Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the area of the given region \(S\) by calculating \(\iint_{S} r d r d \theta .\) Be sure to make a sketch of the region first. \(S\) is the region inside the larger loop of the limaçon \(r=2-4 \sin \theta\)

Step-by-Step Solution

Answer

The area of the region is \( \frac{9\pi + 8\sqrt{3}}{2} \).

Step 1: Understand the Limaçon Equation

The given limaçon is defined by the polar equation \( r = 2 - 4 \sin \theta \). In polar coordinates, \( r \) represents the radius from the origin to the curve, and \( \theta \) represents the angle from the positive x-axis.

Step 2: Analyze the Limaçon to Determine the Region

A limaçon like \( r = 2 - 4 \sin \theta \) has both an inner and an outer loop. We are interested in the region inside the larger loop. This occurs when \( r \) is positive.

Step 3: Sketch the Limaçon

A sketch of the limaçon helps visualize the region. For \( r = 2 - 4 \sin \theta \), as \( \theta \) ranges from \( -\pi/2 \) to \( \pi/2 \), we identify which part represents the larger loop. Ensure the graph starts and ends at the same point to form the loop.

Step 4: Identify the Limits of Integration

To find the area of the larger loop, determine the \( \theta \) values where \( r = 0 \). Solving \( 2 - 4 \sin \theta = 0 \) yields \( \sin \theta = \frac{1}{2} \), giving \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \). The larger loop is within these limits.

Step 5: Set Up the Double Integral

The area of the region is given by the double integral \( \iint_{S} r \, dr \, d\theta \). Here, \( r \) varies from 0 to \( 2 - 4 \sin \theta \) and \( \theta \) varies from \( \frac{\pi}{6} \) to \( \frac{5\pi}{6} \).

Step 6: Evaluate the Integral with respect to \(r\)

The integral, \( \int_{0}^{2-4\sin \theta} r \, dr \), evaluates to \( \left[ \frac{r^2}{2} \right]_{0}^{2-4\sin \theta} = \frac{(2-4\sin \theta)^2}{2} \).

Step 7: Evaluate the Integral with respect to \(\theta\)

Next, integrate \( \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \frac{(2-4\sin \theta)^2}{2} \, d\theta \) to find the total area.

Step 8: Solve the Integral and Find the Area

Calculating the above integral, using trigonometric identities and integration techniques, results in the area \( A = \frac{9\pi + 8\sqrt{3}}{2} \).

Key Concepts

Polar Coordinates

Polar coordinates provide a way to locate points on a plane using angles and distances from a central point, known as the pole (similar to the origin in Cartesian coordinates). Instead of using x and y coordinates, each point is defined by:

• the radius (r), which is the distance from the pole to the point, and
• the angle (\(\theta\)), measured from the positive x-axis.

This system is particularly useful in cases involving circular or rotational symmetry. For instance, complex curves such as the limaçon are easier to represent using polar equations. When we work with these curves, changes in \(\theta\) represent rotations, while changes in \(r\) alter the radial distance, providing a seamless way to describe circular paths.

Limaçon Curve

A limaçon curve can be visualized as a distorted circle. Its shape is influenced heavily by trigonometric functions like sine or cosine. The general form for a limaçon is given by the polar equation: \[ r = a + b \sin \theta \] or \[ r = a + b \cos \theta \] Depending on the relationship between \( a \) and \( b \), a limaçon can have different appearances:

• \( |a| > |b| \) results in an oval-like shape,
• \( |a| = |b| \) forms a cardioid, which looks like a heart, and
• \( |a| < |b| \) results in a limaçon with an inner loop.

In our specific example, the equation \( r = 2 - 4 \sin \theta \) indicates \( |a| < |b| \), meaning our limaçon indeed has an inner loop, as described. The purpose of the exercise was to find the area enclosed by the larger loop, which defines an area of interest in this kind of geometric setup.

Area Calculation

When calculating the area in polar coordinates, we use double integrals. Specifically for polar regions described by equations like our limaçon, the area of a region \( S \) can be computed using: \[ A = \iint_{S} r \, dr \, d\theta \] The integration here is done over both the radius \( r \) and the angle \( \theta \). For the limaçon \( r = 2 - 4 \sin \theta \), the limits of integration for \( r \) range from 0 to the boundary described by \( 2 - 4 \sin \theta \), while \( \theta \) ranges between the critical angles where \( r = 0 \), specifically \( \frac{\pi}{6} \) to \( \frac{5\pi}{6} \).By performing integration in these bounds, we capture the total area of the larger loop of the limaçon, resulting in the computed area as: \[ \frac{9\pi + 8\sqrt{3}}{2} \] This integration process involves breaking down the task into manageable steps, such as first integrating with respect to \( r \), then \( \theta \), while applying trigonometric identities whenever necessary.