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Chapter 5: Problem 312

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] \(R\) 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 moments are \( M_{x} = \frac{108}{5} \) and \( M_{y} = 40.5 \). The center of mass is \((2, 1.067)\).

Step by step solution

01

Understand the Region and Density Function

The lamina occupies a rectangular region \( R \) with vertices at \((0,1),(0,3),(3,3),\) and \((3,1)\). The density function \( \rho(x, y) = x^2 y \) is provided. We need to find the moments about the axes and calculate the center of mass.
02

Define the Integral for Moment about the x-axis (Mx)

The moment about the \( x \)-axis, \( M_{x} \), is given by the double integral \( M_{x} = \iint_{R} y \rho(x, y) \, dA \). This becomes \[ M_{x} = \int_{0}^{3} \int_{1}^{3} y \cdot x^2 y \, dy \, dx \].
03

Compute Mx using a Computer Algebra System (CAS)

Using CAS, calculate the integral \( M_{x} = \int_{0}^{3} \int_{1}^{3} x^2 y^2 \, dy \, dx \). First integrate with respect to \( y \), then \( x \). The result is \( M_{x} = \frac{108}{5} \).
04

Define the Integral for Moment about the y-axis (My)

The moment about the \( y \)-axis, \( M_{y} \), is given by the double integral \( M_{y} = \iint_{R} x \rho(x, y) \, dA \). This becomes \[ M_{y} = \int_{0}^{3} \int_{1}^{3} x^3 y \, dy \, dx \].
05

Compute My using a Computer Algebra System (CAS)

Using CAS, calculate the integral \( M_{y} = \int_{0}^{3} \int_{1}^{3} x^3 y \, dy \, dx \). Integrate first with respect to \( y \), then \( x \). The result is \( M_{y} = \frac{162}{4} \), which simplifies to \( 40.5 \).
06

Calculate the Mass of the Lamina

The total mass \( M \) is given by \( M = \iint_{R} \rho(x, y) \, dA \), resulting in \( M = \int_{0}^{3} \int_{1}^{3} x^2 y \, dy \, dx \) which evaluates to \( \frac{81}{4} \) using CAS.
07

Find the Center of Mass \((\bar{x}, \bar{y})\)

The coordinates of the center of mass are \( \bar{x} = \frac{M_{y}}{M} \) and \( \bar{y} = \frac{M_{x}}{M} \). Calculate \( \bar{x} = \frac{40.5}{20.25} = 2 \) and \( \bar{y} = \frac{108/5}{20.25} = 1.067 \).
08

Plot the Center of Mass

Use CAS to plot the center of mass \((2, 1.067)\) on the rectangular region defined by \( R \). The point \((2, 1.067)\) is inside the region, indicating the center of mass location.

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Key Concepts

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

Moments of Inertia
The moments of inertia, represented as \( M_{x} \) and \( M_{y} \), are important concepts when dealing with a lamina, a thin flat object. These moments provide insight into how mass is distributed in relation to the coordinate axes. For a rectangular region \( R \), \( M_{x} \) denotes the moment about the \( x \)-axis, and \( M_{y} \) represents the moment about the \( y \)-axis. This is critical because:
  • \( M_{x} \) reflects how the lamina will tilt or balance over the \( x \)-axis.
  • \( M_{y} \) shows the potential rotation around the \( y \)-axis.
To find these moments, we employ double integrals that incorporate the area of the region and the density function. These calculations aid in determining the center of mass, a point that embodies the average position of the mass of the region.
Density Function
A density function, \( \rho(x, y) \), specifies how mass is distributed across a region like \( R \). In this exercise, the density is given by \( \rho(x, y) = x^2 y \). This function states:
  • The mass depends on both the \( x \) and \( y \) positions within \( R \).
  • Mass increases as we move in the positive \( x \) direction and is multiplied by the \( y \) position.
The role of this function is crucial as it converts a simple geometric problem into a physical one by adding the concept of varying density. It thereby influences the resulting moments of inertia and the location of the center of mass. The density function helps reflect real-world materials where distribution is not always uniform.
Computer Algebra System (CAS)
A Computer Algebra System, often abbreviated as CAS, is a powerful tool used here to compute complex integrals that are otherwise difficult to solve manually. For this problem:
  • We use CAS to evaluate the double integrals for \( M_{x} \) and \( M_{y} \).
  • These integrals not only provide the moments but also help compute the mass of the lamina.
  • CAS assists in plotting the center of mass accurately on the region \( R \).
The advantage of using CAS lies in its efficiency and accuracy. It allows rapid computation and graphical representation, enabling students to visualize and understand the physical properties of geometric regions integrated with density functions.
Double Integral
In calculus, a double integral enables the calculation of volume under a surface that spans two dimensions. In the context of the lamina and given density:
  • The double integral is used to compute both the mass of the lamina and the moments of inertia.
  • For \( M_{x} \), the integrand is \( y \cdot \rho(x, y) = y \cdot x^2 y = x^2 y^2 \).
  • For \( M_{y} \), it is \( x \cdot \rho(x, y) = x \cdot x^2 y = x^3 y \).
The process usually involves integrating the function first by one variable, then the other over the specified limits. Here, integration is performed over the rectangular region \( R \), producing values for \( M_{x} \), \( M_{y} \), and mass \( M \). These are pivotal for calculating the center of mass of the lamina.
Rectangular Region
The concept of a rectangular region in this problem provides the set boundaries for the area of integration. Specifically, \( R \) is defined by vertices at \((0,1), (0,3), (3,3), (3,1)\). This structured region:
  • Simplifies the integration bounds as the region has constant limits of 0 to 3 for \( x \) and 1 to 3 for \( y \).
  • Facilitates easy application of double integrals.
  • Enables visualization of the center of mass location once calculations are complete.
A rectangular region also helps in understanding how shape affects the distribution of mass and the resulting physical properties. It forms the foundational geometry upon which density and mass are computed, leveraging the simple structure for complex computation.

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Most popular questions from this chapter

The solid \(Q\) of constant density 1 is situated inside the sphere \(x^{2}+y^{2}+z^{2}=16\) and outside the sphere \(x^{2}+y^{2}+z^{2}=1 .\) Show that the center of mass of the solid is not located within the solid.

For the following problems, find the specified area or volume. The volume of the intersection between two spheres of radius \(1,\) the top whose center is \((0,0,0.25)\) and the bottom, which is centered at \((0,0,0)\) For the following problems, find the center of mass of the region.

\(f(x, y, z)=\sqrt{x^{2}+y^{2}}, \quad B\) is bounded above by the half-sphere \(x^{2}+y^{2}+z^{2}=9\) with \(z \geq 0\) and below by the cone \(2 z^{2}=x^{2}+y^{2}\)

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. \(R=\left\\{(x, y) | 9 x^{2}+y^{2} \leq 1, x \geq 0, y \geq 0\right\\} ; \rho(x, y)=\sqrt{9 x^{2}+y^{2}}\)

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