   Chapter 15.4, Problem 9E

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.9. D is bounded by the curves y = e−x, y = 0, x = 0, x = 1; ρ(x, y) = xy

To determine

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

Explanation

Given:

The region D is bounded by the curves y=ex,y=0,x=0,x=1 .

The density function is ρ(x,y)=xy .

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=010exxydydx

Integrate with respect to y and apply the corresponding limit.

m=01(xy22)0exdx=01[x(ex)22x(0)22]dx=01[xe2x20]dx=1201xe2xdx

Integrate with respect to x by applying the technique of integration by parts.

Let u=x .

Then, dv=e2xdx .

Thus, the value of m is obtained as follows.

m=12[(xe2x(2))0101e2x(2)dx]=12[(xe2x2)01+1201e2xdx]=12[(xe2x2)01+12(e2x(2))01]=12[(xe2x2)0114(e2x)01]

Apply the limit of x.

m=12[((1)e2(1)2(0)e2(0)2)14(e2(1)e2(0))]=12[e22e24+14]=12(143e24)=18(13e2)

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

x¯=1mDxρ(x,y)dA=1(18(13e2))010exxy(x)dydx=8(13e2)010exx2ydydx

Integrate with respect to y and apply the corresponding limit.

x¯=8(13e2)01(x2y22)0exdx=8(13e2)01[x2(ex)22x2(0)22]dx=8(13e2)01[x2e2x2]dx=4(13e2)01x2e2xdx

Integrate with respect to x by applying the technique of integration by parts.

Let u=x2 .

Then, dv=e2xdx .

Thus, the value of x¯ is obtained as follows

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