Chapter 8: Applications of Triple Integration
Section 8.3: First Moments
Example 8.3.4
Obtain the centroid of R, the region that is bounded inside by the surface ρ=1+cosφ and outside by the sphere ρ=2. (The variables ρ,φ,θ are spherical coordinates.)
Impose the density δρ,φ,θ=ρ2 on R and find the resulting center of mass.
(See Example 8.1.14.)
Solution
Mathematical Solution
The volume of R, found in Example 8.1.14, is given by
V=∫02 π∫0π∫1+cosφ2ρ2sinφ dρ dφ dθ = 8 π
Table 8.3.4(a) lists the first moments and the coordinates of the centroid.
First Moments
Centroid
Myz=∫02⁢π∫0π∫1+cos⁡φ2sin2φ⁢cos⁡θ⁢ρ3ⅆρⅆφⅆθ=0
x&conjugate0;=MyzV=0V=0
Mxz=∫02⁢π∫0π∫1+cos⁡φ2sin2φ⁢sin⁡θ⁢ρ3ⅆρⅆφⅆθ=0
y&conjugate0;=MxzV=0V=0
Mxy=∫02⁢π∫0π∫1+cos⁡φ2cosφ⁢ρ3⁢sinφⅆρⅆφⅆθ=−3215 π
z&conjugate0;=MxyV=−3215 π8⁢π=−415
Table 8.3.4(a) First moments and the coordinates of the centroid
When the region R supports the density δρ,φ,θ=ρ2, the total mass in R is
m=∫02⁢π∫0π∫1+cos⁡φ2ρ4⁢sin⁡φⅆρⅆφⅆθ = 643 π
Table 8.3.4(b) lists the first moments and the coordinates of the center of mass under this condition.
Center of Mass
Myz=∫02⁢π∫0π∫1+cos⁡φ2sin2φ⁢cos⁡θ⁢ρ5ⅆρⅆφⅆθ = 0
x&conjugate0;=Myzm=0m = 0
Mxz=∫02⁢π∫0π∫1+cos⁡φ2sin2φ⁢sin⁡θ⁢ρ5ⅆρⅆφⅆθ = 0
y&conjugate0;=Mxzm=0m = 0
Mxy=∫02⁢π∫0π∫1+cos⁡φ2cos⁡φ⁢ρ5⁢sin⁡φⅆρⅆφⅆθ = −327 π
z&conjugate0;=Mxym=−327 π643 π = −314
Table 8.3.4(b) First moments and the coordinates of the center of mass
In each integral, the Jacobian ρ2sinφ must be inserted, and where applicable, the "lever arms" x,y,z are replaced respectively by x=ρ sinφ cosθ, y=ρ sinφ sinθ, and z=ρ cosφ.
Maple Solution - Interactive
Based on the CenterOfMass command in the Student MultivariateCalculus package, the task template in Table 8.3.4(c) will find the centroid of R when the density is set to 1.
