/*! 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 61 Consider the thin constant-densi... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 61

Consider the thin constant-density plate \(\\{(r, \theta): 0

Short Answer

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Question: Find the y-coordinate of the center of mass of a thin constant-density plate bounded by the curves \(r = a\) (0 鈮� a 鈮� 1) and \(r = 1\) in the first quadrant in terms of the parameter \(a\) and sketch its graph. Additionally, determine the value of \(a\) for which the center of mass is located on the edge of the plate. Answer: The y-coordinate of the center of mass can be represented by the function \(\bar{y}(a) = \frac{3}{4\pi}\frac{1-a^4}{1-a^3}\), and its graph can be plotted using any graphing software or calculator. The value of \(a\) for which the center of mass is located on the edge of the plate is \(a = 1\).

Step by step solution

01

(1) Set up the integral for the area of the plate and centroid coordinates

We will use the formulas for the area and centroid coordinates in polar coordinates. The area of the plate is given by: $$ A = \frac{1}{2} \int_{0}^{\pi} \int_{a}^{1} r^2 dr d\theta $$ The centroid coordinates are given by: $$ (\bar{x},\bar{y}) = \left(\frac{1}{2A} \int_{0}^{\pi} \int_{a}^{1} r^3 \cos(\theta) dr d\theta, \frac{1}{2A} \int_{0}^{\pi} \int_{a}^{1} r^3 \sin(\theta) dr d\theta\right) $$ In our case, we only need the \(y\)-coordinate of the centroid, which is \(\bar{y}=\frac{1}{2A}\int_{0}^{\pi} \int_{a}^{1} r^3 \sin(\theta) dr d\theta\).
02

(2) Calculate the integral for the area

To find the area of the plate, we will evaluate the double integral: $$ A = \frac{1}{2} \int_{0}^{\pi} \int_{a}^{1} r^2 dr d\theta = \frac{1}{2} \int_{0}^\pi \left[\frac{1}{3} r^3 \right]_a^1 d\theta = \frac{1}{2} \int_{0}^\pi \frac{1}{3} (1-a^3) d\theta $$ Now, integrate with respect to \(\theta\): $$ A = \frac{1}{6}(1-a^3) \left[\theta\right]_0^\pi = \frac{\pi}{6} (1-a^3) $$
03

(3) Calculate the integral for the y-coordinate of the centroid

To find the y-coordinate of the centroid, we will evaluate the double integral: $$ \bar{y} = \frac{1}{2A} \int_{0}^{\pi} \int_{a}^{1} r^3 \sin(\theta) dr d\theta = \frac{3}{\pi (1-a^3)} \int_{0}^\pi \left[\frac{1}{4} r^4 \sin(\theta)\right]_a^1 d\theta = \frac{3}{\pi (1-a^3)} \int_{0}^\pi \frac{1}{4} (1-a^4) \sin(\theta)d\theta $$ Now, integrate with respect to \(\theta\): $$ \bar{y} = \frac{3}{\pi (1-a^3)} \left[-\frac{1}{4}(1-a^4) \cos(\theta)\right]_0^\pi = \frac{3}{\pi (1-a^3)} \frac{1}{4}(1-a^4) $$
04

(4) Determine the function of the y-coordinate of the centroid with respect to \(a\)

Putting it all together, we get the \(y\)-coordinate of the centroid as a function of \(a\): $$ \bar{y}(a) = \frac{3}{4\pi}\frac{1-a^4}{1-a^3} $$
05

(5) Graph the function

To plot the graph, sketch the function \(\bar{y}(a)\) using any graphing software or calculator. The graph shows how the y-coordinate of the center of mass changes as \(a\) varies.
06

(6) Determine the value of \(a\) when the center of mass is on the edge of the plate

For the center of mass to be on the edge of the plate, the y-coordinate of the centroid must be equal to zero. Therefore, we have $$ 0 = \frac{3}{4\pi}\frac{1-a^4}{1-a^3} $$ This equation implies that \(1-a^4=0\), which leads to \(a=1\). Thus, the center of mass is on the edge of the plate when \(a=1\).

