/*! 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 22 Make a sketch of the region and ... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 22

Make a sketch of the region and its bounding curves. Find the area of the region. The region inside the right lobe of \(r=\sqrt{\cos 2 \theta}\)

Short Answer

Expert verified
Answer: The area of the right lobe is \(\frac{1}{2}\) square units.

Step by step solution

01

Analyze the polar equation

To get a better understanding of the given polar equation \(r=\sqrt{\cos 2 \theta}\), we can start by identifying its characteristics. Since the equation is a double-angle equation, it shows symmetry around the origin.
02

Find the bounds of the right lobe

To determine where the equation \(r=\sqrt{\cos 2 \theta}\) becomes zero, we can set \(\cos 2 \theta\) to zero:$$cos 2\theta = 0$$ To solve for \(\theta\), we can find when the cosine function equals zero: $$2\theta = \frac{\pi}{2}, \frac{3\pi}{2}$$ $$\theta = \frac{\pi}{4}, \frac{3\pi}{4}$$ The right lobe exists between these angles: $$\frac{\pi}{4} \leq \theta \leq \frac{3\pi}{4}$$
03

Sketch the region and its bounding curves

Now that we have the bounds, we can sketch the region and its bounding curves. The graph of \(r=\sqrt{\cos 2 \theta}\) forms a four-leaf pattern; however, we are only interested in the right lobe, which is the region enclosed by the graph where \(r\) is positive. This region is symmetrical around the origin, and its bounds lie between \(\frac{\pi}{4}\) and \(\frac{3\pi}{4}\) in the polar plane.
04

Set up the area integral

We can now set up the integral to find the area of the right lobe. Using the formula \(Area = \frac{1}{2}\int_{\alpha}^{\beta} r^2 d\theta\), we obtain the integral: $$Area = \frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} (\sqrt{\cos 2\theta})^2 d\theta$$
05

Calculate the area integral

Now, we can evaluate the integral to find the area of the right lobe using the bounds and our polar equation: $$Area = \frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \cos 2 \theta d\theta$$ To solve this integral, we can use the substitution \(u = 2 \theta\), \(du = 2 d\theta\):\begin{align*} Area &= \frac{1}{4} \int_{\pi/2}^{3\pi/2} \cos u du \\ &= \frac{1}{4} [\sin u]_{\pi/2}^{3\pi/2}\\ &= \frac{1}{4}[-1 - (1)] \\ &= \frac{1}{2} \end{align*} The area of the right lobe of the graph of \(r=\sqrt{\cos 2 \theta}\) is \(\frac{1}{2}\) square units.

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