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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

Find the mass and center of mass of the solid R with the given density function ρ.

40. E is bounded by the parabolic cylinder z = 1 – y2 and the planes x + z = 1, x = 0, and z = 0; ρ (x, y,  z) = 4

To determine

To find: The mass and center of mass of the solid E with density ρ .

Explanation

Given:

E is bounded by the parabolic cylinder z=1y2 and the planes x+z=1,x=0 and z=0 and the density is ρ(x,y,z)=4 .

Calculation:

Mass of the solid , mass(m)=density×volume .

The mass of the solidis, m=liml,m,ni=1lj=1mk=1nρ(xij*,yij*,zij*)ΔA=Eρ(x,y,z)dA .

Here, the density function is given by ρ(x,y,z) and E is the region that is occupied by the solid.

From the given observe that, y varies from 1 to 1, z varies from 0 to 1y2 and x varies from 0 to 1z . Here the density is given by ρ(x,y,z)=4 .

m=1101y201z4dxdzdy

Integrate the above integral with respect to x and apply the limit of it.

m=41101y2[x]01zdzdy=41101y2[1z0]dzdy=41101y2[1z]dzdy

Integrate the above integral with respect to z and apply the limit of it.

m=411[zz22]01y2dy=11[4(1y2)2(1y2)2]dy=11[(1y2)(42+2y2)]dy=11[(1y2)(2+2y2)]dy

On further simplification find the value of m.

m=211[(1y2)(1+y2)]dy=211[(1)2(y2)2]dy=211[(1)2(y2)2]dy=211[1y4]dy

Integrate the above integral with respect to y and apply the limit of it.

m=2[yy55]11=2[((1)(1)55)((1)(1)55)]=2[(115)(1+15)]=2(115+115)

On further simplification find the value of m.

m=2(225)=2(85)=165

Thus, the mass of the solid is 165 .

To calculate the center of mass,

Center of mass located at x¯=Myzm=1mExρ(x,y,z)dV .

x¯=Eρ(x,y,z)xdEm=E4xdEm

x¯=1101y201z4xdxdzdym (1)

Integrate the numerator part with respect to x and apply the limit of it.

Myz=41101y2[x22]01zdzdy=41101y2[((1z)22)((0)22)]dzdy=421101y2(1z)2dzdy=21101y2(z1)2dzdy

Integrate the above integral with respect to z and apply the limit of it.

Myz=211[(z1)33]01y2dy=2311[((1y21)3(01)3)]dy=2311[y6+1]dy=2311[1y6]dy

Integrate the above integral with respect to y and apply the limit of it.

Myz=23[yy77]11=23[((1)(1)77)((1)(1)77)]=23[(117)(1+17)]=23[117+117]

Simplify further.

Myz=23[117+117]=23[227]=23[127]=87

Substitute the value in equation (1),

x¯=Myzm=(87)(165)=87516=514

Center of mass located at y¯=Mzxm=1mEyρ(x,y,z)dV

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