   Chapter 15.4, Problem 10E

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.10. D is enclosed by the curves y = 0 and y = cos x, −π/2 ≤ x ≤ π/2; ρ(x, y) = y

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=0,y=cosx,π2xπ2 .

The density function is ρ(x,y)=y .

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=π2π20cosxydydx

Integrate with respect to y and apply the corresponding limit.

m=π2π2(y22)0cosxdx=π2π2[(cosx)22(0)22]dx=π2π2[cos2x20]dx=12π2π2cos2xdx

Integrate with respect to x.

m=12(12)π2π2(1+cos2x)dx=14[x+sin2x2]π2π2=14[(π2+sin(2(π2))2)(π2+sin(2(π2))2)]=14[(π2+sinπ2)(π2+sin(π)2)]

On further simplification, the value of m becomes,

m=14[(π2+02)(π202)]=14[π2+π2]=14[2π2]=π4

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

x¯=1mDxρ(x,y)dA=1(π4)π2π20cosxy(x)dydx=4ππ2π20cosxxydydx

Integrate with respect to y and apply the corresponding limit.

x¯=4ππ2π2(xy22)0cosxdx=4ππ2π2[x(cosx)22x(0)22]dx=4ππ2π2xcos2x2dx=2ππ2π2xcos2xdx

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

Let u=x .

Then, dv=cos2xdx .

Thus, the value of x¯ is obtained as follows

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