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Chapter 16: Problem 50

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by the lima莽on \(r=2+\cos \theta\)

Short Answer

Expert verified
Based on the step-by-step solution given, the centroid of the region bounded by the lima莽on \(r = 2 + \cos \theta\) with constant density is \((\bar{r},\bar{\theta}) = (5, \frac{13}{12})\).

Step by step solution

01

Identify the bounds for \(r\) and \(\theta\)

The given lima莽on, \(r = 2 + \cos \theta\), determines our region, and as the equation is symmetric with respect to the polar axis (\(\theta = 0\)), we can deduce the bounds for \(r\) and \(\theta\). The minimum value of \(r\) occurs when \(\theta = \pi\), which is \(r = 2 + \cos\pi = 1\). The maximum value of \(r\) occurs when \(\theta = 0\), which is \(r = 2 + \cos0 = 3\). To include the full region, we can let \(\theta\) range from \(0\) to \(2\pi\). So, our bounds are \(1 \leq r \leq 3\) and \(0 \leq \theta \leq 2\pi\).
02

Compute the first moments \(M_r\) and \(M_\theta\)

Now, let's compute the first moments, which are given by: \(M_r = \frac{1}{2}\iint\limits_{R} r^2dA\) and \(M_\theta = \frac{1}{2}\iint\limits_{R} rdA\) Since we have constant density, we compute the integral: \(M_r = \frac{1}{2}\int_{0}^{2\pi} \int_{1}^{3} r^3drd\theta = \frac{1}{8}\int_{0}^{2\pi} (3^4 - 1^4)d\theta = \frac{1}{8}(80\theta\Big|_0^{2\pi}) = 20\pi\) Similarly, the second integral is obtained: \(M_\theta = \frac{1}{2}\int_{0}^{2\pi} \int_{1}^{3} r^2drd\theta = \frac{1}{6}\int_{0}^{2\pi} (3^3 - 1^3)d\theta = \frac{1}{6}(26\theta\Big|_0^{2\pi}) = \frac{13}{3}\pi\)
03

Compute the area of the region \(A\)

To compute the area of the region, use the formula: \(A = \frac{1}{2}\iint\limits_{R} rdA\) So, we have: \(A = \frac{1}{2}\int_{0}^{2\pi} \int_{1}^{3} rdrd\theta = \frac{1}{4}\int_{0}^{2\pi} (3^2 - 1^2)d\theta = \frac{1}{4}(8\theta\Big|_0^{2\pi}) = 4\pi\)
04

Compute the centroid of the region

To find the centroid, use the formula for polar coordinates: \((\bar{r},\bar{\theta}) = \left(\frac{M_r}{A},\frac{M_\theta}{A}\right)\) So, plug in the calculated first moments and area to get: \((\bar{r},\bar{\theta}) = \left(\frac{20\pi}{4\pi},\frac{\frac{13}{3}\pi}{4\pi}\right) = (5,\frac{13}{12})\) Therefore, the centroid of the region bounded by the lima莽on \(r = 2 + \cos \theta\) is \((\bar{r},\bar{\theta}) = (5, \frac{13}{12})\).

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Most popular questions from this chapter

Improper integrals 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: $$\int_{\alpha}^{\beta} \int_{a}^{\infty} g(r, \theta) r d r d \theta=\lim _{b \rightarrow \infty} \int_{\alpha}^{\beta} \int_{a}^{b} g(r, \theta) r d r d \theta$$ Use this technique to evaluate the following integrals. $$\iint_{R} e^{-x^{2}-y^{2}} d A ; R=\left\\{(r, \theta): 0 \leq r<\infty, 0 \leq \theta \leq \frac{\pi}{2}\right\\}$$

Choose the best coordinate system and find the volume of the following solids. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. The solid inside the sphere \(\rho=1\) and below the cone \(\varphi=\pi / 4\) for \(z \geq 0\).

The average distance between points of the disk \(\\{(r, \theta): 0 \leq r \leq a\\}\) and the origin

Use polar coordinates to find the centroid of the following constant-density plane regions. The semicircular disk \(R=\\{(r, \theta): 0 \leq r \leq 2,0 \leq \theta \leq \pi\\}\)

Evaluate the following integrals using the method of your choice. A sketch is helpful. $$\iint_{R} \frac{d A}{4+\sqrt{x^{2}+y^{2}}} ; R=\left\\{(r, \theta): 0 \leq r \leq 2, \frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}\right\\}$$
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