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(a) Sketch the bounded region inside the lemniscate \(r^{2}=4 \cos 2 \theta\) and outside the circle \(r=\sqrt{2}\) (b) Compute the area of the region described in part (a).

Short Answer

Expert verified
Area is calculated using polar integration between intersection angles.

Step by step solution

01

Identify the Region of Interest

We are to consider the region inside the lemniscate defined by the polar curve \( r^{2} = 4 \cos(2\theta) \) and outside the circle \( r = \sqrt{2} \). First, transform the lemniscate equation: \( r = \pm 2\sqrt{\cos(2\theta)} \). The circle has constant radius \( r = \sqrt{2} \). The intersection points are determined by equating the two expressions for \( r \).
02

Find the Intersection Points

Set \( \pm 2\sqrt{\cos(2\theta)} = \sqrt{2} \). Squaring both sides gives \( 4\cos(2\theta) = 2 \). Solving, \( \cos(2\theta) = \frac{1}{2} \), occurs when \( 2\theta = \frac{\pi}{3}, \frac{5\pi}{3} \). Therefore, \( \theta = \frac{\pi}{6}, \frac{5\pi}{6} \). These are the limits of integration for the area.
03

Set Up the Integral for the Area

Calculate the area of the region between the curves by integrating the difference of their areas. The limits of integration are from \( \theta = \frac{\pi}{6} \) to \( \theta = \frac{5\pi}{6} \). The area \( A \) is given by \[ A = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left[ 4\cos(2\theta) - 2 \right] \, d\theta \].
04

Simplify and Integrate

The integral becomes \[ A = \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left[ 2\left(2\cos(2\theta) - 1\right) \right] \, d\theta = \int_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \left( 2\cos(2\theta) - 1 \right) \, d\theta \]. Evaluate \( \int 2\cos(2\theta) \, d\theta = \sin(2\theta) + C \) and \( \int -1 \, d\theta = -\theta + C \).
05

Evaluate the Integral

Compute \( A = \left[ \sin(2\theta) - \theta \right]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} \). Evaluating at the bounds: \( \left( \sin\left(\frac{5\pi}{3}\right) - \frac{5\pi}{6} \right) - \left( \sin\left( \frac{\pi}{3} \right) - \frac{\pi}{6} \right) \). Simplify to find the area.

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Key Concepts

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

Lemniscate
The term 'lemniscate' comes from the Latin word for "ribbon" and is a famous mathematical symbol resembling the infinity symbol (∞). The lemniscate is a type of polar curve and is defined by the equation \( r^2 = 4 \cos(2\theta) \). It is a shape that loops back on itself and is symmetric about the origin.

In polar coordinates, the lemniscate has two loops. To visualize or sketch a lemniscate, it helps to understand how its equation describes these loops. The expression \( 4 \cos(2\theta) \) indicates that the values of \( r^2 \) (and, thus, \( r \)) can change significantly depending on the angle \( \theta \).
For certain values of \( \theta \), the term \( \cos(2\theta) \) can become negative, meaning \( r \) has no real values, which explains the looping nature and gaps within the lemniscate.

The shapes stem from the trigonometric function and its polar coordinates, making it visually appealing and a good topic for understanding curves in mathematics.
Area Between Curves
To find the area between two curves in polar coordinates, we typically calculate the area enclosed by each curve separately and then find the difference. This involves setting up an integral that accounts for the regions bounded by both curves.

In this exercise, we look for the area inside the lemniscate \( r^2=4 \cos(2\theta) \) and outside the circle \( r=\sqrt{2} \). The process involves:
  • Identifying the region of interest by determining the bounds of integration.
  • Knowing that the angle limits where the curves intersect influence the integral's bounds.
Area Calculation
The area bounded between two curves \( r_1(\theta) \) and \( r_2(\theta) \), say from \( \theta = a \) to \( \theta = b \), is given by the formula:\[A = \frac{1}{2} \int_a^b \left( r_1^2(\theta) - r_2^2(\theta) \right) \, d\theta\] This equation essentially calculates the area of each sector formed by the curves and subtracts the smaller area from the larger one to obtain the area in between.
Integral Calculus
Integral Calculus is fundamental in calculating areas under curves, especially when dealing with regions in polar coordinates. Here, integrals allow us to find the precise area enclosed by the lemniscate and circle.

The integral setup in this problem involves:
  • Finding intersection points that actually give the limits for integration.
  • Using the identified limits, apply and evaluate definite integrals to solve for the area.
  • Integrals in polar coordinates often appear as \( \int r^2 \, d\theta \), representing the differential area "dA."
Calculating the Integral
The step-by-step computation involves:
  • Substituting the functions of \( r \) into the integral expression.
  • Evaluating the resulting trigonometric integral.
  • Subtracting the integral of the inner curve from the outer one.
This method highlights how integrals efficiently calculate complex areas by breaking them into smaller computable parts.
Intersection Points in Polar Coordinates
Intersection points are crucial for solving polar coordinate problems, especially when dealing with multiple curves. These points help determine the limits for integration, which are essential for calculating the areas between curves.

Finding Intersection Points
To find where the lemniscate intersects the circle, we equate their respective equations: \( \pm 2\sqrt{\cos(2\theta)} = \sqrt{2} \).
  • Square both sides to eliminate the square roots and solve for \( \theta \).
  • Solving gives values of \( \theta = \frac{\pi}{6} \) and \( \theta = \frac{5\pi}{6} \).
These angles indicate where the two curves meet.

In polar coordinates, intersection points give exact regions of overlap or separation, allowing precise calculations.
Understanding these intersections enables you to effectively apply integration and solve problems involving polar curves.

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