To find the area of the outer loop of the polar curve \( r = 3 + 6\cos\theta \), let's go through the steps to set up and evaluate the integral:

### Step 1: Identify the points of intersection
The given polar equation, \( r = 3 + 6\cos\theta \), represents a limaçon with an inner loop because the coefficient of \( \cos\theta \) (6) is greater than the constant term (3).

To find the points of intersection between the inner and outer loops, set \( r = 0 \):
\[
3 + 6\cos\theta = 0
\]
Solving for \( \cos\theta \), we get:
\[
\cos\theta = -\frac{1}{2}
\]
This occurs at \( \theta = \frac{2\pi}{3} \) and \( \theta = \frac{4\pi}{3} \).

### Step 2: Set up the area integral
The area \( A \) enclosed by the outer loop is given by the integral:
\[
A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta
\]
For the outer loop, we integrate \( r^2 \) from \( \theta = \frac{2\pi}{3} \) to \( \theta = \frac{4\pi}{3} \), since these angles mark the points where \( r \) crosses zero.

So,
\[
A = \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} (3 + 6\cos\theta)^2 \, d\theta
\]

### Step 3: Expand \( (3 + 6\cos\theta)^2 \)
Expanding \( (3 + 6\cos\theta)^2 \) gives:
\[
(3 + 6\cos\theta)^2 = 9 + 36\cos\theta + 36\cos^2\theta
\]

### Step 4: Substitute and simplify using trigonometric identities
Using the identity \( \cos^2\theta = \frac{1 + \cos(2\theta)}{2} \):
\[
36\cos^2\theta = 18 + 18\cos(2\theta)
\]

Substituting into the expression, we get:
\[
(3 + 6\cos\theta)^2 = 9 + 36\cos\theta + 18 + 18\cos(2\theta) = 27 + 36\cos\theta + 18\cos(2\theta)
\]

### Step 5: Integrate term by term
Now, the area \( A \) becomes:
\[
A = \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \left( 27 + 36\cos\theta + 18\cos(2\theta) \right) d\theta
\]
Evaluating each term:

1. **Integral of \( 27 \):**
   \[
   \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} 27 \, d\theta = 27 \cdot \left( \frac{4\pi}{3} - \frac{2\pi}{3} \right) = 27 \cdot \frac{2\pi}{3} = 18\pi
   \]

2. **Integral of \( 36\cos\theta \):**
   \[
   \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} 36\cos\theta \, d\theta = 36 \cdot 0 = 0
   \]
   (The integral of \( \cos\theta \) over a symmetric interval around \( \pi \) is zero.)

3. **Integral of \( 18\cos(2\theta) \):**
   \[
   \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} 18\cos(2\theta) \, d\theta = 18 \cdot 0 = 0
   \]
   (Similarly, the integral of \( \cos(2\theta) \) over this interval is also zero.)

Thus, the area \( A \) is simply:
\[
A = \frac{1}{2} \cdot 18\pi = 9\pi
\]

### Final Answer
The area of the outer loop is:
\[
\boxed{9\pi}
\]

Would you like further details, or do you have any questions?

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**Related Questions:**

1. How would you find the area of the inner loop for this limaçon?
2. What happens to the loop areas if we change the equation to \( r = 3 + 3\cos\theta \)?
3. How can you determine the maximum radius for \( r = 3 + 6\cos\theta \)?
4. What is the geometric interpretation of limaçons in polar coordinates?
5. How does a change in the cosine term's coefficient affect the shape?

**Tip:** In polar coordinates, to find areas bounded by curves, integrate \( \frac{1}{2} r^2 \) over the interval for \( \theta \) that captures the desired section.

Calculate the area of the outer loop of the polar curve r = 3 + 6cosθ using integration. This solution includes key steps like identifying intersection points, setting up the integral, expanding expressions, and applying trigonometric identities. Ideal for high school students in Grades 11-12.

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