Chapter 8: Applications of Triple Integration
Section 8.3: First Moments
Example 8.3.8
Obtain the centroid of R, the region that lies between the paraboloids z=4−x2−y2 and z=3 x2+3 y2.
Impose the density δr ,θ,z=r z cosθ/6 on R and find the resulting center of mass.
(See Example 8.1.22.)
Solution
Mathematical Solution
The volume of R, found in Example 8.1.22, is given by
V=∫02 π∫01∫3 r24− r2r dz dr dθ = 2 π
Table 8.3.8(a) lists the first moments and the coordinates of the centroid.
First Moments
Centroid
Myz=∫02⁢π∫01∫3⁢r24−r2cosθ⁢r2 dz dr dθ = 0
x&conjugate0;=MyzV=0V=0
Mxz=∫02⁢π∫01∫3⁢r24−r2sinθ⁢r2 dz dr dθ = 0
y&conjugate0;=MxzV=0V=0
Mxy=∫02⁢π∫01∫3⁢r24−r2z⁢r dz dr dθ = 143⁢π
z&conjugate0;=MxyV=143 π2 π = 73
Table 8.3.8(a) First moments and the coordinates of the centroid
When the region R supports the density δr,θ,z=r z cosθ/6, the total mass in R is
m=∫02⁢π∫01∫3⁢r24−r2r2⁢z⁢cosθ6 dz dr dθ = 13635⁢3
Table 8.3.8(b) lists the first moments and the coordinates of the center of mass under this condition.
Center of Mass
Myz = ∫02⁢π∫01∫3⁢r24−r2cosθ⁢r3⁢z⁢cosθ6 dz dr dθ = −314
x&conjugate0;=Myzm=−31413635⁢3 = −673256
Mxz = ∫02⁢π∫01∫3⁢r24−r2sinθ⁢r3⁢z⁢cosθ6 dz dr dθ = 37
y&conjugate0;=Mxzm=7313635⁢3 = 671628⁢3
Mxy = ∫02⁢π∫01∫3⁢r24−r2z2⁢r2⁢cosθ6 dz dr dθ = 3256315⁢3
z&conjugate0;=Mxym=3256315⁢313635⁢3 = 84023663
Table 8.3.8(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 8.3.8(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
3 r2
3⁢r2
z2r,θ
4−r2
−r2+4
r1θ
0
r2θ
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∫01∫3⁢r2−r2+4cos⁡θ⁢r2ⅆzⅆrⅆθ∫02⁢π∫01∫3⁢r2−r2+4rⅆzⅆrⅆθ,∫02⁢π∫01∫3⁢r2−r2+4sin⁡θ⁢r2ⅆzⅆrⅆθ∫02⁢π∫01∫3⁢r2−r2+4rⅆzⅆrⅆθ,∫02⁢π∫01∫3⁢r2−r2+4z⁢rⅆzⅆrⅆθ∫02⁢π∫01∫3⁢r2−r2+4rⅆzⅆrⅆθ
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
0,0,73
Table 8.3.8(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 8.3.8(d) will find the center of mass of R for a given density.
r z cosθ/6
r⁢z⁢cos⁡16⁢θ
∫02⁢π∫01∫3⁢r2−r2+4cos⁡θ⁢r3⁢z⁢cos⁡16⁢θⅆzⅆrⅆθ∫02⁢π∫01∫3⁢r2−r2+4r2⁢z⁢cos⁡16⁢θⅆzⅆrⅆθ,∫02⁢π∫01∫3⁢r2−r2+4sin⁡θ⁢r3⁢z⁢cos⁡16⁢θⅆzⅆrⅆθ∫02⁢π∫01∫3⁢r2−r2+4r2⁢z⁢cos⁡16⁢θⅆzⅆrⅆθ,∫02⁢π∫01∫3⁢r2−r2+4z2⁢r2⁢cos⁡16⁢θⅆzⅆrⅆθ∫02⁢π∫01∫3⁢r2−r2+4r2⁢z⁢cos⁡16⁢θⅆzⅆrⅆθ
5272⁢13,−arctan⁡2⁢3+π,407153
Table 8.3.8(d) Centroid computed by task template that implements the CenterOfMass command
Maple Solution - Coded
In Table 8.3.8(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..4−r2,r=0..1,θ=0..2 π,coordinates=cylindricalr,θ,z,output=integral
∫02⁢π∫01∫0−r2+4cos⁡θ⁢r2ⅆzⅆrⅆθ∫02⁢π∫01∫0−r2+4rⅆzⅆrⅆθ,∫02⁢π∫01∫0−r2+4sin⁡θ⁢r2ⅆzⅆrⅆθ∫02⁢π∫01∫0−r2+4rⅆzⅆrⅆθ,∫02⁢π∫01∫0−r2+4z⁢rⅆzⅆrⅆθ∫02⁢π∫01∫0−r2+4rⅆzⅆrⅆθ
C≔CenterOfMass1,z=0..4−r2,r=0..1,θ=0..2 π,coordinates=cylindricalr,θ,z
0,0,3721
Table 8.3.8(e) Centroid in cylindrical coordinates
The coordinates C = 0,0,3721 are cylindrical; the Cartesian equivalents, obtained by making the appropriate substitutions, are then
evalr cosθ,r sinθ,z,Equater,θ,z,C = 0,0,3721
In Table 8.3.8(f), the center of mass is obtained via the CenterOfMass command from the Student MultivariateCalculus package.
Define the density δ.
δ≔r z cosθ/6:
CenterOfMassδ,z=0..4−r2,r=0..1,θ=0..2 π,coordinates=cylindricalr,θ,z,output=integral
∫02⁢π∫01∫0−r2+4cos⁡θ⁢r3⁢z⁢cos⁡16⁢θⅆzⅆrⅆθ∫02⁢π∫01∫0−r2+4r2⁢z⁢cos⁡16⁢θⅆzⅆrⅆθ,∫02⁢π∫01∫0−r2+4sin⁡θ⁢r3⁢z⁢cos⁡16⁢θⅆzⅆrⅆθ∫02⁢π∫01∫0−r2+4r2⁢z⁢cos⁡16⁢θⅆzⅆrⅆθ,∫02⁢π∫01∫0−r2+4z2⁢r2⁢cos⁡16⁢θⅆzⅆrⅆθ∫02⁢π∫01∫0−r2+4r2⁢z⁢cos⁡16⁢θⅆzⅆrⅆθ
CM≔CenterOfMassδ,z=0..4−r2,r=0..1,θ=0..2 π,coordinates=cylindricalr,θ,z
673256⁢13,−arctan⁡2⁢3+π,84023663
Table 8.3.8(f) Center of mass in cylindrical coordinates
The coordinates
CM = 673256⁢13,−arctan⁡2⁢3+π,84023663
are cylindrical; the Cartesian equivalents, obtained by making the appropriate substitutions, are then
cm≔evalr cosθ,r sinθ,z,Equater,θ,z,CM
−673256,671628⁢3,84023663
evalfcm = −0.02057739558,0.07128218926,2.293748294
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