   Chapter 15.4, Problem 8E

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.8. D is bounded by y = x + 2 and y = x2; ρ(x, y) = kx2

To determine

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

Explanation

Given:

The region D is bounded by y=x+2,y=x2.

The density function is ρ(x,y)=kx2.

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 and obtain the value of x.

That is, x varies from 1 to 2. Then, the total mass of the lamina is,

m=Dρ(x,y)dA=12x2x+2kx2dydx

Integrate with respect to y and apply the corresponding limit.

m=k12(x2y)x2x+2dx=k12[x2(x+2)x2(x2)]dx=k12[x3+2x2x4]dx

Integrate with respect to x and apply the corresponding limit.

m=k[x44+2x33x55]12=k[((2)44+2(2)33(2)55)((1)44+2(1)33(1)55)]=k[(164+2(8)3325)(1423+15)]=k(154+183335)

=63k20

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

x¯=1mDxρ(x,y)dA=1(63k20)12x2x+2kx2(x)dydx=20k63k12x2x+2x3dydx=206312x2x+2x3dydx

Integrate with respect to y and apply the corresponding limit.

x¯=206312(x3y)x2x+2dx=206312[x3(x+2)x3(x2)]dx=206312[x4+2x3x5]dx

Integrate with respect to x and apply the corresponding limit.

x¯=2063(x55+2x44x66)12=2063[((2)55+(2)42(2)66)((1)55+(1)42(1)66)]=2063[(325+162646)(15+1216)]=2063(335+152636)

Further simplify the terms as shown below

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