Question

Sketch the graph of the given equation and find the area of the region bounded by it. $$ r=2+\cos \theta $$

Short answer

The graph is a dimpled limaçon and the bounded area is \( \frac{9\pi}{2} \).

Step-by-step solution

Convert Polar Equation to Cartesian Form

The given polar equation is \( r = 2 + \cos \theta \). In polar coordinates, \( x = r \cos \theta \) and \( y = r \sin \theta \). Start by expressing \( r \cos \theta \) in terms of \( x \): \( r \cos \theta = x \). Thus, \( r = 2 + \frac{x}{r} \). Multiply both sides by \( r \) to isolate terms: \( r^2 = 2r + x \). Since \( r^2 = x^2 + y^2 \), we have \( x^2 + y^2 = 2\sqrt{x^2 + y^2} + x \).

Identify the Shape of the Graph

Recognize the form of the equation \( r = 2 + \cos \theta \), which indicates that the graph is a limaçon. This type of polar graph is known as a cardioid when the function takes the form \( r = a + a \cos \theta \). Here, the equation resembles a limaçon with a dimple, as it follows \( r = a + b \cos \theta \), where \( a = 2 \) and \( b = 1 \).

Determine Intersection Points

To find where the curve intersects itself or the origin, set \( r = 0 \). Solve \( 2 + \cos \theta = 0 \) to get \( \cos \theta = -2 \). Since \( \cos \theta \) ranges from -1 to 1, there are no values of \( \theta \) for which \( \cos \theta = -2 \). Therefore, the curve does not intersect the origin, but loops without touching or crossing the origin.

Graph the Polar Curve

Plot key points by evaluating \( \theta \) at common angles: \( \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \). For example, at \( \theta = 0 \), \( r = 2 + 1 = 3 \); at \( \theta = \pi \), \( r = 2 - 1 = 1 \). Sketch the cardioid, noting it has a dimple at \( \theta = \pi \). The shape is symmetric with the y-axis.

Calculate the Area Enclosed by the Curve

The area \( A \) enclosed by a polar curve \( r(\theta) \) can be calculated using the formula \[ A = \frac{1}{2} \int_{\alpha}^{\beta} [r(\theta)]^2 \, d\theta \]For \( r = 2 + \cos \theta \), integrate from \( \theta = 0 \) to \( \theta = 2\pi \):\[ A = \frac{1}{2} \int_{0}^{2\pi} (2 + \cos \theta)^2 \, d\theta \]Simplify and calculate the integral:\[ A = \frac{1}{2} \int_{0}^{2\pi} (4 + 4\cos \theta + \cos^2 \theta) \, d\theta \]Using trigonometric identities and completing the integral gives \( A = \frac{9\pi}{2} \).

Key concepts

Graphing Polar Equations

Graphing polar equations involves plotting points where each point is determined by a distance, \( r \), from the origin and an angle, \( \theta \), from the polar axis. Polar coordinates provide a unique way to represent curves that appear complex in Cartesian formats. This exercise uses the equation \( r = 2 + \cos \theta \) as an example.
To sketch the graph of \( r = 2 + \cos \theta \), recognize that \( \theta \) can take on values from \( 0 \) to \( 2\pi \). This creates a full cycle of the curve known as a limaçon, a type of spiral curve. Breaking it down by evaluating \( \theta \) at key angles like \( 0, \pi/2, \pi, \) and \( 3\pi/2 \) helps in plotting significant points. For instance, at \( \theta = 0 \), \( r \) equals 3, which places a point at a distance of 3 units along the polar axis. Working through additional angles reveals the full limaçon shape. Each point, determined by the angle and the calculated radius, guides the curve's overall structure.

Area of Polar Regions

Finding the area of polar regions involves integration, where the integral computes the region enclosed by the polar curve. In this exercise, the curve defined by \( r = 2 + \cos \theta \) creates a bounded area that can be calculated using a specialized polar area formula.
The formula used is \[A = \frac{1}{2} \int_{\alpha}^{\beta} [r(\theta)]^2 \, d\theta \]where \( \alpha \) and \( \beta \) are the lower and upper bounds for \( \theta \). For a complete traversal of the limaçon, integration occurs from \( 0 \) to \( 2\pi \). The equation becomes \[A = \frac{1}{2} \int_{0}^{2\pi} (2 + \cos \theta)^2 \, d\theta \]This integral accounts for the square of the curve as it rotates around the origin, calculating the space it encloses. Be sure to integrate each term separately and apply trigonometric identities where needed to simplify calculations.

Limaçon Curve

The limaçon curve is a fascinating type of polar graph that arises from equations of the form \( r = a + b\cos \theta \) or \( r = a + b \sin \theta \). This specific curve, \( r = 2 + \cos \theta \), is a limaçon with an inner dimple.
There are several variations of limaçons, depending on the values of \( a \) and \( b \):

• When \( a > b \), the curve forms a dimpled limaçon.
• When \( a = b \), it forms a cardioid.
• When \( a < b \), it forms a looped limaçon.

This curve exhibits symmetry due to its cosine foundation, which reflects across the vertical axis. The dimple occurs on the side where \( \theta \) equals \( \pi \) in this example, corresponding to the minimal \( r \) value.

Integral Calculus

Integral calculus is a cornerstone of mathematics, used here to determine the area of complex shapes like those described in polar coordinates. This approach calculates the accumulation of values over an interval, and is particularly useful for functions that describe curves.
For the limaçon curve represented by \( r = 2 + \cos \theta \), integral calculus computes the total area the curve bounds. Setting up the integral \[A = \frac{1}{2} \int_{0}^{2\pi} (2 + \cos \theta)^2 \, d\theta \]involves evaluating the space from the start angle to the end angle across one full period of the function.
Simplifying the expression inside the integral — using techniques like u-substitution, symmetry, and trigonometric identities — streamlines the process, making it more manageable. Solving the integral involves computing each term throughout the range and is a key example of using integral calculus to analyze and define areas of complex, bounded regions.