/*! 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 25 Make a sketch of the region and ... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 25

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside the lima莽on \(r=2+\cos \theta\)

Short Answer

Expert verified
Answer: The area of the region inside the given lima莽on is \(\frac{9\pi}{2}\).

Step by step solution

01

Sketch the lima莽on

To sketch the lima莽on with the given polar equation \(r = 2 + \cos\theta\), we can use an online mathematical plotter or graphing calculator like Desmos, GeoGebra, or WolframAlpha. Upon graphing the equation, we can see that the lima莽on forms a loop shape with an inner circular region and an outer circular region. Observe this and note that the polar plot indicates a single region enclosed by the curve.
02

Determine the bounds

The range of \(\theta\) required to complete a full loop of the lima莽on is from \(0\) to \(2\pi\). As such, these will be our limits of integration: \(\alpha = 0\) \(\beta = 2\pi\)
03

Set up the integral and find the area

Now that we have our bounds of integration, we can set up the area integral and evaluate it: \(A = \frac{1}{2} \int_{0}^{2\pi} (2+\cos\theta)^2 d\theta\) We will expand the expression inside the integral: \((2 + \cos\theta)^2 = 4 + 4 \cos\theta + \cos^2\theta\) We can write \(\cos^2\theta\) as \(\frac{1 + \cos(2\theta)}{2}\) and then substitute this back into the integral: \(A = \frac{1}{2} \int_{0}^{2\pi} (4 + 4 \cos\theta + \frac{1 + \cos(2\theta)}{2}) d\theta\) Now we integrate with respect to \(\theta\): \(A = \frac{1}{2} \left[ 4\theta + 4\sin\theta + \frac{1}{2}\theta + \frac{1}{4}\sin(2\theta) \right]_0^{2\pi}\)
04

Calculate the area

Now that we have our integral evaluated, plug in the limits of integration to find the value of the area: \(A = \frac{1}{2}\left[ (4\cdot 2\pi + 4\sin{2\pi} + \frac{1}{2}\cdot 2\pi + \frac{1}{4}\sin(4\pi)) - (4\cdot 0 + 4\sin{0} + \frac{1}{2}\cdot 0 + \frac{1}{4}\sin(0)) \right]\) \(A = \frac{1}{2}(8\pi + 0 + \pi + 0) = \frac{1}{2}(9\pi)\) \(A = \frac{9\pi}{2}\) So, the area of the region inside the given lima莽on is \(\frac{9\pi}{2}\).

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