Problem 2 - Advanced Engineering Mathematics

Question

In Problems, use a double integral in polar coordinates to find the area of the region bounded by the graphs of the given polar equations. $$ r \quad 2+\cos \theta $$

Step-by-Step Solution

Answer

The area is \( 9\pi \).

Step 1: Determine the Region

The polar equation given is \( r = 2 + \cos \theta \). This equation describes a limaçon with an inner loop. We need to determine the range of \( \theta \) for which this curve is traced. The inner loop occurs where \( 2 + \cos \theta = 0 \), solving gives \( \cos \theta = -2 \), which is not possible. Therefore, the inner loop does not exist, and \( \theta \) ranges from \( 0 \) to \( 2\pi \).

Step 2: Set Up the Double Integral

The area \( A \) of the region in polar coordinates is given by the integral \( A = \int_0^{2\pi} \int_0^{2+\cos \theta} r \, dr \, d\theta \). This integral accounts for all points \( r, \theta \) in the limaçon from \( 0 \) to \( 2+\cos \theta \).

Step 3: Evaluate the Inner Integral

First, evaluate the inner integral with respect to \( r \): \( \int_0^{2+\cos \theta} r \, dr = \left[ \frac{1}{2}r^2 \right]_0^{2+\cos \theta} = \frac{1}{2}(2+\cos \theta)^2 \).

Step 4: Evaluate the Outer Integral

Now, evaluate the outer integral with respect to \( \theta \): \( A = \int_0^{2\pi} \frac{1}{2}(2+\cos \theta)^2 \, d\theta \). Expand \( (2+\cos \theta)^2 \) to \( 4 + 4\cos \theta + \cos^2 \theta \).

Step 5: Simplify and Integrate

The integral becomes \( A = \frac{1}{2} \int_0^{2\pi} (4 + 4\cos \theta + \cos^2 \theta) \, d\theta \). Split it as three integrals: \( 2\int_0^{2\pi} 4 \, d\theta + 2\int_0^{2\pi} 4\cos \theta \, d\theta + \frac{1}{2}\int_0^{2\pi} (1 + \cos 2\theta) \, d\theta \).\( 4\cos \theta \) integrates to zero over the period \( 0 \) to \( 2\pi \).

Step 6: Final Integration

Evaluate \( 2\int_0^{2\pi} 4 \, d\theta = 16\pi \) and for \( \frac{1}{2}\int_0^{2\pi} (1 + \cos 2\theta) \), substitute: \( \frac{1}{2}(2\pi + 0) = \pi \). Hence, \( A = 8\pi + \pi = 9\pi \).

Key Concepts

Limaçon

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.

Area 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.

Integral Evaluation

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.