance between the curves $x = 8 - y$ and $x = y$, so:

$\text{height} = (8 - y) - y = 8 - 2y$

The limits of integration are determined by the points where the two curves intersect. Setting $x = y$ and $x = 8 - y$, we solve for $y$:

$y = 8 - y \implies 2y = 8 \implies y = 4$

Thus, the region is bounded between $y = 0$ and $y = 4$.

Step 3: Write the Definite Integral

Now we can set up the definite integral:

$V = 2\pi \int_{0}^{4} y \cdot (8 - 2y) \, dy$

Step 4: Evaluate the Integral

Let's expand the integrand:

$y \cdot (8 - 2y) = 8y - 2y^2$

Now, integrate term by term:

$V = 2\pi \int_{0}^{4} (8y - 2y^2) \, dy$

$V = 2\pi \left[ \frac{8y^2}{2} - \frac{2y^3}{3} \right]_{0}^{4}$

$V = 2\pi \left[ 4y^2 - \frac{2y^3}{3} \right]_{0}^{4}$

Now, evaluate at the bounds:

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

$V = 2\pi \left[ 4(16) - \frac{2(64)}{3} \right]$

$V = 2\pi \left[ 64 - \frac{128}{3} \right]$

$V = 2\pi \left[ \frac{192}{3} - \frac{128}{3} \right] = 2\pi \left( \frac{64}{3} \right)$

$V = \frac{128\pi}{3}$

Final Answer

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

$V = \frac{128\pi}{3}$

Would you like a more detailed breakdown of any step?

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Here are 5 related questions for practice:

1. How would the setup change if the region were revolved around the y-axis?
2. What if the region was bounded by different lines, like $y = 2x$ instead of $y = x$?
3. Can you solve for the volume if the region was revolved around the line $y = -2$?
4. How would the washer method be used for this same problem?
5. What is the geometric interpretation of the result?

Tip: In the shell method, always ensure you're using the correct radius and height relative to the axis of rotation.