/*! 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 23 Find the center of mass of the f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 23

Find the center of mass of the following plane regions with variable density. Describe the distribution of mass in the region. The triangular plate in the first quadrant bounded by \(x+y=4\) with \(\rho(x, y)=1+x+y\)

Short Answer

Expert verified
Answer: The center of mass of the triangular region is at point (2,2).

Step by step solution

01

Calculate the mass of the region

We will calculate the mass by integrating the density function over the entire region: \(M = \iint_R \rho(x,y) dA\) First we need to find the limits of the integral to define the region: \(0 \leq y \leq 4-x\) \(0 \leq x \leq 4\) The region being a triangle in the first quadrant, we can calculate the double integral as follows: \(M = \int_0^4 \int_0^{4-x} (1+x+y) dydx\)
02

Calculate the moments

Next, we will calculate the moments of the region as follows: 1. Moment about x-axis: \(\bar{y} = \frac{1}{M} \iint_R y \rho(x, y) dA\) 2. Moment about y-axis: \(\bar{x} = \frac{1}{M} \iint_R x \rho(x, y) dA\) Keeping the same limits as before, we can write the double integrals: \(\bar{y}M = \int_0^4 \int_0^{4-x} y(1+x+y) dydx\) \(\bar{x}M = \int_0^4 \int_0^{4-x} x(1+x+y) dydx\)
03

Evaluate the double integrals

Now we will evaluate each double integral one by one: 1. Calculating the mass \(M\): \(M = \int_0^4 \int_0^{4-x} (1+x+y) dydx = \int_0^4 (8 - \frac{1}{2}x^3) dx = 28\) 2. Calculating the moment about x-axis \(\bar{y}M\): \(\bar{y}M = \int_0^4 \int_0^{4-x} y(1+x+y) dydx = \int_0^4 (\frac{8}{3} - \frac{1}{4}x^4) dx = \frac{224}{3}\) 3. Calculating the moment about y-axis \(\bar{x}M\): \(\bar{x}M= \int_0^4 \int_0^{4-x} x(1+x+y) dydx = \int_0^4 (\frac{8}{3} - \frac{1}{4}x^4) dx = \frac{224}{3}\)
04

Calculate the center of mass

Finally, we will calculate the center of mass by dividing the moments by the mass: \(\bar{x} = \frac{\bar{x}M}{M} = \frac{224/3}{28} = 2\) \(\bar{y} = \frac{\bar{y}M}{M} = \frac{224/3}{28} = 2\) Thus, the center of mass of the triangular region is \((\bar{x},\bar{y}) = (2,2)\).

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 91影视!

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

In Section 10.3 it was shown that the area of a region enclosed by the polar curve \(r=g(\theta)\) and the rays \(\theta=\alpha\) and \(\theta=\beta,\) where \(\beta-\alpha \leq 2 \pi,\) is \(A=\frac{1}{2} \int_{\alpha}^{\beta} r^{2} d \theta .\) Prove this result using the area formula with double integrals.

Explain why or why not ,Determine whether the following statements are true and give an explanation or counterexample. a. A thin plate of constant density that is symmetric about the \(x\) -axis has a center of mass with an \(x\) -coordinate of zero. b. A thin plate of constant density that is symmetric about both the \(x\) -axis and the \(y\) -axis has its center of mass at the origin. c. The center of mass of a thin plate must lie on the plate. d. The center of mass of a connected solid region (all in one piece) must lie within the region.

Gravitational field due to spherical shell A point mass \(m\) is a distance \(d\) from the center of a thin spherical shell of mass \(M\) and radius \(R .\) The magnitude of the gravitational force on the point mass is given by the integral $$F(d)=\frac{G M m}{4 \pi} \int_{0}^{2 \pi} \int_{0}^{\pi} \frac{(d-R \cos \varphi) \sin \varphi}{\left(R^{2}+d^{2}-2 R d \cos \varphi\right)^{3 / 2}} d \varphi d \theta$$ where \(G\) is the gravitational constant. a. Use the change of variable \(x=\cos \varphi\) to evaluate the integral and show that if \(d>R,\) then \(F(d)=\frac{G M m}{d^{2}},\) which means the force is the same as if the mass of the shell were concentrated at its center. b. Show that if \(d

Let \(D\) be the solid bounded by the ellipsoid \(x^{2} / a^{2}+y^{2} / b^{2}+z^{2} / c^{2}=1,\) where \(a>0, b>0,\) and \(c>0\) are real numbers. Let \(T\) be the transformation \(x=\)au, \(y=b v, z=c w\) Evaluate \(\iiint_{D}|x y z| d A\)

Water in a gas tank Before a gasoline-powered engine is started, water must be drained from the bottom of the fuel tank. Suppose the tank is a right circular cylinder on its side with a length of \(2 \mathrm{ft}\) and a radius of 1 ft. If the water level is 6 in above the lowest part of the tank, determine how much water must be drained from the tank.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation
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.