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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 E with the given density function ρ.

41. E. is the cube given by 0 ≤ xa, 0 ≤ y a, 0 ≤ z ≤ a; ρ(x, y, z) = x2 + y2 + z2

To determine

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

Explanation

Given:

E is cube given by 0xa, 0ya and 0za and the density is ρ(x,y,z)=x2+y2+z2.

Calculation:

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

The mass of the solid is, 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 x varies from 0 to a, y varies from 0 to a and z varies from 0 to a. Here the density is given by ρ(x,y,z)=x2+y2+z2.

m=0a0a0a(x2+y2+z2)dxdydz

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

m=0a0a[x33+y2x+z2x]0adydz=0a0a[((a)33+y2(a)+z2(a))((0)33+y2(0)+z2(0))]dydz=0a0a[(a33+y2a+z2a)]dydz

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

m=0a0a[a33y+ay33+az2y]0adz=0a[(a33(a)+a(a)33+az2(a))(a33(0)+a(0)33+az2(0))]dz=0a(a43+a43+a2z2)dz=0a(2a43+a2z2)dz

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

m=[2a43z+a2z33]0a=[(2a43(a)+a2(a)33)(2a43(0)+a2(0)33)]0a=[(2a53+a53)(00)]=a5

Thus, the mass of the solid is a5.

To find the center of mass,

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

x¯=Eρ(x,y,z)xdEm=E(x2+y2+z2)xdEm

x¯=0a0a0a(x3+y2x+z2x)dxdzdym (1)

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

Myz=0a0a[x44+y2x22+z2x22]0adzdy=0a0a[((a)44+y2(a)22+z2(a)22)((0)33+y2(0)+z2(0))]dydz=0a0a[(a44+a2y22+a2z22)]dydz

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

Myz=0a[a44y+a22y33+a2z22y]0adz=0a[(a44(a)+a22(a)33+a2z22(a))(a44(0)+a22(0)33+a2z22(0))]dz=0a[(a54+a56+a3z22)(0+0+0)]dz=0a[a54+a56+a3z22]dz

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

Myz=[a54z+a56z+a32z33]0a=[(a54(a)+a56(a)+a32(a)33)(a54(0)+a56(0)+a32(0)33)]0a=[(a64+a66+a66)(0+0+0)]=7a612

Substitute the value in equation (1),

x¯=Myzm=(7a612)a5=7a6121a5=7a12

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

y¯=Eρ(x,y,z)ydEm=E(x2+y2+z2)ydEm

y¯=0a0a0a(x2y+y3+z2y)dxdydzm (1)

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

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Chapter 15 Solutions

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