Chapter 6: Applications of Double Integration
Section 6.5: First Moments
Example 6.5.6
Calculate the coordinates of the center of mass of the lamina whose shape is that of the cardioid r=1+cosθ and whose density is ρ=1+2 x2+3 y2/10.
Solution
Mathematical Solution
Figure 6.5.6(a) shows the region R in red, and the surface ρ, in blue. The green dot represents the center of mass x&conjugate0;,y&conjugate0;=449/418,0. The relevant calculations are tabulated to the left of the figure.
m=∫02 π∫01+cosθr⋅ρ ⅆr ⅆθ = 209320 π
Mx=∫02 π∫01+cosθr2⋅ρ sinθ ⅆr ⅆθ = 0
My=∫02 π∫01+cosθr2⋅ρ cosθ ⅆr ⅆθ = 449640π
x&conjugate0;=My/m=449/418
y&conjugate0;=Mx/m=0
Figure 6.5.6(a) Center of mass, R, and ρ
Maple Solution - Interactive
Table 6.5.6(a) contains a solution via the task template that implements the CenterOfMass command from the Student MultivariateCalculus package. In polar coordinates, the density is ρ=1+r2 2 cos2θ+3 sin2θ/10.
Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Center of Mass≻Polar
Center of Mass for Planar Region in Polar Coordinates
Density:
1+r2 2 cos2θ+3 sin2θ/10
110+110⁢r2⁢2⁢cos⁡θ2+3⁢sin⁡θ2
Region: r1θ≤r ≤r2θ,a≤θ≤b
r1θ
0
r2θ
1+cosθ
1+cos⁡θ
a
b
2 π
2⁢π
Moments÷Mass:
Inert Integral - dr dθ
StudentMultivariateCalculusCenterOfMass, r=..,θ=..,coordinates=polarr, θ,output=integral
∫02⁢π∫01+cos⁡θr2⁢110+110⁢r2⁢2⁢cos⁡θ2+3⁢sin⁡θ2⁢cos⁡θⅆrⅆθ∫02⁢π∫01+cos⁡θr⁢110+110⁢r2⁢2⁢cos⁡θ2+3⁢sin⁡θ2ⅆrⅆθ,∫02⁢π∫01+cos⁡θr2⁢110+110⁢r2⁢2⁢cos⁡θ2+3⁢sin⁡θ2⁢sin⁡θⅆrⅆθ∫02⁢π∫01+cos⁡θr⁢110+110⁢r2⁢2⁢cos⁡θ2+3⁢sin⁡θ2ⅆrⅆθ
Explicit values for r&conjugate0; and θ&conjugate0;
StudentMultivariateCalculusCenterOfMass,r=..,θ=..,coordinates=polarr, θ
449418,0
Plot:
StudentMultivariateCalculusCenterOfMass,r=..,θ=.., coordinates=polarr, θ,output=plot,caption=
Table 6.5.6(a) Calculation of center of mass via task template
The figure produced by the option "output = plot" has had constrained scaling imposed via the Context Panel for the graph. The graph itself shows the region R in red, and the function ρ=r in blue. The centroid is represented by the green dot.
The polar coordinates of the centroid are therefore r&conjugate0;,θ&conjugate0;=, which, in Cartesian coordinates would be r&conjugate0; cosθ&conjugate0;,r&conjugate0; sinθ&conjugate0;=449/418,0. In other words, the task template calculates the mass and first moments in Cartesian coordinates, but returns the center of mass in polar coordinates.
A solution from first principles is detailed in Table 6.5.6(b).
Define the density ρ in polar coordinates
Context Panel: Assign Name
ρ=1+r2 2 cos2θ+3 sin2θ/10→assign
Obtain m, the total mass in region R
Calculus palette: Iterated double-integral template
Context Panel: Evaluate and Display Inline
Context Panel: Assign to a Name≻m
∫02 π∫01+cosθr⋅ρ ⅆr ⅆθ = 209320⁢π→assign to a namem
Obtain Mx, the total moments about the x-axis
Context Panel: Assign to a Name≻Mx
∫02 π∫01+cosθr2⋅ρ sinθ ⅆr ⅆθ = 0→assign to a nameMx
Obtain My, the total moments about the y-axis
Context Panel: Assign to a Name≻My
∫02 π∫01+cosθr2⋅ρ cosθ ⅆr ⅆθ = 449640⁢π→assign to a nameMy
Obtain x&conjugate0;=My/m
My/m = 449418
Obtain y&conjugate0;=Mx/m
Mx/m = 0
Table 6.5.6(b) Calculation of the centroid from first principles
Maple Solution - Coded
Total mass
Define the density function ρ.
ρ≔1+r2 2 cos2θ+3 sin2θ/10:
Use the Int and value commands.
q≔Intr⋅ρ,r=0..1+cosθ,θ=0..2 π
∫02⁢π∫01+cos⁡θr⁢110+110⁢r2⁢2⁢cos⁡θ2+3⁢sin⁡θ2ⅆrⅆθ
m≔valueq
209320⁢π
First Moments
q≔Intr2⋅ρ⋅sinθ,r=0..1+cosθ,θ=0..2 π
∫02⁢π∫01+cos⁡θr2⁢110+110⁢r2⁢2⁢cos⁡θ2+3⁢sin⁡θ2⁢sin⁡θⅆrⅆθ
Mx≔valueq
q≔Intr2⋅ρ⋅cosθ,r=0..1+cosθ,θ=0..2 π
∫02⁢π∫01+cos⁡θr2⁢110+110⁢r2⁢2⁢cos⁡θ2+3⁢sin⁡θ2⁢cos⁡θⅆrⅆθ
My≔valueq
449640⁢π
Coordinates of the center of mass x&conjugate0;,y&conjugate0;
Mym,Mxm = 449418,0
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