Chapter 7: Problem 195
For the following exercises, determine a definite integral that represents the area. Region enclosed by the inner loop of \(r=2-3 \sin \theta\)
Short Answer
Expert verified
The area of the inner loop is a definite integral from \(\arcsin(\frac{2}{3})\) to \(\pi - \arcsin(\frac{2}{3})\).
Step by step solution
01
Identify the curve type
The given polar equation is \( r=2-3\sin\theta \), which is a limaçon. Based on the coefficients of the equation, it's a limaçon with an inner loop.
02
Find points of intersection with the pole
To find points where the curve intersects the pole, set \( r=0 \): \( 0=2-3\sin\theta \). Solving for \( \theta \), we get \( \sin\theta = \frac{2}{3} \). Thus, the values of theta where the curve intersects the pole are \( \theta = \arcsin\left(\frac{2}{3}\right) \) and \( \theta = \pi - \arcsin\left(\frac{2}{3}\right) \).
03
Determine the integral limits for the inner loop
The inner loop is enclosed between the two intersections with the pole. Therefore, the limits for the integral are from \( \theta = \arcsin\left(\frac{2}{3}\right) \) to \( \theta = \pi - \arcsin\left(\frac{2}{3}\right) \).
04
Set up the integral for the area of the region
The area \( A \) enclosed by the curve in polar coordinates is given by \( A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \). Thus, the definite integral is: \[A = \frac{1}{2} \int_{\arcsin(\frac{2}{3})}^{\pi - \arcsin(\frac{2}{3})} (2-3\sin\theta)^2 \, d\theta\]
05
Simplify the expression inside the integral
Expand \((2-3\sin\theta)^2\) to get the expression for the integrand. \[(2-3\sin\theta)^2 = 4 - 12\sin\theta + 9\sin^2\theta\]
06
Solve the integral
Integrate each term separately:- \( \int \sin\theta \, d\theta = -\cos\theta \)- \( \int \sin^2\theta \, d\theta = \int \frac{1}{2}(1-\cos(2\theta)) \, d\theta = \frac{1}{2}\theta - \frac{1}{4}\sin(2\theta) \).This gives: \[A = \frac{1}{2} \left[4\theta + 12\cos\theta + \frac{9}{2}\theta - \frac{9}{4}\sin(2\theta) \right]_{\arcsin(\frac{2}{3})}^{\pi - \arcsin(\frac{2}{3})}\]
07
Evaluate the definite integral
Substitute the limits into the evaluated integral to find the area of the inner loop. This step will require a calculator to approximate the value.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polar Coordinates
Polar coordinates are a two-dimensional coordinate system used to specify the location of a point by the angle and the distance from a fixed point, known as the pole. This system is different from the Cartesian coordinate system, where points are defined by
- their x-coordinate (horizontal distance from the origin)
- their y-coordinate (vertical distance from the origin)
- \(r\): the radial distance from the pole to the point, and
- \(\theta\): the angular coordinate or angle measured in radians from the positive x-axis.
³¢¾±³¾²¹Ã§´Ç²Ô
The ³¢¾±³¾²¹Ã§´Ç²Ô is a type of polar curve that can have an inner loop, a dimple, or no interior loops depending on its equation's coefficients. The general form of a ³¢¾±³¾²¹Ã§´Ç²Ô in polar coordinates is given by:
- \( r = a + b\sin\theta \)
- OR \( r = a + b\cos\theta \)
Integral Limits
Determining integral limits in polar coordinates involves finding the angles \(\theta\) at which the polar curve intersects the pole. For our ³¢¾±³¾²¹Ã§´Ç²Ô curve \( r = 2 - 3\sin\theta \), the integral limits are established by setting the equation equal to zero and solving for \(\theta\):
- Set \(r = 0\)
- Find \(\theta\) where \(2 - 3\sin\theta = 0\), which simplifies to \(\sin\theta = \frac{2}{3}\)
- Thus, the solutions are \(\theta = \arcsin\left(\frac{2}{3}\right)\) and \(\theta = \pi - \arcsin\left(\frac{2}{3}\right)\)
Area Calculation
To calculate the area enclosed by a region in polar coordinates, we use the formula for the area \(A\) of a region bounded by a polar curve from angle \(\theta_1\) to \(\theta_2\):\[ A = \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d\theta \]For the ³¢¾±³¾²¹Ã§´Ç²Ô \(r = 2 - 3\sin\theta \), we must square the expression and integrate over the found limits:
- Integrand: Expand \((2-3\sin\theta)^2 = 4 - 12\sin\theta + 9\sin^2\theta\)
- Integrate each term:
- \( \int \sin\theta \, d\theta = -\cos\theta \)
- \( \int \sin^2\theta \, d\theta = \frac{1}{2} \theta - \frac{1}{4}\sin(2\theta) \)