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

The Shell Method is useful when rotating around an axis that is *parallel* to the axis of integration, in this case, the x-axis. For the problem given, we have the equation \( y = 1 - x \), and we are revolving the region around the x-axis. 

### Step 1: Identify the region
The equation \( y = 1 - x \) is a straight line with y-intercept 1 and slope -1. The region of interest lies between \( y = 1 - x \), the x-axis (i.e., \( y = 0 \)), and the limits where the curve intersects the axes.

The line intersects the x-axis when \( y = 0 \), so:
\[
1 - x = 0 \quad \Rightarrow \quad x = 1
\]
Thus, the region is bounded between \( x = 0 \) and \( x = 1 \).

### Step 2: Set up the shell method formula
The Shell Method formula for rotating around the x-axis is:
\[
V = 2\pi \int_{a}^{b} \text{(radius)} \times \text{(height)} \, dx
\]

- **Radius**: The distance from a point \( x \) to the axis of rotation, which is simply \( y \) (since we're rotating around the x-axis).
- **Height**: The value of \( x \) at each point, which is \( x \).

Thus, for this problem, the radius is \( y = 1 - x \) and the height is \( x \).

### Step 3: Set up the integral
The bounds of integration are from \( x = 0 \) to \( x = 1 \), and the integral becomes:
\[
V = 2\pi \int_{0}^{1} (1 - x) x \, dx
\]

### Step 4: Evaluate the integral
Now, expand the integrand:
\[
V = 2\pi \int_{0}^{1} (x - x^2) \, dx
\]

Next, compute the integral:
\[
V = 2\pi \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{1}
\]

Evaluate the expression at the limits:
\[
V = 2\pi \left( \left( \frac{1^2}{2} - \frac{1^3}{3} \right) - \left( \frac{0^2}{2} - \frac{0^3}{3} \right) \right)
\]
\[
V = 2\pi \left( \frac{1}{2} - \frac{1}{3} \right)
\]

Simplify:
\[
V = 2\pi \left( \frac{3}{6} - \frac{2}{6} \right) = 2\pi \times \frac{1}{6} = \frac{\pi}{3}
\]

### Final Answer:
The volume of the solid generated by revolving the region around the x-axis is:
\[
V = \frac{\pi}{3}
\]

---

Would you like more details on any step? Here are 5 related questions for further exploration:

1. How would the volume change if the region was revolved around the y-axis instead?
2. Can the Disk/Washer Method be applied to solve this same problem?
3. How does the Shell Method differ from the Disk Method in terms of applications?
4. What is the general formula for the Shell Method, and when is it best applied?
5. How can you extend this method to more complex curves?

**Tip**: When using the Shell Method, always double-check the orientation of the axis of rotation to set up the radius and height correctly!

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

Learn how to calculate the volume of a solid generated by revolving the region defined by y = 1 − x around the x-axis using the Shell Method. This step-by-step solution involves setting up and evaluating a definite integral and is suitable for students in Grades 11-12 or college-level calculus.

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