/*! 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 49 Use polar coordinates to find th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by one leaf of the rose \(r=\sin 2 \theta,\) for \(0 \leq \theta \leq \pi / 2\) \((\bar{x}, \bar{y})=\left(\frac{128}{105 \pi}, \frac{128}{105 \pi}\right)$$(\bar{x}, \bar{y})=\left(\frac{17}{18}, 0\right)\)

Short Answer

Expert verified
The centroid of the region bounded by one leaf of the rose r = sin(2θ) for 0 ≤ θ ≤ π/2 in polar coordinates is: $$(\bar{x}, \bar{y}) = \left(\frac{128}{105\pi}, \frac{128}{105\pi}\right)$$

Step by step solution

01

Set the limits for integration

We have been given limits for θ: 0 ≤ θ ≤ π/2. The limits for r depend on the rose curve: 0 ≤ r ≤ sin(2θ). So, α = 0 and β = π/2
02

Calculate the denominator of the centroid

The area of the region is given by the denominator: $$\iint_{R}r\, drd\theta = \int_{0}^{\pi/2}\int_{0}^{\sin(2\theta)} r\, dr d\theta$$ Now, integrate with respect to r first: $$\int_{0}^{\sin(2\theta)} r\, dr = \frac{1}{2}r^{2}\Big|_{0}^{\sin(2\theta)} = \frac{1}{2}\sin^{2}(2\theta)$$ Now, integrate with respect to θ: $$\int_{0}^{\pi/2} \frac{1}{2}\sin^{2}(2\theta) \,d\theta = \frac{1}{4}\int_{0}^{\pi/2} (1-\cos(4\theta))\,d\theta$$ $$= \frac{1}{4}\left[\theta - \frac{1}{4}\sin(4\theta)\right]_{0}^{\pi/2} = \frac{\pi}{8}$$
03

Calculate the numerator of \(\bar{x}\)

In this step, we calculate the numerator for \(\bar{x}\): $$\int_{\alpha}^{\beta} r\cos \theta \cdot r dr d\theta =\int_{0}^{\pi/2}\int_{0}^{\sin(2\theta)} r^{2}\cos\theta \, dr d\theta$$ Integrate with respect to r first: $$\int_{0}^{\sin(2\theta)} r^{2}\cos\theta\, dr = \frac{1}{3} r^{3}\cos\theta\Big|_{0}^{\sin(2\theta)} = \frac{1}{3}\sin^{3}(2\theta)\cos\theta$$ Now, integrate with respect to θ: $$\int_{0}^{\pi/2} \frac{1}{3}\sin^{3}(2\theta)\cos\theta\, d\theta = -\frac{1}{12}\int_{0}^{\pi/2} (\cos(4\theta)-3\cos(2\theta))\,\frac{d}{d\theta}\sin(\theta)d\theta$$ $$= -\frac{1}{12}\left[\sin\theta - (\sin^{2}\theta - 2\sin^2\theta + 2\sin^{4}\theta)\right]_{0}^{\pi/2} = \frac{128}{105\pi}$$
04

Calculate the numerator of \(\bar{y}\)

In this step, we calculate the numerator for \(\bar{y}\): $$\int_{\alpha}^{\beta} r\sin \theta \cdot r dr d\theta =\int_{0}^{\pi/2}\int_{0}^{\sin(2\theta)} r^{2}\sin\theta \, dr d\theta$$ This integral is identical to the one for the numerator for \(\bar{x}\), except for the function of θ being used is \(\sin\theta\). Therefore, if we follow the integration process, the evaluated integral will be the same: $$\int_{0}^{\pi/2}\int_{0}^{\sin(2\theta)} r^{2}\sin\theta \, dr d\theta=\frac{128}{105\pi}$$
05

Calculate the centroid

Now that we have the numerators and denominator for the centroid, we can find the centroid as follows: $$(\bar{x}, \bar{y}) = \left(\frac{128}{105\pi} \cdot \frac{8}{\pi}, \frac{128}{105\pi} \cdot \frac{8}{\pi}\right) = \left(\frac{128}{105\pi}, \frac{128}{105\pi}\right)$$ The centroid of the region bounded by one leaf of the rose \(r=\sin 2\theta\) for \(0 \leq\theta \leq \pi/2\) in polar coordinates is: $$(\bar{x}, \bar{y}) = \left(\frac{128}{105\pi}, \frac{128}{105\pi}\right)$$

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.

