/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 Find the areas of the regions. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the areas of the regions. Inside the circle \(r=-2 \cos \theta\) and outside the circle \(r=1\)

Short Answer

Expert verified
The area of the region is \(\frac{\pi}{2}\).

Step by step solution

01

Understand the Problem

We need to find the area of the region that lies inside the circle given by the equation \(r = -2 \cos \theta\) and outside the circle \(r = 1\). The equation \(r = -2 \cos \theta\) represents a circle with its center at \((-1,0)\) and radius of 1. The equation \(r = 1\) represents another circle centered at the origin with radius 1.
02

Set Up the Integration Bounds

The circles intersect when \(-2\cos \theta = 1\). Solving \(-2\cos \theta = 1\) gives \(\cos \theta = -\frac{1}{2}\). This occurs at \(\theta = \frac{2\pi}{3}\) and \(\theta = \frac{4\pi}{3}\). These will be the limits of integration.
03

Determine the Integral

The area of the region inside one curve and outside another is calculated using the formula for polar area: \(\frac{1}{2} \int_{\theta_1}^{\theta_2} (r_1^2 - r_2^2)\,d\theta\). Here, \(r_1 \) is \(-2 \cos \theta\) and \(r_2 \) is 1.
04

Calculate the Integral

Integrate the function: \[\text{Area} = \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} ((-2\cos \theta)^2 - 1^2)\,d\theta \]This simplifies to: \[\text{Area} = \frac{1}{2} \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} (4\cos^2 \theta - 1)\,d\theta \]
05

Solve the Integral

Use the double angle identity \(\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}\) to simplify the integral:\[\text{Area} = \frac{1}{2} \left[ 2\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \cos^2 \theta\,d\theta - \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} 1\,d\theta \right] = \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} (2 + \cos(2\theta) - \frac{1}{2})\,d\theta \]
06

Evaluate the Integrals

Compute the individual integrals:1. \(\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \cos(2\theta)\,d\theta\): Use substitution to find that this integral evaluates to zero over its symmetric bounds.2. \(\int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} 2\,d\theta - \int_{\frac{2\pi}{3}}^{\frac{4\pi}{3}} \frac{1}{2}\,d\theta\): Results in the integral \(\frac{3\pi}{6}\times 2 - \frac{3\pi}{6} \times \frac{1}{2} = \frac{3\pi}{3} - \frac{3\pi}{12} = \frac{9\pi}{12} - \frac{3\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2}\).
07

Combine Results for Final Area

The total area of the region is \(\frac{\pi}{2}\).

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Key Concepts

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

Polar Area Calculation
Calculating areas in polar coordinates is a bit different from the Cartesian system. Instead of using rectangles, in polar coordinates, we consider sweeps of sectors that stretch from one direction to another. Imagine a pie wedge being carved out with varying lengths! For polar coordinates, the formula to determine the area of a region between
  • two curves,
  • from an angle \( \theta_1 \) to \( \theta_2 \)
is given by: \[ \text{Area} = \frac{1}{2} \int_{\theta_1}^{\theta_2} \left(r_1^2 - r_2^2\right) \,d\theta \]Here's how it works:
  • The function \( r_1 \) corresponds to the curve further from the origin,
  • while \( r_2 \) is the closer curve.
By using this formula, you can get the area where one curve circles around and the other tightly nests within. Remember, each of these curving paths is essentially a radius sweeping through angles to cut out chunks of space.
Integral Bounds
Integral bounds are crucial for slicing the precise portions of geometric graphs we need. In polar coordinates, we're working with angles and trying to figure out the range of \( \theta \) between which our desired region exists. Unlike rectangular coordinates, where we use linear boundaries, here we have to:
  • Identify the angles at which two curves intersect, because this is where boundaries change.
  • Find these specific values of \( \theta \) by equating the radius functions of intersecting curves.
For instance, given circle equations \( r = -2 \cos \theta \) and \( r = 1 \), setting these equal tells us where their overlap begins and ends, yielding critical angle points like \( \theta = \frac{2\pi}{3} \) and \( \theta = \frac{4\pi}{3} \). These angles are our integration boundaries when calculating polar areas.
Double Angle Identity
The double angle identity is a neat trigonometric trick. It's especially handy for simplifying integrals involving squares of cosine or sine. When faced with expressions like \(\cos^2 \theta\), recognizing \[ \cos^2 \theta = \frac{1 + \cos(2\theta)}{2} \]allows us to transition from a squared term, which is cumbersome to integrate, into a mix of constants and \(\cos(2\theta)\). It's easier to tackle algebraically.
  • This identity helps us in reducing the degree or power of trig functions.
  • Once simplified, integration becomes more straightforward.
In our example, simplifying the square term by applying this identity allows us to concentrate on the cosine of double angles, turning an otherwise complex integration into a manageable task.
Circle Equations
Circle equations in polar coordinates present circles in a different light compared to the Cartesian plane. Here, the equation \( r = 1 \) represents a perfect circle centered at the origin with radius 1.
  • As for \( r = -2 \cos \theta \), this creates a circle with the center shifting.
  • Its center lies at \((-1,0)\), and the altered equation takes advantage of the symmetries in cosine to carve out a distinctive path around a new center.
Understanding these parameters helps visualize the shapes that circles take in polar form. Recognizing the signs and trigonometric functions can also provide insights into how these equations manipulate and manipulate the classic circular shape to create offsets and view the plane in unique new ways.

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Most popular questions from this chapter

Graph the limaçons in Exercises. Limaçon ("lee-ma-sahn") is Old French for "snail." You will understand the name when you graph the limaçons in Exercise \(25 .\) Equations for limaçons have the form \(r=a \pm b \cos \theta\) or \(r=a \pm b \sin \theta .\) There are four basic shapes. Limaçons with an inner loop a. \(r=\frac{1}{2}+\cos \theta\) b. \(r=\frac{1}{2}+\sin \theta\)

Find the lengths of the curves. $$\text { The curve } r=a \sin ^{2}(\theta / 2), \quad 0 \leq \theta \leq \pi, \quad a>0$$

Use a CAS to perform the following steps for the given curve over the closed interval. a. Plot the curve together with the polygonal path approximations for \(n=2,4,8\) partition points over the interval. (See Figure \(10.16 .)\) b. Find the corresponding approximation to the length of the curve by summing the lengths of the line segments. c. Evaluate the length of the curve using an integral. Compare your approximations for \(n=2,4,8\) with the actual length given by the integral. How does the actual length compare with the approximations as \(n\) increases? Explain your answer. $$x=\frac{1}{3} t^{3}, \quad y=\frac{1}{2} t^{2}, \quad 0 \leq t \leq 1$$

Replace the polar equations in Exercises \(27-52\) with equivalent Cartesian equations. Then describe or identify the graph. $$r=\cot \theta \csc \theta$$

Length is independent of parametrization To illustrate the fact that the numbers we get for length do not depend on the way we parametrize our curves (except for the mild restrictions preventing doubling back mentioned earlier), calculate the length of the semicircle \(y=\sqrt{1-x^{2}}\) with these two different parametrizations: a. \(x=\cos 2 t, \quad y=\sin 2 t, \quad 0 \leq t \leq \pi / 2\) b. \(x=\sin \pi t, \quad y=\cos \pi t, \quad-1 / 2 \leq t \leq 1 / 2\)

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