   Chapter 15.4, Problem 7E

Chapter
Section
Textbook Problem

Find the mass and center of mass of the lamina that occupies the region D and has the given density function ρ.7. D is bounded by y = 1 − x2 and y = 0; ρ(x, y) = ky

To determine

To find: The total mass and the center of mass of the lamina.

Explanation

Given:

The region D is bounded by y=1x2,y=0.

The density function is ρ(x,y)=ky.

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

Calculation:

Solve the given equations, it is observed that x varies from 1 to 1. Then, the total mass of the lamina is,

m=Dρ(x,y)dA=1101x2kydydx.

Integrate with respect to y and apply the corresponding limit.

m=k11(y22)01x2dx=k11[(1x2)22(0)22]dx=k11[122x22+x420]dx=k11(12x2+x42)dx

Integrate with respect to x and apply the corresponding limit.

m=k[x2x33+x52(5)]11=k[((1)2(1)33+(1)510)((1)2(1)33+(1)510)]=k[(1213+110)(12+13110)]=k(123+15)

=8k15

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

x¯=1mDxρ(x,y)dA=1(8k15)1101x2ky(x)dydx=15k8k1101x2xydydx

Integrate with respect to y and apply the corresponding limit.

x¯=15811(xy22)01x2dx=15811[x(1x2)22x(0)22]dx=15811[x(12x2+x4)20]dx=15811[x22x32+x52]dx

=15811[x2x3+x52]dx

Integrate with respect to x and apply the corresponding limit

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