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.