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Find the specified quantities for the solids described below:

The center of mass of the region from Exercise 49, assuming that the density at every point is proportional to the point’s distance from the z-axis.

Short Answer

Expert verified

Center of mass is(x¯,y¯,z¯)=0,0,94153+40π.

Step by step solution

01

Given Information

The equation of sphere with radius 2is x2+y2+z2=4and for outside the cylinderx2+y2=1.

02

Evaluating of Center of Mass for x, y coordinates

Center of Mass as per given conditions is given by

x¯=MyzM=Exρdxdydzρdxdydz

y=MxzM=EyρdxdydzEρdxdydz

z¯=MxyMEzρdxdydzEρdxdydz

Also

z¯=MxyM=EzρdVEρdVwhere ρ=kx2+y2=kr

It is given that density is proportional to point distance from zaxis

Hence, x¯=0,y¯=0

03

Evaluation of Center of Mass of z coordinate

It is given by

z¯=EzρdVEρdV

=θ=02πr=12z=0z=4-r2z(kr)rdrdθdzθ=02πr=12z=04-r2(kr)rdrdθdz

=12θ=02πr=1r=2kr24-r2drdθθ=02πr=12kr24-r2drdθ

Use r=2sinαdr=2cosαdαin denominator

If r=1α=π6

If r=2α=π2

Integral becomes

=k2θ=02π4r33-r55r=1r=2dθθ=02πα=π/6π/2k(2sinα)2(2cosα)2dα

z¯=k2θ=02π323-325-43-15dθθ=02πα=π/6π/2(4k)sin22αdα

=47k30(θ)θ=0θ=2π(4k)θ=02πdθ12α-sin4α4α=π/6π/2

=94kπ30(4k)(2π)π2-π6-0-sin(2π/3)4

z¯=94kπ30(4k)12(2π)π3+38

=94kπ30(4kπ)24(8π+33)

=94kπ30×244kπ(8π+33)

=94153+40π

Hence, Center of Mass is(x¯,y¯,z¯)=0,0,94153+40π

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