/*! 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 36 Find the area between a large lo... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the area between a large loop and the enclosed small loop of the curve \( r = 1 + 2 \cos 3 \theta \).

Short Answer

Expert verified
The area between the loops of the curve is the difference between the total area and the inner loop area.

Step by step solution

01

Understand the Polar Equation

The equation provided is a polar equation: \( r = 1 + 2 \cos 3\theta \). This equation suggests a limaçon shape with a loop, where the parametric angles will describe different points on the curve.
02

Determine the Loop Points

For limaçon curves defined by \( r = 1 + 2 \cos 3\theta \), find the values of \( \theta \) where \( r = 0 \). This indicates the points where the curve intersects itself or contains loops. Solve \( 1 + 2 \cos 3\theta = 0 \). This results in \( \cos 3\theta = -\frac{1}{2} \), leading to \( 3\theta = \frac{2\pi}{3} + 2k\pi \) and \( 3\theta = \frac{4\pi}{3} + 2k\pi \). Solve for \( \theta \) to find the loop limits: \( \theta = \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9} \), and \( \frac{10\pi}{9} \).
03

Expression for Area Calculation

Note that the area \( A \) between two radii in polar coordinates is calculated using: \[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \] For loops, integrate within the bounds found in Step 2. Specifically, compute the area from one loop \( \theta_1 = \frac{4\pi}{9} \) to \( \theta_2 = \frac{8\pi}{9} \).
04

Integrate for Inner Loop

Compute the area \( A_\text{inner} \) for the smaller loop: \[ A_\text{inner} = \frac{1}{2} \int_{\frac{4\pi}{9}}^{\frac{8\pi}{9}} (1 + 2\cos 3\theta)^2 \, d\theta \] Simplify and solve this integral.
05

Integrate for Outer Loop

To find the entire area enclosed, compute over a full period of \( \theta \): \[ A_\text{outer} = \frac{1}{2} \int_{0}^{2\pi} (1 + 2\cos 3\theta)^2 \, d\theta \]. Simplify and evaluate for one full cycle.
06

Calculate the Final Area Between Loops

Subtract the inner loop area from the total area to find the area between the large and small loops. \[ A_\text{between} = A_\text{outer} - A_\text{inner} \]. Complete the calculations using the evaluated integrals from Steps 4 and 5.

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Key Concepts

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

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The limaçon is an interesting type of curve that you commonly encounter in polar coordinates. In polar equations, it is expressed in the form \( r = a + b \cos \theta \) or \( r = a + b \sin \theta \). This specific form, \( r = 1 + 2 \cos 3\theta \), creates a limaçon with a loop. It combines a central point with loops that can appear inside or outside the main body, depending on the values of \( a \) and \( b \).

When you come across terms like 'inner loop' and 'outer loop', they refer to the unique shapes that can be formed within a single limaçon. A limaçon can resemble a cardioid, a looped limaçon, or even a dimpled limaçon. Understanding these formations is crucial because the loop shapes significantly impact the area calculations in polar coordinate systems.
  • The limaçon with loops happens when \(|b| > |a|\).
  • If \(a = b\), the limaçon takes on a cardioid shape.
  • When \(|a| > |b|\), the curve forms a dimpled or convex limaçon.
Recognizing these appearances helps in deciphering the nature of complex curves, especially when you're tasked with finding areas between different parts of the curves.
Parametric Angles
The concept of parametric angles is a cornerstone in understanding and plotting curves in polar coordinates. In the equation \( r = 1 + 2 \cos 3\theta \), the term \(3\theta\) signifies that the curve is parametrized with a factor of 3. This modulation means for each complete cycle of \(\theta\) from 0 to \(2\pi\), the angle term cycles three times through its possible range.

This results in multiple intersections with itself, forming the loops within the limaçon. When solving for \(\theta\) such that \( r = 0 \), we're essentially finding the angles at which the curve forms its loops — points where the curve intersects itself. From \(1 + 2 \cos 3\theta = 0\), solving gives \(\theta = \frac{2\pi}{9}, \frac{4\pi}{9}, \frac{8\pi}{9}, \text{and} \frac{10\pi}{9}\). These values are crucial as they mark the transitions between different sections of the curve, helping in establishing the integration limits for area calculations.
  • Parametric angles define how many times a curve crosses itself.
  • Use these angles to set integration limits for area calculation.
  • These determine the shape and self-intersection properties of the limaçon.
Understanding these angles is essential for performing precise calculations in polar-integral geometry like area computation of loops.
Area Calculation in Polar Coordinates
Calculating areas in polar coordinates requires a different approach than in Cartesian coordinates. In polar form, the equation for finding the area enclosed by a curve is \[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \]. This calculation involves squaring the polar function that describes the radius \(r\) as a function of the angle \(\theta\) and then integrating it over the desired interval.

For the limaçon given in the exercise, the area between the large and small loops can be found by first determining the complete area enclosed by the curve and then subtracting the area of the inner loop. By integrating \((1 + 2 \cos 3\theta)^2\) between specific values of \(\theta\), we can calculate the precise areas of these loops.
  • Compute the area of the inner loop, using the integral from \(\frac{4\pi}{9}\) to \(\frac{8\pi}{9}\).
  • Find the total area by integrating over a full cycle from \(0\) to \(2\pi\).
  • Subtract the inner loop area from the total to find the area between the loops.
This method highlights how polar coordinates can be used for complex geometric calculations, making it simple to decompose and solve regions encircled by unusual curves such as the limaçon.

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