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.