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Chapter 11: Problem 55

Find the area of the regions bounded by the following curves. The complete three-leaf rose \(r=2 \cos 3 \theta\)

Short Answer

Expert verified
Answer: The area of the regions bounded by the curve is \(\pi\).

Step by step solution

01

1. Find the Limits of Integration

To find the limits of integration for \(\theta\), we need to find the points where \(r=0\). To do this, solve the equation \(r=2\cos(3\theta)=0\) for \(\theta\). $$ 0 = 2\cos(3\theta) \\ \cos(3\theta) = 0 $$ As we know, if \(\cos(x) = 0\), then \(x = \frac{(2n+1)\pi}{2}\), where \(n\) is an integer. Therefore, $$ 3\theta = \frac{(2n+1)\pi}{2} \\ \theta = \frac{(2n+1)\pi}{6} $$ For the first petal, we can use \(n=0\) and \(n=1\). So, the limits of integration are \(\theta = 0\) and \(\theta = \frac{\pi}{3}\).
02

2. Calculate the Area of One Petal

To calculate the area of one petal, we will use the following polar coordinate area formula: $$ A_\text{petal} = \frac{1}{2}\int_{\theta_1}^{\theta_2} r^2 d\theta $$ We have the limits of integration: \(\theta_1=0\) and \(\theta_2=\frac{\pi}{3}\), and the function \(r=2\cos(3\theta)\). Thus, the integral is: $$ A_\text{petal} = \frac{1}{2}\int_{0}^{\frac{\pi}{3}} (2\cos(3\theta))^2 d\theta $$
03

3. Calculate the Integral

Now, we'll calculate the integral: $$ A_\text{petal} = \frac{1}{2}\int_{0}^{\frac{\pi}{3}} 4\cos^2(3\theta) d\theta $$ Using the double angle formula, we can rewrite \(\cos^2(3\theta)\) as \(\frac{1+\cos(6\theta)}{2}\): $$ A_\text{petal} = \frac{1}{2}\int_{0}^{\frac{\pi}{3}} 4\left(\frac{1+\cos(6\theta)}{2}\right) d\theta $$ Now, we simplify and evaluate the integral: $$ A_\text{petal} = \frac{1}{2}\int_{0}^{\frac{\pi}{3}} (2+2\cos(6\theta)) d\theta \\ A_\text{petal} = \frac{1}{2}\left[2\theta + \frac{\sin(6\theta)}{3}\right]_0^\frac{\pi}{3} $$
04

4. Evaluate the Integral and Calculate the Total Area

By plugging in the limits of integration, we can find the area of one petal: $$ A_\text{petal} = \frac{1}{2}\left[2\left(\frac{\pi}{3}\right) + \frac{\sin(2\pi)}{3} - \left(2(0) + \frac{\sin(0)}{3}\right)\right] \\ A_\text{petal} = \frac{1}{2}\left[\frac{2\pi}{3}\right] \\ A_\text{petal} = \frac{\pi}{3} $$ As there are 3 petals, we multiply the area of one petal by 3: $$ A_\text{total} = 3 \times A_\text{petal} \\ A_\text{total} = 3 \times \frac{\pi}{3} \\ A_\text{total} = \pi $$ So, the area of the regions bounded by the curve \(r=2\cos3\theta\) is \(\pi\).

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Most popular questions from this chapter

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The hyperbola \(x^{2} / 4-y^{2} / 9=1\) has no \(y\) -intercepts. b. On every ellipse, there are exactly two points at which the curve has slope \(s,\) where \(s\) is any real number. c. Given the directrices and foci of a standard hyperbola, it is possible to find its vertices, eccentricity, and asymptotes. d. The point on a parabola closest to the focus is the vertex.

Graph the following conic sections, labeling the vertices, foci, directrices, and asymptotes (if they exist). Use a graphing utility to check your work. $$r=\frac{12}{3-\cos \theta}$$

Sketch the graph of the following ellipses. Plot and label the coordinates of the vertices and foci, and find the lengths of the major and minor axes. Use a graphing utility to check your work. $$\frac{x^{2}}{5}+\frac{y^{2}}{7}=1$$

Use a graphing utility to graph the parabolas \(y^{2}=4 p x,\) for \(p=-5,-2,-1,1,2,\) and 5 on the same set of axes. Explain how the shapes of the curves vary as \(p\) changes.

Sketch the graph of the following parabolas. Specify the location of the focus and the equation of the directrix. Use a graphing utility to check your work. $$8 y=-3 x^{2}$$
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