   Chapter 15.4, Problem 6E

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.6. D is the triangular region enclosed by the lines y = 0, y = 2x, and x + 2y = 1; ρ(x, y) = x

To determine

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

Explanation

Given:

The region D is the triangle enclosed by the lines y=0,y=2x,x+2y=1

The density function is ρ(x,y)=x.

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:

The total mass of the lamina is,

m=Dρ(x,y)dA=025y212yxdxdy.

Integrate with respect to x and apply the corresponding limit.

m=025(x22)y212ydy=025[(12y)22(y2)22]dy=025[124y2+4y22y22(4)]dy=025(122y+15y28)dy

Integrate with respect to y and apply the corresponding limit.

m=[y22y22+15y38(3)]025=((25)2(25)2+15(25)324)((0)2(0)2+15(0)324)=((2)5(2)425+15(8)24(125))(00+0)=15425+125

=225

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

x¯=1mDxρ(x,y)dA=1(225)025y212yx(x)dxdy=252025y212yx2dxdy

Integrate with respect to x and apply the corresponding limit.

x¯=252025(x33)y212ydy=252025[((12y)33)((y2)33)]dy=252025[(136y3+12y238y33)(y33(8))]dy=252025[132y+4y28(8)y33(8)y33(8)]dy

=252025(132y+4y265y324)dy

Integrate with respect to y and apply the corresponding limit

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