The given problem involves sketching the polar curve defined by $r = 3 + 2\cos(\theta)$, and then finding the area enclosed by this curve.

Step 1: Understanding the Polar Equation

The polar equation given is: $r = 3 + 2\cos(\theta).$ This type of equation represents a limaçon with an inner loop, as the coefficient in front of the cosine term (2) is smaller than the constant term (3).

Step 2: Plotting the Curve

The curve exhibits symmetry about the polar axis ($x$-axis), as $\cos(\theta)$ is an even function. As $\theta$ varies from $0$ to $2\pi$, $r$ traces out the limaçon.

Step 3: Calculating the Enclosed Area

The area $A$ enclosed by a polar curve $r(\theta)$ from $\theta = a$ to $\theta = b$ is given by: $A = \frac{1}{2} \int_a^b r^2 \, d\theta.$ For a complete loop, we set $a = 0$ and $b = 2\pi$, thus: $A = \frac{1}{2} \int_0^{2\pi} (3 + 2\cos(\theta))^2 \, d\theta.$

Step 4: Expanding $(3 + 2\cos(\theta))^2$

Expanding the expression: $(3 + 2\cos(\theta))^2 = 9 + 12\cos(\theta) + 4\cos^2(\theta).$ We can integrate each term separately: $A = \frac{1}{2} \int_0^{2\pi} \left(9 + 12\cos(\theta) + 4\cos^2(\theta)\right) \, d\theta.$

Step 5: Integrating Each Term

1. $\int_0^{2\pi} 9 \, d\theta = 9(2\pi) = 18\pi$.
2. $\int_0^{2\pi} 12\cos(\theta) \, d\theta = 0$ (since the integral of $\cos(\theta)$ over a full period is zero).
3. For $\int_0^{2\pi} 4\cos^2(\theta) \, d\theta$, use the identity $\cos^2(\theta) = \frac{1 + \cos(2\theta)}{2}$: $\int_0^{2\pi} 4\cos^2(\theta) \, d\theta = 4 \int_0^{2\pi} \frac{1 + \cos(2\theta)}{2} \, d\theta = 2 \int_0^{2\pi} 1 \, d\theta + 2 \int_0^{2\pi} \cos(2\theta) \, d\theta.$ This simplifies to: $2(2\pi) + 0 = 4\pi.$

Step 6: Adding the Results

Thus, the total area is: $A = \frac{1}{2} (18\pi + 4\pi) = \frac{1}{2} \times 22\pi = 11\pi.$

The area enclosed by the curve is $11\pi$ square units.

Do you have any questions or would you like further details?

Here are five related questions to consider:

1. What happens to the area if the equation changes to $r = 3 + 4\cos(\theta)$?
2. How would you find the area enclosed by a polar curve if the integration bounds are not from $0$ to $2\pi$?
3. What is the significance of the coefficients in the equation $r = a + b\cos(\theta)$?
4. How do different trigonometric identities simplify polar integrals?
5. Can the equation $r = 3 + 2\cos(\theta)$ be converted into Cartesian coordinates?

Tip: When integrating polar equations, remember that $r$ is squared in the area formula. This accounts for the radial distance's effect on the area calculation.