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.