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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
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)

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Chapter 15 Solutions

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