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 = 1 − 

x
,      y = x + 1,      y = 0

To find the volume of the solid generated by revolving the given plane region about the x-axis using the shell method, we need to set up the integral for the volume and then evaluate it.

### Step 1: Identify the bounds and region
The given equations are:
- \( y = 1 - x \)
- \( y = x + 1 \)
- \( y = 0 \) (the x-axis)

### Step 2: Determine the points of intersection
To find the limits of integration, we need to determine the points where the curves intersect.

1. Set \( 1 - x = x + 1 \) to find the intersection points.
   \[
   1 - x = x + 1
   \]
   \[
   1 - 1 = 2x
   \]
   \[
   0 = 2x
   \]
   \[
   x = 0
   \]

Thus, the two lines intersect at \( x = 0 \).

### Step 3: Set up the shell method integral
The shell method formula for volume when rotating around the x-axis is:
\[
V = 2\pi \int_{a}^{b} y \cdot \text{(radius)} \, dy
\]

Where:
- \( y \) is the height of the shell.
- The radius is the distance from the x-axis to the curve (in terms of \( y \)).

Since we're revolving around the x-axis, we solve each curve for \( x \) in terms of \( y \):

1. For \( y = 1 - x \), solve for \( x \):
   \[
   x = 1 - y
   \]
2. For \( y = x + 1 \), solve for \( x \):
   \[
   x = y - 1
   \]

### Step 4: Evaluate the integral
The limits of integration are from \( y = 0 \) to \( y = 1 \) because the region of interest lies between these two values.

The volume \( V \) is given by:
\[
V = 2\pi \int_{0}^{1} \left[(1 - y) - (y - 1)\right] y \, dy
\]
Simplify the integrand:
\[
V = 2\pi \int_{0}^{1} \left[2 - 2y\right] y \, dy
\]
Distribute \( y \):
\[
V = 2\pi \int_{0}^{1} \left(2y - 2y^2\right) \, dy
\]

### Step 5: Integrate and calculate
Now integrate term by term:
\[
V = 2\pi \left[\int_{0}^{1} 2y \, dy - \int_{0}^{1} 2y^2 \, dy\right]
\]
\[
V = 2\pi \left[\left. y^2 \right|_0^1 - \left. \frac{2y^3}{3} \right|_0^1 \right]
\]
Evaluate the integrals:
\[
V = 2\pi \left[1^2 - 0^2 - \frac{2(1)^3}{3} + \frac{2(0)^3}{3}\right]
\]
\[
V = 2\pi \left[1 - \frac{2}{3}\right]
\]
\[
V = 2\pi \cdot \frac{1}{3}
\]
\[
V = \frac{2\pi}{3}
\]

### Final Answer
The volume of the solid generated by revolving the region about the x-axis is \( \frac{2\pi}{3} \) cubic units.

Would you like more details or have any questions?

### Related Questions:
1. How would the integral change if the region were revolved around the y-axis instead?
2. What if the line \( y = 1 - x \) were replaced with \( y = 2 - x \)?
3. How does the shell method compare with the disk method for this problem?
4. What would be the volume if the region was revolved around the line \( y = -1 \)?
5. How can you set up the integral using the disk/washer method instead?

### Tip:
When setting up integrals for volumes of revolution, carefully determine whether to use the shell method or the disk/washer method based on the axis of rotation and the function given.

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 = 1 − x, y = x + 1, y = 0

Learn how to calculate the volume of a solid generated by revolving the region bounded by y = 1 - x, y = x + 1, and y = 0 about the x-axis using the shell method. This step-by-step guide covers integral setup, solving for intersections, and evaluating the definite integral. Suitable for college-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation