Find the area of the region enclosed by the given curves. y = 5 cos(x),  y = 5 − 5 cos(x),  0 ≤ x ≤ 𝜋

To find the area of the region enclosed by the curves \( y = 5 \cos(x) \) and \( y = 5 - 5 \cos(x) \) over the interval \( 0 \leq x \leq \pi \), we follow these steps:

### 1. **Determine the points of intersection**:
   The curves intersect where \( y = 5 \cos(x) \) equals \( y = 5 - 5 \cos(x) \). 

   \[
   5 \cos(x) = 5 - 5 \cos(x)
   \]
   
   Simplify and solve for \( \cos(x) \):
   
   \[
   10 \cos(x) = 5
   \]
   
   \[
   \cos(x) = \frac{1}{2}
   \]
   
   The solution to \( \cos(x) = \frac{1}{2} \) within the interval \( 0 \leq x \leq \pi \) is:
   
   \[
   x = \frac{\pi}{3}
   \]
   
### 2. **Set up the integral for the area**:
   The area \( A \) of the region enclosed by the curves is given by the integral of the difference between the top curve and the bottom curve over the interval from \( 0 \) to \( \pi \). The curves switch roles at \( x = \frac{\pi}{3} \), so we break the integral into two parts:

   \[
   A = \int_{0}^{\frac{\pi}{3}} \left( [5 - 5 \cos(x)] - [5 \cos(x)] \right) dx + \int_{\frac{\pi}{3}}^{\pi} \left( [5 \cos(x)] - [5 - 5 \cos(x)] \right) dx
   \]

### 3. **Simplify the integrals**:
   Simplifying the integrals:
   
   For the first integral:
   \[
   \int_{0}^{\frac{\pi}{3}} \left( 5 - 10 \cos(x) \right) dx
   \]
   
   For the second integral:
   \[
   \int_{\frac{\pi}{3}}^{\pi} \left( 10 \cos(x) - 5 \right) dx
   \]

### 4. **Compute the integrals**:
   Compute each integral separately.
   
   - For the first integral:
   
   \[
   \int_{0}^{\frac{\pi}{3}} \left( 5 - 10 \cos(x) \right) dx = \left[ 5x - 10 \sin(x) \right]_{0}^{\frac{\pi}{3}}
   \]
   
   Evaluate:
   \[
   = 5\left(\frac{\pi}{3}\right) - 10 \sin\left(\frac{\pi}{3}\right) - \left( 5(0) - 10\sin(0) \right)
   \]
   
   \[
   = \frac{5\pi}{3} - 10 \left(\frac{\sqrt{3}}{2}\right)
   \]
   
   \[
   = \frac{5\pi}{3} - 5\sqrt{3}
   \]
   
   - For the second integral:
   
   \[
   \int_{\frac{\pi}{3}}^{\pi} \left( 10 \cos(x) - 5 \right) dx = \left[ 10 \sin(x) - 5x \right]_{\frac{\pi}{3}}^{\pi}
   \]
   
   Evaluate:
   \[
   = \left(10 \sin(\pi) - 5\pi \right) - \left( 10 \sin\left(\frac{\pi}{3}\right) - 5\left(\frac{\pi}{3}\right)\right)
   \]
   
   \[
   = -5\pi - \left( 10\left(\frac{\sqrt{3}}{2}\right) - \frac{5\pi}{3} \right)
   \]
   
   \[
   = -5\pi - 5\sqrt{3} + \frac{5\pi}{3}
   \]
   
   \[
   = -\frac{10\pi}{3} - 5\sqrt{3}
   \]

### 5. **Add the results**:
   The total area is:
   
   \[
   A = \left(\frac{5\pi}{3} - 5\sqrt{3}\right) + \left(-\frac{10\pi}{3} - 5\sqrt{3}\right)
   \]
   
   \[
   A = -\frac{5\pi}{3} - 10\sqrt{3}
   \]

Since the area cannot be negative, take the absolute value:

\[
A = \frac{5\pi}{3} + 10\sqrt{3}
\]

Thus, the area of the region enclosed by the given curves is \( \frac{5\pi}{3} + 10\sqrt{3} \).

Would you like further details or have any questions?

Here are some related questions you might find interesting:

1. How do you determine if a region's area will be positive or negative?
2. What are the key steps in identifying the bounds of integration for enclosed regions?
3. Can the concept of area between curves be applied to functions in polar coordinates?
4. How does the behavior of trigonometric functions affect the shape of the enclosed region?
5. What happens if we extend the interval beyond \(0\) to \(\pi\)?

**Tip:** When finding areas between curves, always ensure the correct identification of which function is on top (greater value) over the interval of interest.

Discover how to calculate the area enclosed by the curves y = 5 cos(x) and y = 5 - 5 cos(x) over the interval 0 ≤ x ≤ π. This detailed solution involves calculus and trigonometry concepts, including integration of functions and determination of intersection points.

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