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Part 1
Find the area of the following region.
The region common to the circles requals=negative 2 cosine theta−2cosθ and requals=11.

The problem involves finding the area of the region common to two polar curves:

1. \( r = -2\cos(\theta) \)
2. \( r = 11 \)

Let's break down the process of solving this.

### 1. Understanding the curves:
- The first equation, \( r = -2\cos(\theta) \), represents a circle in polar coordinates. In rectangular coordinates, this is a circle with center at \( (1, 0) \) and radius 1. This can be deduced by converting it to Cartesian form, or recognizing the standard form of polar circles of the type \( r = a + b\cos(\theta) \).
  
- The second equation, \( r = 11 \), is a circle with center at the origin and radius 11. In polar coordinates, \( r = 11 \) is a constant radial distance from the origin.

### 2. Identifying the region of interest:
We are interested in the area common to these two circles, which means finding the area where they overlap.

### 3. Determine the limits of integration:
We need to find the points of intersection between the two curves. This occurs when the radial distances are equal, i.e., when:
\[
-2\cos(\theta) = 11
\]
Solving this equation for \( \theta \):

\[
\cos(\theta) = -\frac{11}{2}
\]
This equation has no real solution because the cosine function ranges between \(-1\) and \(1\), and \(-\frac{11}{2} = -5.5\) is outside this range. This means the two curves do not intersect directly. 

Therefore, the area of the common region will be the portion of the smaller circle that lies inside the larger circle.

### 4. Area Calculation:
Since the two circles do not intersect, the smaller circle is entirely inside the larger circle. The area of the common region is just the area of the smaller circle, \( r = -2\cos(\theta) \). The negative sign in the equation does not affect the area.

For a polar curve, the area is given by:
\[
A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta
\]

Here, the limits for \( \theta \) range from \( 0 \) to \( 2\pi \), and \( r = -2\cos(\theta) \), so the area is:

\[
A = \frac{1}{2} \int_0^{2\pi} (-2\cos(\theta))^2 \, d\theta
\]

Simplifying:

\[
A = \frac{1}{2} \int_0^{2\pi} 4\cos^2(\theta) \, d\theta
\]

Now, use the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \) to simplify the integral:

\[
A = 2 \int_0^{2\pi} \frac{1 + \cos(2\theta)}{2} \, d\theta
\]

This becomes:

\[
A = 2 \left( \int_0^{2\pi} \frac{1}{2} \, d\theta + \int_0^{2\pi} \frac{\cos(2\theta)}{2} \, d\theta \right)
\]

The second term integrates to zero because \( \int_0^{2\pi} \cos(2\theta) \, d\theta = 0 \), so:

\[
A = 2 \times \int_0^{2\pi} \frac{1}{2} \, d\theta = 2 \times \left[ \frac{\theta}{2} \right]_0^{2\pi} = 2 \times \frac{2\pi}{2} = 2\pi
\]

### 5. Conclusion:
The area of the region common to the two circles is \( \boxed{2\pi} \).

---

Would you like to explore details of this problem, or do you have any further questions?

Here are some follow-up questions to extend your understanding:

1. How does the polar curve \( r = a + b\cos(\theta) \) generally describe a circle?
2. What changes occur if the second curve is shifted away from the origin?
3. How can we visualize the area common to two intersecting circles in Cartesian coordinates?
4. What is the significance of the cosine identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \) in polar integration?
5. How can we use polar coordinates to find areas involving ellipses and parabolas?

**Tip:** When solving problems with polar curves, converting them to Cartesian coordinates can sometimes simplify identifying geometric shapes like circles or ellipses.

Find the area of the following region. The region common to the circles r = -2cosθ and r = 11.

Learn how to find the area of the region common to two polar circles: r = -2cos(θ) and r = 11. This problem involves polar coordinates, trigonometric integration, and calculating the area of a polar curve. Suitable for students in Grades 10-12.

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