Tools≻Tasks≻Browse: Calculus - Multivariate≻Integration≻Center of Mass≻Spherical
Center of Mass for 3D Region in Spherical Coordinates
(φ is the colatitude, measured down from the z-axis)
Density:
1
Region: ρ1φ,θ≤ρ≤ρ2φ,θ,φ1θ≤φ≤φ2θ,a≤θ≤b
ρ1φ,θ
1+cosφ
1+cos⁡φ
ρ2φ,θ
2
φ1θ
0
φ2θ
π
a
b
2 π
2⁢π
Moments ÷ Mass:
Inert Integral - dρ dφ dθ
StudentMultivariateCalculusCenterOfMass,ρ=..,φ=..,θ=..,coordinates=sphericalρ,φ,θ,output=integral
∫02⁢π∫0π∫1+cos⁡φ2sin⁡φ2⁢cos⁡θ⁢ρ3ⅆρⅆφⅆθ∫02⁢π∫0π∫1+cos⁡φ2ρ2⁢sin⁡φⅆρⅆφⅆθ,∫02⁢π∫0π∫1+cos⁡φ2sin⁡φ2⁢sin⁡θ⁢ρ3ⅆρⅆφⅆθ∫02⁢π∫0π∫1+cos⁡φ2ρ2⁢sin⁡φⅆρⅆφⅆθ,∫02⁢π∫0π∫1+cos⁡φ2cos⁡φ⁢ρ3⁢sin⁡φⅆρⅆφⅆθ∫02⁢π∫0π∫1+cos⁡φ2ρ2⁢sin⁡φⅆρⅆφⅆθ
Explicit values for ρ&conjugate0;,φ&conjugate0;, and θ&conjugate0;, the center of mass given in spherical coordinates:
StudentMultivariateCalculusCenterOfMass,ρ=..,φ=..,θ=..,coordinates=sphericalρ,φ,θ
415,π,0
Table 8.3.4(c) Centroid computed by task template that implements the CenterOfMass command
Based on the CenterOfMass command in the Student MultivariateCalculus package, the task template in Table 8.3.4(d) will find the center of mass of R for a given density.
ρ2
∫02⁢π∫0π∫1+cos⁡φ2sin⁡φ2⁢cos⁡θ⁢ρ5ⅆρⅆφⅆθ∫02⁢π∫0π∫1+cos⁡φ2ρ4⁢sin⁡φⅆρⅆφⅆθ,∫02⁢π∫0π∫1+cos⁡φ2sin⁡φ2⁢sin⁡θ⁢ρ5ⅆρⅆφⅆθ∫02⁢π∫0π∫1+cos⁡φ2ρ4⁢sin⁡φⅆρⅆφⅆθ,∫02⁢π∫0π∫1+cos⁡φ2cos⁡φ⁢ρ5⁢sin⁡φⅆρⅆφⅆθ∫02⁢π∫0π∫1+cos⁡φ2ρ4⁢sin⁡φⅆρⅆφⅆθ
314,π,0
Table 8.3.4(d) Centroid computed by task template that implements the CenterOfMass command
Maple Solution - Coded
In Table 8.3.4(e), the centroid of R is obtained via the CenterOfMass command from the Student MultivariateCalculus package, provided the density is set equal to 1.
Initialize
Install the Student MultivariateCalculus package.
withStudent:-MultivariateCalculus:
Apply the CenterOfMass command from the Student MultivariateCalculus package
CenterOfMass1,ρ=1+cosφ..2,φ=0..π,θ=0..2 π,coordinates=sphericalρ,φ,θ,output=integral
C≔CenterOfMass1,ρ=1+cosφ..2,φ=0..π,θ=0..2 π,coordinates=sphericalρ,φ,θ
Table 8.3.4(e) Centroid in spherical coordinates
The coordinates C = 415,π,0 are spherical; the Cartesian equivalents, obtained by making the appropriate substitutions, are then
evalρ sinφcosθ,ρ sinφsinθ,ρ cosφ,Equateρ,φ,θ,C = 0,0,−415
In Table 8.3.4(f), the center of mass is obtained via the CenterOfMass command from the Student MultivariateCalculus package.
Define the density δ.
δ≔ρ2:
CenterOfMassδ,ρ=1+cosφ..2,φ=0..π,θ=0..2 π,coordinates=sphericalρ,φ,θ,output=integral
CM≔simplifyCenterOfMassδ,ρ=1+cosφ..2,φ=0..π,θ=0..2 π,coordinates=sphericalρ,φ,θ
Table 8.3.4(f) Center of mass in cylindrical coordinates
The coordinates CM = 314,π,0 are spherical; the Cartesian equivalents, obtained by making the appropriate substitutions, are then
evalρ sinφcosθ,ρ sinφsinθ,ρ cosφ,Equateρ,φ,θ,CM
0,0,−314
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