Chapter 8: Applications of Triple Integration
Section 8.3: First Moments
Example 8.3.5
Obtain the centroid of R, the region that is inside the cylinder x2+y2=9 and between the planes z=0 and y+z=5.
Impose the density δr,θ,z=r2+z sinθ/4 on R and find the resulting center of mass.
(See Example 8.1.15.)
Solution
Mathematical Solution
The volume of R, found in Example 8.1.15, is given by
V=∫02 π∫03∫05−r sinθr dz dr dθ = 45 π
Table 8.3.5(a) lists the first moments and the coordinates of the centroid.
First Moments
Centroid
Myz=∫02⁢π∫03∫05−r⁢sin⁡θcosθ⁢r2 dz dr dθ = 0
x&conjugate0;=MyzV=0V=0
Mxz=∫02⁢π∫03∫05−r⁢sin⁡θr2⁢sinθ dz dr dθ = −814⁢π
y&conjugate0;=MxzV=−814⁢π45 π=−920
Mxy=∫02⁢π∫03∫05−r⁢sin⁡θz⁢r dz dr dθ = 9818⁢π
z&conjugate0;=MxyV=9818⁢π45 π=10940
Table 8.3.5(a) First moments and the coordinates of the centroid
When the region R supports the density δr,θ,z=r2+z sinθ/4, the total mass in R is
m=∫02⁢π∫03∫05−r⁢sin⁡θr⁢r2+z⁢sinθ4 dz dr dθ = 131322175
Table 8.3.5(b) lists the first moments and the coordinates of the center of mass under this condition.
Center of Mass
Myz = ∫02⁢π∫03∫05−r⁢sin⁡θcosθ⁢r2⁢r2+z⁢sinθ4 dz dr dθ = −99295825025
x&conjugate0;=Myzm=−99295825025131322175 = −1654933129841
Mxz = ∫02⁢π∫03∫05−r⁢sin⁡θr2⁢sinθ⁢r2+z⁢sinθ4 dz dr dθ = −2125053625025
y&conjugate0;=Mxzm=−2125053625025131322175 = −35417563129841
Mxy = ∫02⁢π∫03∫05−r⁢sin⁡θz⁢r⁢r2+z⁢sinθ4 dz dr dθ = 13025229350050
z&conjugate0;=Mxym=13025229350050131322175 = 4341743112519364
Table 8.3.5(b) First moments and the coordinates of the center of mass
In each integral, the Jacobian r must be inserted, and where applicable, the "lever arms" x and y are replaced respectively by x=r cosθ, y=r sinθ.
Maple Solution - Interactive
Based on the CenterOfMass command in the Student MultivariateCalculus package, the task template in Table 8.3.5(c) will find the centroid of R when the density is set to 1.
Tools≻Tasks≻Browse: Calculus - Multivariate≻Integration≻Center of Mass≻Cylindrical
Center of Mass for 3-D Region in Cylindrical Coordinates
Density:
1
Region: z1r,θ≤z≤z2r,θ,r1θ≤r≤r2θ,a≤θ≤b
z1r,θ
0
z2r,θ
5− r sinθ
5−r⁢sin⁡θ
r1θ
r2θ
3
a
b
2 π
2⁢π
Moments ÷ Mass:
Inert Integral - dz dr dθ
StudentMultivariateCalculusCenterOfMass,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z,output=integral
∫02⁢π∫03∫05−r⁢sin⁡θcos⁡θ⁢r2ⅆzⅆrⅆθ∫02⁢π∫03∫05−r⁢sin⁡θrⅆzⅆrⅆθ,∫02⁢π∫03∫05−r⁢sin⁡θr2⁢sin⁡θⅆzⅆrⅆθ∫02⁢π∫03∫05−r⁢sin⁡θrⅆzⅆrⅆθ,∫02⁢π∫03∫05−r⁢sin⁡θz⁢rⅆzⅆrⅆθ∫02⁢π∫03∫05−r⁢sin⁡θrⅆzⅆrⅆθ
Explicit values for r&conjugate0;, θ&conjugate0;, and z&conjugate0;, the center of mass given in cylindrical coordinates:
StudentMultivariateCalculusCenterOfMass,z=..,r=..,θ=..,coordinates=cylindricalr,θ,z
920,−12⁢π,10940
Table 8.3.5(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 8.3.5(d) will find the center of mass of R for a given density.
r2+z sinθ/4
r2+z⁢sin⁡14⁢θ
∫02⁢π∫03∫05−r⁢sin⁡θcos⁡θ⁢r2⁢r2+z⁢sin⁡14⁢θⅆzⅆrⅆθ∫02⁢π∫03∫05−r⁢sin⁡θr⁢r2+z⁢sin⁡14⁢θⅆzⅆrⅆθ,∫02⁢π∫03∫05−r⁢sin⁡θr2⁢sin⁡θ⁢r2+z⁢sin⁡14⁢θⅆzⅆrⅆθ∫02⁢π∫03∫05−r⁢sin⁡θr⁢r2+z⁢sin⁡14⁢θⅆzⅆrⅆθ,∫02⁢π∫03∫05−r⁢sin⁡θz⁢r⁢r2+z⁢sin⁡14⁢θⅆzⅆrⅆθ∫02⁢π∫03∫05−r⁢sin⁡θr⁢r2+z⁢sin⁡14⁢θⅆzⅆrⅆθ
11593089069⁢3257016970918738385,arctan⁡3541756165493−π,4341743112519364
Table 8.3.5(d) Centroid computed by task template that implements the CenterOfMass command
Maple Solution - Coded
In Table 8.3.5(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,z=0..5− r sinθ,r=0..3,θ=0..2 π,coordinates=cylindricalr,θ,z,output=integral
C≔CenterOfMass1,z=0..5− r sinθ,r=0..3,θ=0..2 π,coordinates=cylindricalr,θ,z
Table 8.3.5(e) Centroid in cylindrical coordinates
The coordinates C = 920,−12⁢π,10940 are cylindrical; the Cartesian equivalents, obtained by making the appropriate substitutions, are then
evalr cosθ,r sinθ,z,Equater,θ,z,C = 0,−920,10940
In Table 8.3.5(f), the center of mass is obtained via the CenterOfMass command from the Student MultivariateCalculus package.
Define the density δ.
δ≔r2+z sinθ/4:
CenterOfMassδ,z=0..5− r sinθ,r=0..3,θ=0..2 π,coordinates=cylindricalr,θ,z,output=integral
CM≔simplifyCenterOfMassδ,z=0..5− r sinθ,r=0..3,θ=0..2 π,coordinates=cylindricalr,θ,z
Table 8.3.5(f) Center of mass in cylindrical coordinates
The coordinates
CM = 11593089069⁢3257016970918738385,arctan⁡3541756165493−π,4341743112519364
are cylindrical; the Cartesian equivalents, obtained by making the appropriate substitutions, are then
cm≔simplifyevalr cosθ,r sinθ,z,Equater,θ,z,CM
−1654933129841,−35417563129841,4341743112519364
evalfcm = −0.05287584896,−1.131608922,3.468022098
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