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The term "limaçon" refers to a specific type of curve that can be plotted in polar coordinates. A limaçon has the general form \[ r = a + b \cos \theta \] or \[ r = a + b \sin \theta \] where \( a \) and \( b \) are constants. Depending on the values of \( a \) and \( b \), the limaçon can appear with various distinctive shapes, including those with an inner loop, a dimple, or a cardioid shape.

• If \( |b| < |a| \), it's a limaçon with a dimple.
• If \( a = b \), it becomes a cardioid, which is heart-shaped.
• If \( |b| > |a| \), it presents an inner loop.

In our problem, the given polar equation is \[ r = 2 + \cos \theta \].Here, \( a = 2 \) and \( b = 1 \). Since \( |b| < |a| \), this indicates that the graph is a dimpled limaçon without an inner loop. The curve is traced completely as \( \theta\) ranges from \( 0 \) to \( 2\pi \). Understanding these parameters helps in visualizing and solving area problems in polar coordinates.

Finding the area enclosed by a curve described in polar coordinates requires using a specific formula derived from calculus. For the curve \[ r = f(\theta) \],the area \( A \) can be calculated with a double integral\[ A = \int_{\alpha}^{\beta} \int_{0}^{f(\theta)} r \, dr \, d\theta \].Here, \( \alpha \) to \( \beta \) define the range of \( \theta \) over which the curve is traced.

• The function \( f(\theta) \) provides the radial distance from the origin.
• The inner integral \( \int_0^{f(\theta)} r \, dr \) calculates the area of small radial segments for each \( \theta \).
• Integrating over \( \theta \) then sums all these segments to provide the total area.

For our limaçon curve, the area is given by \[ A = \int_0^{2\pi} \int_0^{2+\cos \theta} r \, dr \, d\theta \],which integrates the radial component \( r \) first, and then accumulates these values by varying \( \theta \) from \( 0 \) to \( 2\pi \). This method is powerful for handling more complex curves in polar systems.

Solving the double integral involves a series of steps to ensure the correct area is calculated. For polar coordinates, let's break down the integral \( A = \int_0^{2\pi} \int_0^{2+\cos \theta} r \, dr \, d\theta \):

1. **Evaluate the Inner Integral:** - Integrate \( \int_0^{2+\cos \theta} r \, dr \), effectively finding half the sum of squares: \[ \frac{1}{2}r^2 \] evaluated from \( 0 \) to \( 2+\cos \theta \) gives \( \frac{1}{2}(2+\cos \theta)^2 \).2. **Expand and Simplify:** - Expand \( (2+\cos \theta)^2 \) into \( 4 + 4\cos \theta + \cos^2 \theta \). - Simplify by noting that some trigonometric integrals will evaluate over their period of integration.3. **Evaluate the Outer Integral:** - The integral becomes a sum of parts: \( \frac{1}{2} \int_0^{2\pi} (4 + 4\cos \theta + \cos^2 \theta) \, d\theta \). - Breaking this down:

• \( 2\int_0^{2\pi} 4 \, d\theta = 16\pi \)
• \( 2\int_0^{2\pi} 4\cos \theta \, d\theta = 0 \) since the cosine function averages to zero over one period.
• The trickiest part involves \( \cos^2 \theta \): Use the identity \( \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \) to integrate.
• The area approximation becomes: \( A = 8\pi + \pi = 9\pi \).

Understanding each step in evaluating these integrals in polar coordinates allows for precise calculation of areas bounded by complex curves. Keeps the integration organized and make use of identities to manage trigonometric components.