Sketch the following region and use a double integral to find the area.
The region inside both of the circles requals2 and requals4 cosine theta.\

To solve this problem, we will find the area of the region inside both circles using polar coordinates. The two given equations are:
1. \( r = 2 \) — a circle with radius 2 centered at the origin.
2. \( r = 4 \cos(\theta) \) — a circle with radius 2 centered at \( (2, 0) \).

### Step 1: Understand the region
The goal is to find the region common to both circles. To do this, we first need to analyze where these two curves intersect.

#### Intersection points:
Set the two equations equal to each other:
\[
2 = 4 \cos(\theta)
\]
Solve for \( \theta \):
\[
\cos(\theta) = \frac{1}{2}
\]
This implies that:
\[
\theta = \frac{\pi}{3}, \quad \theta = -\frac{\pi}{3}
\]
Thus, the curves intersect at \( \theta = \frac{\pi}{3} \) and \( \theta = -\frac{\pi}{3} \).

### Step 2: Set up the double integral
We'll compute the area of the region using a double integral in polar coordinates. The area in polar coordinates is given by:
\[
A = \int_{\theta_1}^{\theta_2} \int_{r_{\text{inner}}(\theta)}^{r_{\text{outer}}(\theta)} r \, dr \, d\theta
\]
Here:
- The limits for \( \theta \) are from \( -\frac{\pi}{3} \) to \( \frac{\pi}{3} \).
- The inner curve is \( r = 4 \cos(\theta) \).
- The outer curve is \( r = 2 \).

So, the area becomes:
\[
A = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \int_{4 \cos(\theta)}^{2} r \, dr \, d\theta
\]

### Step 3: Solve the double integral
1. First, integrate with respect to \( r \):
\[
\int_{4 \cos(\theta)}^{2} r \, dr = \left[ \frac{r^2}{2} \right]_{4 \cos(\theta)}^{2} = \frac{1}{2} \left( 2^2 - (4 \cos(\theta))^2 \right)
\]
\[
= \frac{1}{2} \left( 4 - 16 \cos^2(\theta) \right) = 2 - 8 \cos^2(\theta)
\]

2. Now, integrate with respect to \( \theta \):
\[
A = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left( 2 - 8 \cos^2(\theta) \right) d\theta
\]
The first term is straightforward:
\[
\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 2 \, d\theta = 2 \left( \frac{\pi}{3} - \left(-\frac{\pi}{3}\right) \right) = 2 \times \frac{2\pi}{3} = \frac{4\pi}{3}
\]
For the second term, we use the identity \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \):
\[
\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 8 \cos^2(\theta) \, d\theta = 8 \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{1 + \cos(2\theta)}{2} \, d\theta
\]
\[
= 4 \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 1 \, d\theta + 4 \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \cos(2\theta) \, d\theta
\]
The first integral is straightforward:
\[
4 \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 1 \, d\theta = 4 \times \frac{2\pi}{3} = \frac{8\pi}{3}
\]
The second integral is zero because \( \cos(2\theta) \) is symmetric and integrates to zero over symmetric limits:
\[
\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \cos(2\theta) \, d\theta = 0
\]

Thus, we have:
\[
8 \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \cos^2(\theta) \, d\theta = \frac{8\pi}{3}
\]

### Step 4: Final area
Putting everything together:
\[
A = \frac{4\pi}{3} - \frac{8\pi}{3} = \boxed{0}
\]

This indicates an error in the selection of limits or the geometry. Instead, visualize the region to refine

Sketch the following region and use a double integral to find the area. The region inside both of the circles r = 2 and r = 4 cos(θ).

This problem involves finding the area of the region inside two polar curves: r = 2 and r = 4 cos(θ). Using double integrals in polar coordinates, we walk through the intersection points, set up the integral, and calculate the area step-by-step. Suitable for students in advanced high school or college-level calculus.

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