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Chapter 14: Problem 41

Computing areas Sketch each region and use integration to find its area. The region bounded by all leaves of the rose \(r=2 \cos 3 \theta.\)

Short Answer

Expert verified
Answer: The total area bounded by the rose curve is \(2\pi\).

Step by step solution

01

Sketch the curve

Plot the polar coordinates curve given by \(r = 2\cos(3\theta)\). The shape of this curve is known as a rose with three leaves.
02

Determine the interval of integration

As we can see from the graph, the curve has symmetry with respect to the origin. There are three leaves, so each leaf covers an angle of \(\frac{2\pi}{3}\). To calculate the area of one leaf, let's find when the curve intersects the origin. Set \(r = 0\) to find these points: \(0 = 2\cos(3\theta)\) Divide both sides by 2: \(\cos(3\theta) = 0\) Find the values of \(\theta\) between 0 and \(\frac{2\pi}{3}\): \(3\theta = \frac{\pi}{2}, \frac{3\pi}{2}\) Divide by 3: \(\theta = \frac{\pi}{6}, \frac{\pi}{2}\) These are the bounds we will use for our integral to calculate the area of one leaf.
03

Find the area using integration

Now, we are going to find the area of one leaf using integration over the range we found in Step 2 and then multiply it by 3 (since there are 3 leaves) to find the total area. The area of one leaf will be given by the following integral: Area = \(\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} [2\cos(3\theta)]^2 d\theta\) Let's evaluate the integral and then multiply it by \(\frac{3}{1}\) to find the total area: \(A = \frac{3}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} 4\cos^2(3\theta) d\theta\) Use the power-reduction formula for \(\cos^2(3\theta)\), which is \(\cos^2(3\theta) = \frac{1+\cos(6\theta)}{2}\): \(A = 6 \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{1+\cos(6\theta)}{2} d\theta\) Now, we can split the integral into two parts separate the terms and integrate: \(A = 6 \left[\frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} d\theta + \frac{1}{4} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos(6\theta) d\theta \right]\) Find the antiderivatives: \(A = 6 \left[\frac{1}{2} \theta + \frac{1}{24} \sin(6\theta) \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\) Now, plug in the bounds and subtract: \(A = 6 \left[\left(\frac{1}{2}\frac{\pi}{2} + \frac{1}{24}\sin(3\pi) \right) - \left(\frac{1}{2}\frac{\pi}{6} + \frac{1}{24}\sin(\pi) \right) \right]\) Simplify: \(A = 6 \left[\frac{\pi}{4} - \frac{\pi}{12}\right] = 6 \left[\frac{\pi}{3}\right]\) Finally, we get the total area: \(A = 2\pi\)

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Key Concepts

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

Polar Coordinates
When working with polar coordinates, imagine a point's location described by its distance from the origin and an angle instead of the typical Cartesian coordinates. This coordinate system can be especially powerful when dealing with curves like circles and spirals. In the context of polar coordinates, the function \(r = 2 \cos (3\theta)\) represents a "rose" curve, known for its petal-like symmetry. Each point on the curve is defined by a radius \(r\) and an angle \(\theta\), which together help in sketching intricate patterns easily. By understanding how polar coordinates function, tackling problems within this coordinate system becomes less daunting.
Integration Techniques
Integration in polar coordinates can be a bit different compared to Cartesian coordinates though the fundamental principles apply. To find the area enclosed by a curve like our three-leafed rose, we use a specialized form of integration. We set up the integral by squaring the radius function, \([2\cos(3\theta)]^2\), using the area formula for polar curves:
  • \(\frac{1}{2} \int [r(\theta)]^2 d\theta\)
This formula is crucial, as it helps calculate area by integrating over the angle \(\theta\). Picking the correct interval of integration, based on the symmetry and intersections of the curve, allows us to find the area of one petal, which can then be multiplied to find the total area. Mastering this technique is key to analyzing complex curves efficiently.
Power-Reduction Formula
The power-reduction formula serves as a handy tool, especially when dealing with integrals involving trigonometric functions raised to powers. For this exercise, we encountered the term \(\cos^2(3\theta)\). By applying the power-reduction formula:
  • \(\cos^2(3\theta) = \frac{1 + \cos(6\theta)}{2}\)
we simplify the integration process significantly. The power-reduction formula transforms tedious integration tasks into more manageable steps by reducing powers, making it easier to find antiderivatives and ultimately the integral itself.
Symmetry in Polar Curves
Symmetry greatly simplifies the analysis of polar curves. For the rose curve \(r = 2 \cos(3\theta)\), symmetry plays a critical role. The curve's symmetry about the origin allows us to deduce relationships and simplifies the integral calculations. Each petal of the rose spans an equal interval of \(\theta\), namely \(\frac{2\pi}{3}\), which repeats three times to cover the full curve. By understanding and recognizing these symmetries, when calculating areas or other characteristics, you save time and effort effectively analyzing the full curve with less calculation.

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Most popular questions from this chapter

The occurrence of random events (such as phone calls or e-mail messages) is often idealized using an exponential distribution. If \(\lambda\) is the average rate of occurrence of such an event, assumed to be constant over time, then the average time between occurrences is \(\lambda^{-1}\) (for example, if phone calls arrive at a rate of \(\lambda=2 /\) min, then the mean time between phone calls is \(\lambda^{-1}=\frac{1}{2}\) min). The exponential distribution is given by \(f(t)=\lambda e^{-\lambda t},\) for \(0 \leq t<\infty\)

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Let \(T\) be the transformation \(x=u^{2}-v^{2}\) \(y=2 u v\) a. Show that the lines \(u=a\) in the \(u v\) -plane map to parabolas in the \(x y\) -plane that open in the negative \(x\) -direction with vertices on the positive \(x\) -axis. b. Show that the lines \(v=b\) in the \(u v\) -plane map to parabolas in the \(x y\) -plane that open in the positive \(x\) -direction with vertices on the negative \(x\) -axis. c. Evaluate \(J(u, v)\) d. Use a change of variables to find the area of the region bounded by \(x=4-y^{2} / 16\) and \(x=y^{2} / 4-1\) e. Use a change of variables to find the area of the curved rectangle above the \(x\) -axis bounded by \(x=4-y^{2} / 16\) \(x=9-y^{2} / 36, x=y^{2} / 4-1,\) and \(x=y^{2} / 64-16\) f. Describe the effect of the transformation \(x=2 u v\) \(y=u^{2}-v^{2}\) on horizontal and vertical lines in the \(u v\) -plane.

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