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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.

Sketch the curve for r = 3 + 2cos(θ) and find the area it encloses.

Sketch the limaçon curve defined by r = 3 + 2cos(θ) and calculate the area enclosed by this polar equation. This solution involves integrating the square of the polar function over a full cycle of θ, demonstrating key concepts in polar coordinates and calculus.

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