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The problem involves using the **washer method** to find the volume of a solid generated by revolving the shaded region around the x-axis. Let's break it down step by step.

The graph shows two curves:
- The upper curve: \( y = 1 \) (a horizontal line).
- The lower curve: \( y = \cos(x) \).

We are asked to find the volume of the region between these two curves, which is revolved around the x-axis, bounded between \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \).

### Washer Method Formula:
The general formula for the volume \( V \) using the washer method is:
\[
V = \pi \int_{a}^{b} \left[ R_{\text{outer}}(x)^2 - R_{\text{inner}}(x)^2 \right] \, dx
\]
Where:
- \( R_{\text{outer}}(x) \) is the distance from the x-axis to the outer curve.
- \( R_{\text{inner}}(x) \) is the distance from the x-axis to the inner curve.

### Step-by-Step Solution:

1. **Outer Radius \( R_{\text{outer}}(x) \)**:
   The outer curve is the line \( y = 1 \), so the outer radius is always 1.
   \[
   R_{\text{outer}}(x) = 1
   \]

2. **Inner Radius \( R_{\text{inner}}(x) \)**:
   The inner curve is \( y = \cos(x) \), so the inner radius is given by \( R_{\text{inner}}(x) = \cos(x) \).

3. **Bounds**:
   The region is bounded between \( x = -\frac{\pi}{2} \) and \( x = \frac{\pi}{2} \).

4. **Volume Integral**:
   Now, applying the washer method formula:
   \[
   V = \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[ 1^2 - \cos(x)^2 \right] \, dx
   \]
   Simplify the integrand:
   \[
   V = \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( 1 - \cos^2(x) \right) \, dx
   \]
   Use the identity \( \sin^2(x) = 1 - \cos^2(x) \):
   \[
   V = \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2(x) \, dx
   \]
   
5. **Solve the Integral**:
   We can use the half-angle identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \):
   \[
   V = \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 - \cos(2x)}{2} \, dx
   \]
   This simplifies to:
   \[
   V = \frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( 1 - \cos(2x) \right) \, dx
   \]
   Now, we can integrate term by term:
   \[
   V = \frac{\pi}{2} \left[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(2x) \, dx \right]
   \]
   The first integral is straightforward:
   \[
   \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 \, dx = \pi
   \]
   The second integral involves \( \cos(2x) \), and its integral is:
   \[
   \int \cos(2x) \, dx = \frac{\sin(2x)}{2}
   \]
   Evaluating this at the bounds:
   \[
   \left[ \frac{\sin(2x)}{2} \right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 0
   \]
   Thus, the volume is:
   \[
   V = \frac{\pi}{2} \times \pi = \frac{\pi^2}{2}
   \]

### Final Answer:
\[
V = \frac{\pi^2}{2}
\]

Would you like further details or have any questions?

### Related Questions:
1. How is the washer method different from the disk method for finding volumes?
2. What would change if the region were revolved around the y-axis instead?
3. How can you use the shell method to solve this problem?
4. What if the region were bounded by \( y = 2 \) instead of \( y = 1 \)?
5. How would you handle the problem if the lower curve was \( y = \sin(x) \) instead of \( y = \cos(x) \)?

### Tip:
Always double-check the limits of integration when setting up a volume problem, as incorrect bounds are a common mistake.

Use the washer method to find the volume of the solid generated by revolving the shaded region about the x-axis.

Calculate the volume of a solid generated by revolving the shaded region between y = cos(x) and y = 1 around the x-axis using the washer method. This problem covers integral calculus and trigonometric identities, suitable for college-level calculus students.

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