/*! 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 300 In the following exercises, the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 300

In the following exercises, the region \(R\) occupied by a lamina is shown in a graph. Find the mass of \(R\) with the density function \(\rho .\) \(R\) is the rectangular region with vertices \((0,1),(0,3),(3,3),\) and \((3,1) ; \quad \rho(x, y)=x^{2} y\)

Short Answer

Expert verified
The mass of the region is 36.

Step by step solution

01

Define the region

The region \( R \) is a rectangle with vertices at \((0,1), (0,3), (3,3), \) and \((3,1)\). This means the rectangle extends from \(x = 0\) to \(x = 3\) and from \(y = 1\) to \(y = 3\).
02

Express the mass as a double integral

The mass \( M \) of region \( R \) with density function \( \rho(x, y) = x^2 y \) can be found using the double integral: \( M = \int_{x=0}^{3} \int_{y=1}^{3} x^2 y \: dy \: dx \).
03

Integrate with respect to y

Calculate the inner integral \( \int_{y=1}^{3} x^2 y \: dy \). Treating \(x^2\) as a constant with respect to \(y\), the integral becomes \( x^2 \cdot \frac{y^2}{2} \bigg|_{1}^{3} = x^2 \cdot \left( \frac{3^2}{2} - \frac{1^2}{2} \right) = x^2 \cdot \frac{9 - 1}{2} = 4x^2 \).
04

Integrate with respect to x

Substitute the result from the \( y \) integration to get \( \int_{x=0}^{3} 4x^2 \: dx \). Integrate to obtain \(4 \cdot \frac{x^3}{3} \bigg|_{0}^{3} = \frac{4}{3} \cdot (3^3 - 0) = \frac{4}{3} \cdot 27 = 36 \).
05

Final result

After calculating the integrals, the mass \( M \) of the region \( R \) is computed as \( 36 \).

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!

Key Concepts

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

Mass of Lamina
The mass of a lamina is a vital concept in physics and calculus, representing the total mass of a thin plate, or lamina, with varying density across its region. In this context, the lamina is a two-dimensional object where the mass is distributed according to its location.
To compute the mass of the lamina, you need to use a double integral. A double integral helps in summing up all the tiny masses over the entire region. The regional mass is calculated by integrating the density function over the specified region.
  • Start by expressing the double integral to capture all the small elements of mass.
  • The calculation involves integrating the density function over both axes (x and y) over the specified limits of the region.
  • This approach provides the comprehensive mass by accumulating small packets of mass determined by the density function.
Overall, understanding how to set up and solve a double integral is key to finding the mass in these problems.
Density Function
A density function is crucial when dealing with physical properties such as mass or charge per unit area. In this scenario, the density function defines how densely the mass is packed at each point in the region. Here, the density function is represented as \( \rho(x, y) = x^2 y \).
This density function varies with both the x and y coordinates, meaning that the density is not uniform across the region. In our problem:
  • The function \( x^2 \) indicates that as x increases, the density increases quadratically, implying mass is denser further from the y-axis.
  • The factor \( y \) indicates linear variation with y, adding more mass to points higher up in the region.
Therefore, it's essential to integrate this function over the desired region to measure how these variations contribute to the total mass of the lamina.
Region of Integration
The region of integration is the specified area over which you perform the integration to find the total mass in this case. In our example, the region \( R \) is a rectangle bounded by the vertices \((0,1), (0,3), (3,3), \) and \((3,1)\). This shapes a rectangular boundary for numerical evaluation using integration.
To frame this in simple steps:
  • Recognize the bounds for x and y from the coordinates of the vertices; x ranges from 0 to 3, and y ranges from 1 to 3.
  • These limits form the boundaries for the double integral, guiding the 2D integration process.
  • Within these numeric limits, the density function is integrated, accounting for every unit of the region and hence weighing the effect of all mass contained within \( R \).
Understanding the limits and the shape of the region is essential for setting up correct integrals, determining the final evaluation of the mass of the lamina accurately.

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 the following exercises, consider a lamina occupying the region \(R\) and having the density function \(\rho\) given in the first two groups of Exercises. a. Find the moments of inertia \(I_{x}, I_{y},\) and \(I_{0}\) about the \(x\) -axis, \(y\) -axis, and origin, respectively. b. Find the radii of gyration with respect to the \(x\) -axis, \(y\) -axis, and origin, respectively. \(R\) is the region bounded by \(y=\frac{1}{x}, y=\frac{2}{x}, y=1,\) and \(y=2 ; \rho(x, y)=4(x+y)\)

Consider a lamina occupying the region \(R\) and having the density function \(\rho\) given in the preceding group of exercises. Use a computer algebra system (CAS) to answer the following questions. a. Find the moments \(M_{x}\) and \(M_{y}\) about the \(x\) -axis and \(y\) -axis, respectively. b. Calculate and plot the center of mass of the lamina. c. [T] Use a CAS to locate the center of mass on the graph of \(R .\) [T] \(\quad R=\left\\{(x, y) | 9 x^{2}+y^{2} \leq 1, x \geq 0, y \geq 0\right\\}\) \(\rho(x, y)=\sqrt{9 x^{2}+y^{2}}\)

Evaluate the following integrals. $$ \iiint_{D} \frac{y}{3 x^{2}+1} d A, D=\\{(x, y) | 0 \leq x \leq 1,-x \leq y \leq x\\} $$

In the following exercises, evaluate the triple integral \(\quad \iiint_{B} f(x, y, z) d V\) over the solid \(B\) \(269 . f(x, y, z)=1\) \(B=\left\\{(x, y, z) | x^{2}+y^{2}+z^{2} \leq 90, z \geq 0\right\\}\)

In the following exercises, consider a lamina occupying the region \(R\) and having the density function \(\rho\) given in the first two groups of Exercises. a. Find the moments of inertia \(I_{x}, I_{y},\) and \(I_{0}\) about the \(x\) -axis, \(y\) -axis, and origin, respectively. b. Find the radii of gyration with respect to the \(x\) -axis, \(y\) -axis, and origin, respectively. \(R\) is the triangular region with vertices \((0,0),(0,3),\) and \((6,0) ; \rho(x, y)=x y\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Statistics

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.