Problem 57 - Calculus Early Transcendentals

Question

Find the area of the regions bounded by the following curves. \text { The limaçon } r=4-2 \cos \theta

Step-by-Step Solution

Answer

Answer: The area of the region is \(\frac{22\pi}{3} - \frac{32\sqrt{3}}{3}\).

Step 1: 1. Sketch the limaçon

First, sketch the limaçon by plotting a few points to find the shape. Generally, a limaçon can have a loop, be a simple closed curve, or have a dimple. For \(r = 4-2\cos\theta\), we can try some angles to understand the shape: For \(\theta = 0\), we have \(r = 4-2(1) = 2\). For \(\theta = \frac{\pi}{2}\), we have \(r = 4-2(0) = 4\). For \(\theta = \pi\), we have \(r = 4-2(-1) = 6\). For \(\theta = \frac{3\pi}{2}\), we have \(r = 4-2(0) = 4\). From these points and additional ones, we can see that the limaçon is a dimpled curve.

Step 2: 2. Find the range of \(\theta\)

Next, we need to find the limits of integration. To do so, we need to determine the points where the limaçon starts and ends, or the points where \(r = 0\): \(0 = 4 - 2\cos\theta \Rightarrow \cos\theta = 2\) This equation has two solutions: \(\theta = \frac{2\pi}{3}\) and \(\theta = \frac{4\pi}{3}\). So, the range of \(\theta\) for this limaçon is \(\left[\frac{2\pi}{3}, \frac{4\pi}{3}\right]\).

Step 3: 3. Set up the integral

Now, we will set up the integral to find the area. We need to integrate the area element in polar coordinates, which is \(\frac{1}{2}r^2d\theta\). So, our integral will look like this: \(\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \frac{1}{2} (4-2\cos\theta)^2 d\theta\)

Step 4: 4. Integrate the function

To integrate the function, first expand \((4-2\cos\theta)^2\): \((4-2\cos\theta)^2 = 16 - 16\cos\theta + 4\cos^2\theta\) Now integrate each term: \(\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \frac{1}{2} (16 - 16\cos\theta + 4\cos^2\theta) d\theta = \frac{1}{2}\left[ 16\theta - 16\sin\theta + 2\theta\cos\theta + 2\sin\theta\right]_{\frac{2\pi}{3}}^{\frac{4\pi}{3}}\)

Step 5: 5. Evaluate the integral

Finally, evaluate the integral by plugging in the limits of integration: \(A = \frac{1}{2}\left[ 16\left(\frac{4\pi}{3}\right) - 16\sin\left(\frac{4\pi}{3}\right) + 2\left(\frac{4\pi}{3}\right)\cos\left(\frac{4\pi}{3}\right) + 2\sin\left(\frac{4\pi}{3}\right) - 16\left(\frac{2\pi}{3}\right) + 16\sin\left(\frac{2\pi}{3}\right) - 2\left(\frac{2\pi}{3}\right)\cos\left(\frac{2\pi}{3}\right) - 2\sin\left(\frac{2\pi}{3}\right)\right]\) After simplifying this expression, we get: \(A = \frac{22\pi}{3} - \frac{32\sqrt{3}}{3}\) So, the area of the regions bounded by the limaçon \(r = 4 - 2\cos\theta\) is \(\frac{22\pi}{3} - \frac{32\sqrt{3}}{3}\).

Key Concepts

Polar Coordinates

Polar coordinates provide a unique way of describing points and shapes in a plane using a combination of distance and angle from a fixed point, known as the pole. This is different from the typical Cartesian coordinates which use x and y coordinates.

• **The radial coordinate**: Denoted as \(r\), this represents the distance from the pole to the point.
• **The angular coordinate**: Denoted as \(\theta\), this is the angle measured from the positive x-axis to the line segment connecting the pole and the point.

Describing curves like limaçons in polar coordinates simplifies the understanding of their geometry. For example, the curve described by the equation \(r = 4 - 2\cos\theta\) forms a limaçon. Such equations reveal how the radius changes with the angle, allowing us to sketch and understand different curve shapes by plotting several points and observing their pattern. You'll notice curves may loop, form dimples, or simply enclose an area, which is often more complex in Cartesian coordinates.

Integral Calculation

Calculating an integral involves finding the area under a curve, which in case of polar coordinates becomes slightly different. Here, we consider the area swept out by a radius as \(\theta\) changes.

• **Integral Representation**: The area in polar coordinates is represented by the integral \(\int \frac{1}{2}r^2 d\theta\).
• **Function of \(\theta\)**: In our exercise, \(r = 4 - 2\cos\theta\) translates to calculating \(\int \frac{1}{2}(4-2\cos\theta)^2 d\theta\).

This situation differs from Cartesian integrals since each d\(\theta\) corresponds to an infinitesimal sector of a circle, rather than a sliver of a rectangle. Solving these requires expanding and integrating each term separately, providing insights into how different parts of the curve contribute to the total area.

Area of Curves

Finding the area bound by curves in polar coordinates is a fascinating task. By using the specific formula \(\frac{1}{2} \int (r)^2 d\theta\), we integrate over the specified range of \(\theta\).

• **Determining Limits**: For the limaçon, these limits are found where \(r = 0\), giving us \(\theta = \frac{2\pi}{3}\) and \(\theta = \frac{4\pi}{3}\).
• **Evaluating the Integral**: Plugging these into the integral simplifies to compute the whole enclosed area.

Understanding these areas reveals how features like dimples or loops in curves can impact total bounded space. The outcome, in our example \(A = \frac{22\pi}{3} - \frac{32\sqrt{3}}{3}\), accounts for all the intricacies in the curve structure and gives insight into how tightly or loosely the curve envelops its center point, which is essential in many real-world application scenarios.