   Chapter 15.4, Problem 12E

Chapter
Section
Textbook Problem

Find the center of mass of the lamina in Exercise 11 if the density at any point is proportional to the square of 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 disk x2+y21 in the first quadrant.

The density function is proportional to the square of its 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 0 to 1 and θ varies from 0 to π2 . Then, by the equation (1) the total mass of the lamina is,

m=Dρ(x,y)dA=0π201k(x2+y2)dydx=0π201kr2(r)drdθ=k0π201r3drdθ .

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

m=k0π2dθ01r3dr=k[θ]0π2[r44]01=k(π20)(144044)=k(π2)(140)

= kπ8

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(kπ8)0π201k(x2+y2)(x)dydx=8kkπ0π201r2(rcosθ)(r)drdθ=8π0π201r4cosθdrdθ

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

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