   Chapter 15.4, Problem 14E

Chapter
Section
Textbook Problem

Find the center of mass of the lamina in Exercise 13 if the density at any point is inversely 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 inversely 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=k0π12drdθ

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

m=k0πdθ12dr=k[θ]0π[r]12=k(π0)(21)=k(π)(1)

= kπ

In order to get the coordinates of the center of mass, find x¯ and y¯

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