   Chapter 15.4, Problem 13E

Chapter
Section
Textbook Problem

The boundary of a lamina consists of the semicircles y = 1 − x 2   and   y = 4 − x 2 together with the portions of the x-axis that join them. Find the center of mass of the lamina if the density at any point is proportional to its distance from the origin.

To determine

To find: The center of mass of the lamina occupied by the given disk D.

Explanation

Given:

The region D is the semi annular region of the covered by y=1x2,y=4x2 .

The density function is proportional to the distance from the origin, that is ρ(x,y)=k(x2+y2) .

Formula used:

The total mass of the lamina is, m=limk,li=1kj=1lρ(xij*,yij*)ΔA=Dρ(x,y)dA .

Here, the density function is given by ρ(x,y) and D  is the region that is occupied by the lamina.

The center of mass of the lamina that occupies the given region D is (x¯,y¯) .

Here, x¯=Mym=1mDxρ(x,y)dA and y¯=Mxm=1mDyρ(x,y)dA

If f is a polar rectangle R given by 0arb,αθβ, where 0βα2π , then, Rf(x,y)dA=αβabf(rcosθ,rsinθ)rdrdθ (1)

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (2)

Calculation:

Convert into polar coordinates to make the problem easier. So, from the given conditions it is observed that r varies from 1 to 2 and θ varies from 0 to π . Then, by the equation (1) the total mass of the lamina is,

m=Dρ(x,y)dA=0π12k(x2+y2)dydx=0π12kr(r)drdθ=k0π12r2drdθ

Integrate with respect to r and θ by using the property (2).

m=k0πdθ12r2dr=k[θ]0π[r33]12=k(π0)(233133)=k(π)(8313)

= 7kπ3

In order to get the coordinates of the center of mass, find x¯ and y¯ . Therefore, by the equation (1),

x¯=1mDxρ(x,y)dA=1(7kπ3)0π12k(x2+y2)(x)dydx=3k7kπ0π12r(rcosθ)(r)drdθ=37π0π12r3cosθdrdθ

Integrate with respect to r and θ by using the property (2)

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Find more solutions based on key concepts 