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.

*x* + *y* = 8,      *y* = *x*,      *y* = 0

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?

---

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.

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. x + y = 8, y = x, y = 0

Explore how to find the volume of a solid generated by revolving the region defined by x + y = 8, y = x, and y = 0 about the x-axis using the shell method. This detailed solution covers essential calculus concepts including setting up and evaluating definite integrals. Ideal for students in Grades 11-12 or college-level calculus courses.

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