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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

A lamina occupies the region inside the circle x2 + y2 = 2y but outside the circle x2 + y2 = 1. Find the center of mass 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 lies inside the circle x2+y2=2y and outside the circle x2+y2=1 .

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)

Calculation:

Convert into polar coordinates to make the problem easier. So, from the given conditions it is observed that r varies from 1 to 2sinθ and θ varies from π6 to 5π6 . Then, by the equation (1), the total mass of the lamina is,

m=Dρ(x,y)dA=π65π612sinθk(x2+y2)dydx=π65π612sinθkr(r)drdθ=kπ65π612sinθdrdθ

Integrate with respect to r and θ .

m=kπ65π6[r]12sinθdθ=kπ65π6[2sinθ1]dθ=k[2cosθθ]π65π6=k[(2cos5π65π6)(2cosπ6π6)]

Further simplify the terms as shown below.

m=k(2(32)5π6+2(32)+π6)=k(234π6)=k(232π3)=2k(3π3)

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=12k(3π3)π65π612sinθk(x2+y2)(x)dydx=k2k(3π3)π65π612sinθ1r(rcosθ)(r)drdθ=12(3π3)π65π612sinθrcosθdrdθ

Integrate with respect to r.

x¯=12(3π3)π65π6cosθ[r22]12sinθdθ=12(3π3)π65π6cosθ[(2sinθ)22(1)22]dθ=12(3π3)π65π6cosθ[4sin2θ212]dθ=12(3π3)π65π6cosθ[4(1cos2θ)212]dθ

Further simplify the terms as shown below.

x¯=12(3π3)π65π6cosθ[44cos2θ12]dθ=12(3π3)π65π6cosθ[34cos2θ2]dθ=14(3π3)π65π6(3cosθ4cos3θ)dθ=14(3π3)π65π6cos3θdθ

Integrate with respect to θ

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