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Key Concepts

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

Polar Coordinates
Polar coordinates are a two-dimensional coordinate system that expresses each point uniquely in terms of a distance from a reference point and an angle from a reference direction.
In this system, a point is defined by
  • Radius (\(r\))
  • Angle (\(\theta\))
The reference point is known as the pole and is analogous to the origin in the Cartesian coordinate system, while the polar axis is the line from the pole in the direction of the angle zero.
For the given problem, the region is defined between two semicircles, showcasing the flexibility of polar coordinates for circular shapes. This makes evaluating integrals and understanding geometric properties more intuitive compared to Cartesian coordinates.
Integrals
Integrals play a crucial role in calculating areas, centroids, and other geometric properties. In this problem, a double integral aids in finding the area of the plate in polar coordinates.
The setup of the integral for the area is:\[A = \frac{1}{2} \int_{0}^{\pi} \int_{a}^{1} r^2 dr d\theta\]Here, the integral covers all the points (\(r\), \(\theta\)) in the region. The inner integral integrates with respect to \(r\), while the outer integrates with respect to \(\theta\).
Integrals in polar coordinates help account for both distance and angle, providing accurate coverage of circular regions.
Centroid
The centroid refers to the center of mass of a geometric object having uniform density. In this exercise, the centroid's \(y\)-coordinate needs to be computed.
The formula for the \(y\)-coordinate is:\[\bar{y} = \frac{1}{2A}\int_{0}^{\pi} \int_{a}^{1} r^3 \sin(\theta) dr d\theta\]This integral respects the geometry of the plate, integrating the vertical distances, weighted by the area, across the object.
  • The denominator, \(2A\), scales the integral appropriately for the size of the region.
  • The integral itself accommodates the change in radial and angular directions.
This \(y\)-coordinate captures the average location of mass vertically in polar form, illustrating the power of integration and geometry together.
Constant-Density Plate
A constant-density plate implies the mass distribution is uniform throughout the plate. This simplifies calculations as density does not vary across the region.
For a plate with constant density \(\rho\), the mass forms directly relate to the area, making calculations straightforward.
In the problem, given the plate's geometric bounds and uniform density, the area itself plays a pivotal role in determining properties like the center of mass without extra correction factors. This assumption reduces the complexity of the mathematics involved and focuses purely on the plate's shape and size.

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Most popular questions from this chapter

Sketch the following regions \(R\). Then express \(\iint_{R} g(r, \theta) d A\) as an iterated integral over \(R\) in polar coordinates. The region bounded by the spiral \(r=2 \theta,\) for \(0 \leq \theta \leq \pi,\) and the \(x\) -axis

Evaluate the following integrals using polar coordinates. Assume \((r, \theta)\) are polar coordinates. A sketch is helpful. $$\int_{0}^{3} \int_{0}^{\sqrt{9-x^{2}}} \sqrt{x^{2}+y^{2}} d y d x$$

Find the coordinates of the center of mass of the following solids with variable density. The solid bounded by the cone \(z=9-r\) and \(z=0\) with \(\rho(r, \theta, z)=1+z\)

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. $$\iint_{R} e^{-x^{2}-y^{2}} d A ; R=\left\\{(r, \theta): 0 \leq r<\infty, 0 \leq \theta \leq \frac{\pi}{2}\right\\}$$

Write iterated integrals in spherical coordinates for the following regions in the specified orders. Sketch the region of integration. Assume \(g\) is continuous on the region. $$\begin{aligned}&\int_{0}^{2 \pi} \int_{x / 6}^{\pi / 2} \int_{\cos \varphi}^{2} g(\rho, \varphi, \theta) \rho^{2} \sin \varphi d \rho d \varphi d \theta \text { in the orders } d \rho d \theta d \varphi\\\&\text { and } d \theta d \rho d \varphi\end{aligned}$$
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