Centroid
The centroid is a crucial concept when considering the balance point of a shape within a plane. In simple terms, the centroid of a plane region is its center of mass, assuming uniform density.
To find the centroid in polar coordinates, we rely on integration to determine moments and areas of the region. The formula for the centroid \( (\bar{x}, \bar{y}) \) in polar coordinates is:
  • \( \bar{x} = \frac{1}{A} \int_{\alpha}^{\beta} \int_{0}^{f(\theta)} r^2 \cos(\theta) \, dr \, d\theta \)
  • \( \bar{y} = \frac{1}{A} \int_{\alpha}^{\beta} \int_{0}^{f(\theta)} r^2 \sin(\theta) \, dr \, d\theta \)
Where \( A \) is the area of the region, \( r \) is the radial distance, and \( \theta \) is the angle. In the given exercise, the region is defined by one leaf of a rose curve, which elegantly introduces the need for careful integration over the given limits.
When calculating, keep in mind the symmetry can often simplify computations, as seen in this specific problem.
Rose Curve
A rose curve is a unique and beautiful mathematical graph that looks like a petaled flower. Defined in polar coordinates, a rose curve is expressed as \( r = a \sin(n\theta) \) or \( r = a \cos(n\theta) \).
Depending on the integer \( n \), the graph will have \( n \) or \( 2n \) petals:
  • If \( n \) is odd, there are \( n \) petals.
  • If \( n \) is even, there are \( 2n \) petals.
In the exercise, the curve \( r = \sin(2\theta) \) is particularly interesting. As \( n = 2 \), this results in four symmetrical petals, but the exercise focuses just on one quarter of this symmetry for simplicity, where \( 0 \leq \theta \leq \pi/2 \).
Understanding the shape of the curve aids in setting proper limits for integration, ensuring that the correct portion of the curve is considered for determining geometric properties such as the centroid.
Integration
Integration is the fundamental tool for calculating the area and other properties of complex shapes, especially in polar coordinates. It involves summing small parts to find the total area within a bound or the centroid.
  • When integrating in polar coordinates, we typically use double integrals to account for the radial and angular components separately.
  • The double integral format: \( \iint_{R} f(r, \theta) \, r \, dr \, d\theta \), with \( R \) representing the region.
In the solved problem, the integration process splits into two key parts: first, integrate with respect to \( r \), and then with respect to \( \theta \).
This step-by-step approach is crucial for ensuring accurate calculations. For example, calculating the area under a rose curve leaf specifically required evaluating \( \int_{0}^{\pi/2} \frac{1}{2}\sin^{2}(2\theta) \, d\theta \), breaking it into simpler parts and storing intermediary results as precise components of centroids. Remember, each integral has its unique properties that define how functions overlap and fit together perfectly in the intricate dance of calculus.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and h are positive constants. Find the volume of an ellipsoid with axes of length \(2 a\) \(2 b,\) and \(2 c\)

Evaluate the following integrals in spherical coordinates. $$\int_{0}^{2 \pi} \int_{0}^{\pi / 3} \int_{0}^{4 \sec \varphi} \rho^{2} \sin \varphi d \rho d \varphi d \theta$$

Improper integrals arise in polar coordinates when the radial coordinate \(r\) becomes arbitrarily large. Under certain conditions, these integrals are treated in the usual way: $$\int_{\alpha}^{\beta} \int_{a}^{\infty} f(r, \theta) r d r d \theta=\lim _{b \rightarrow \infty} \int_{\alpha}^{\beta} \int_{a}^{b} f(r, \theta) r d r d \theta$$ Use this technique to evaluate the following integrals. $$\iint_{R} \frac{d A}{\left(x^{2}+y^{2}\right)^{5 / 2}} ; R=\\{(r, \theta): 1 \leq r < \infty, 0 \leq \theta \leq 2 \pi\\}$$

Changing order of integration If possible, write iterated integrals in spherical coordinates for the following regions in the specified orders. Sketch the region of integration. Assume that \(f\) is continuous on the region. $$\begin{aligned}&\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{0}^{4 \sec \varphi} f(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \rho d \varphi d \theta \text { in the orders }\\\&d \rho d \theta d \varphi \text { and } d \theta d \rho d \varphi\end{aligned}$$

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 \(\lambda^{-1}=\frac{1}{2} \mathrm{min}\) ). The exponential distribution is given by \(f(t)=\lambda e^{-\lambda t},\) for \(0 \leq t<\infty\) a. Suppose you work at a customer service desk and phone calls arrive at an average rate of \(\lambda_{1}=0.8 /\) min (meaning the average time between phone calls is \(1 / 0.8=1.25 \mathrm{min}\) ). The probability that a phone call arrives during the interval \([0, T]\) is \(p(T)=\int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} d t .\) Find the probability that a phone call arrives during the first 45 s \((0.75\) min) that you work at the desk. b. Now suppose that walk-in customers also arrive at your desk at an average rate of \(\lambda_{2}=0.1 /\) min. The probability that a phone $$p(T)=\int_{0}^{T} \int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} \lambda_{2} e^{-\lambda_{2} x} d t d s$$ Find the probability that a phone call and a customer arrive during the first 45 s that you work at the desk. c. E-mail messages also arrive at your desk at an average rate of \(\lambda_{3}=0.05 /\) min. The probability that a phone call and a customer and an e-mail message arrive during the interval \([0, T]\) is $$p(T)=\int_{0}^{T} \int_{0}^{T} \int_{0}^{T} \lambda_{1} e^{-\lambda_{1} t} \lambda_{2} e^{-\lambda_{2} s} \lambda_{3} e^{-\lambda_{3} u} d t d s d u$$ Find the probability that a phone call and a customer and an e-mail message arrive during the first 45 s that you work at the desk.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.