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Find the area of the region that lies inside the first curve and outside the second curve. $$ r=3 \cos \theta, \quad r=1+\cos \theta $$

Short Answer

Expert verified
The area is approximately \(8.47\).

Step by step solution

01

Set up the Equations

We are given two polar equations: \( r_1 = 3 \cos \theta \) and \( r_2 = 1 + \cos \theta \). We want to find the area inside the first curve and outside the second curve. First, we need to determine their points of intersection to establish bounds for integration.
02

Find the Points of Intersection

To find the points of intersection, set \( r_1 \) equal to \( r_2 \): \[ 3\cos\theta = 1 + \cos\theta \]Rearrange the equation: \[ 2\cos\theta = 1 \]This simplifies to \( \cos\theta = \frac{1}{2} \). Hence the solutions in the range \([0, 2\pi]\) are \( \theta = \frac{\pi}{3} \) and \( \theta = \frac{5\pi}{3} \).
03

Set up the Integral for Area

The area \( A \) enclosed between two polar curves from \( \theta = a \) to \( \theta = b \) is given by:\[ A = \frac{1}{2} \int_{a}^{b} (r_1^2 - r_2^2) \, d\theta \]Substitute the known values:\[ A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} ((3\cos\theta)^2 - (1 + \cos\theta)^2) \, d\theta \]
04

Simplify the Expression

Compute the squares inside the integral:\[ (3\cos\theta)^2 = 9\cos^2\theta \] and \[ (1 + \cos\theta)^2 = 1 + 2\cos\theta + \cos^2\theta \]Thus, the integral becomes:\[ A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (9\cos^2\theta - 1 - 2\cos\theta - \cos^2\theta) \, d\theta \]Simplifying further:\[ A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (8\cos^2\theta - 2\cos\theta - 1) \, d\theta \]
05

Integrate using Trigonometric Identities

Rewrite \( \cos^2\theta \) using a trigonometric identity: \( \cos^2\theta = \frac{1 + \cos 2\theta}{2} \).Substitute into the integral:\[ A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} \left( 8\left(\frac{1 + \cos 2\theta}{2}\right) - 2\cos\theta - 1 \right) \, d\theta \]This simplifies to:\[ A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (4 + 4\cos 2\theta - 2\cos\theta - 1) \, d\theta \]\[ A = \frac{1}{2} \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} (3 + 4\cos 2\theta - 2\cos\theta) \, d\theta \]
06

Solve the Integrals

Separate the integrals:\[ A = \frac{1}{2} \left( \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} 3 \, d\theta + \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} 4\cos 2\theta \, d\theta - \int_{\frac{\pi}{3}}^{\frac{5\pi}{3}} 2\cos\theta \, d\theta \right) \]Compute each integral:- The first integral is simply: \(3(\theta)\) evaluated from \(\frac{\pi}{3}\) to \(\frac{5\pi}{3}\).- For the second integral, the antiderivative of \(\cos 2\theta\) is \(\frac{1}{2}\sin 2\theta\).- For the third integral, the antiderivative of \(\cos \theta\) is \(\sin \theta\).
07

Evaluate the Integrals

Evaluate each part:- \(3(\theta)\) from \(\frac{\pi}{3}\) to \(\frac{5\pi}{3}\) gives \[ 3\cdot \left(\frac{5\pi}{3} - \frac{\pi}{3}\right) = 3 \cdot \pi \]- \(4\cdot \left(\frac{1}{2} \sin 2\theta \right)\) from \(\frac{\pi}{3}\) to \(\frac{5\pi}{3}\) gives \[ 2(\sin\frac{10\pi}{3} - \sin\frac{2\pi}{3}) \approx 0 \]- \(-2\sin\theta\) from \(\frac{\pi}{3}\) to \(\frac{5\pi}{3}\) gives \[-2 (\sin\frac{5\pi}{3} - \sin\frac{\pi}{3}) = 2 \cdot (\sin\frac{\pi}{3} + \sin\frac{5\pi}{3}) = -2(\sin\frac{\pi}{3}) \approx -3.46 \]
08

Calculate the Total Area

Combine the results: \[ A = \frac{1}{2} \left( 3\pi + 0 + 3.46 \right) \approx \frac{1}{2} (3\pi + 3.46) \]Finally, simplify this to find the total area.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polar Curves
Polar curves are a way of representing mathematical functions in the polar coordinate system, where each point is described by a distance from the origin (known as the radial coordinate, \(r\)) and an angle (known as the angular coordinate, \(\theta\)). This is different from the Cartesian coordinate system, which uses \(x\) and \(y\) coordinates to describe a point in the plane. In polar coordinates, the position of a point is derived from rotating an angle \(\theta\) from the positive x-axis and moving a distance \(r\) from the origin. For instance, in the exercise, the polar equations given are \(r_1 = 3 \cos \theta\) and \(r_2 = 1 + \cos \theta\). These equations define curves by relating the radius \(r\) to the angle \(\theta\). When graphed, these curves can form shapes like circles and loops that are otherwise complex to express in Cartesian coordinates.Exploring polar curves can offer insights into symmetry, as some curves are symmetric with respect to the x-axis, y-axis, or even the origin itself. Recognizing these properties can be very helpful when sketching polar graphs or setting up integration.
Integration Bounds
In the context of polar coordinates, integration bounds refer to the values of \(\theta\) over which you want to calculate the value, like area or length, enclosed by polar curves. Finding these bounds accurately is crucial for the correct evaluation of an integral. When given two polar curves, it is necessary first to determine where these curves intersect, as these points define the boundaries for integration. For the given exercise, the points of intersection are found by setting the equations \(3 \cos \theta = 1 + \cos \theta\) equal to each other and solving for \(\theta\). Solving gives \(\theta = \frac{\pi}{3}\) and \(\theta = \frac{5\pi}{3}\), which are used as the limits for integration. These bounds not only help in computing the exact area between curves but also in any calculations involving polar functions. Identifying these accurately ensures that the computations involve the desired sections of the curves.
Trigonometric Identities
Trigonometric identities play a crucial role when working with polar coordinates, as they often simplify the integration process. In polar calculus, simplifying expressions involving trigonometric functions like \(\cos^2\theta\) or \(\sin^2\theta\) is a common task.One key identity used in the exercise is \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\). This identity helps reduce powers of cosine to a linear form, making the integration process more straightforward. Employing these identities can transform complicated expressions into manageable forms, allowing for the integration of polar areas to be conducted with ease.Understanding a handful of these identities can therefore significantly enhance your ability to solve integration problems involving polar functions. Other useful identities include \(\sin^2 \theta = \frac{1 - \cos 2\theta}{2}\) and the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), which often arise in different steps of solving calculus problems.
Area between Curves
Finding the area between polar curves involves integration, much as it does with Cartesian coordinates, but with a polar twist. The formula for finding the area enclosed between two polar curves \(r_1\) and \(r_2\) from \(\theta = a\) to \(\theta = b\) is:\[A = \frac{1}{2} \int_{a}^{b} (r_1^2 - r_2^2) \, d\theta\]This formula accounts for the radial nature of polar coordinates, effectively integrating the difference in squares of the radial distances from the origin as they vary with \(\theta\).In the exercise, substituting \(r_1 = 3 \cos \theta\) and \(r_2 = 1 + \cos \theta\) into this integral between \(\theta = \frac{\pi}{3}\) and \(\theta = \frac{5\pi}{3}\) helps find the area that lies inside \(r_1\) but outside \(r_2\). By setting up the polar integral and computing it, you capture the space between these often irregular and complex-shaped polar graphs, hence finding the desired area enclosed.

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