Chapter 6: Applications of Double Integration
Section 6.5: First Moments
Example 6.5.10
Determine the coordinates of the center of mass of the lamina whose density is ρ = 3 + 2 x + 3 y , and whose shape is that of R , the region bounded by the ellipse x 2 + 4 y 2 = 1 .
Solution
Mathematical Solution
Figure 6.5.10(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; = 1 / 6 , 1 / 16 . The relevant calculations are tabulated to the left of the figure, where ρ = 3 + r 2 cos θ + 3 sin θ and φ θ = 1 / 4 − 3 cos 2 θ .
m = ∫ 0 2 π ∫ 0 φ θ r ⋅ ρ ⅆ r ⅆ θ = 3 2 π
M x = ∫ 0 2 π ∫ 0 φ θ r 2 ⋅ ρ sin θ ⅆ r ⅆ θ = 3 32 π
M y = ∫ 0 2 π ∫ 0 φ θ r 2 ⋅ ρ cos θ ⅆ r ⅆ θ = π 4
x &conjugate0; = M y / m = 1 / 6
y &conjugate0; = M x / m = 1 / 16
Figure 6.5.10(a) Center of mass, R , and ρ
Maple Solution - Interactive
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The total mass is calculated in Table 6.5.10(a) via a task template that integrates over an ellipse via the modified int command in the Student VectorCalculus package.
Tools≻Tasks≻Browse:
Calculus - Vector≻Integration≻Multiple Integration≻2-D≻Over an Ellipse
Integrate f x , y over an Ellipse
f x , y =
Equation of ellipse:
From θ = to θ =
Table 6.5.10(a) Total mass calculated with a task template that integrates over an ellipse
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The ellipse is expressed in polar coordinates as
1 2 tan θ 2 + 1 1 4 + tan θ 2
= sec 2 θ 1 + 4 tan 2 θ
= 1 / cos 2 θ + 4 sin θ 2
= 1 / cos 2 θ + 4 1 − cos 2 θ
= 1 / 4 − 3 cos 2 θ
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Similarly, the first moments M x = 3 π / 32 and M y = π / 4 could be obtained with this same task template. Alternatively, as shown in Table 6.5.10(b), calculate the coordinates of the center of mass via the task template that implements (in polar coordinates) the CenterOfMass command from the Student MultivariateCalculus package. The density is then ρ = 3 + r 2 cos θ + 3 sin θ .
Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Center of Mass≻Polar
Center of Mass for Planar Region in Polar Coordinates
Density:
>
3 + r 2 cos θ + 3 sin θ
3 + r 2 cos θ + 3 sin θ
(1)
Region: r 1 θ ≤ r ≤ r 2 θ , a ≤ θ ≤ b
r 1 θ
r 2 θ
a
b
Moments ÷ Mass:
Inert Integral - dr d θ
>
Student MultivariateCalculus CenterOfMass , r = .. , θ = .. , coordinates = polar r , θ , output = integral
∫ 0 2 π ∫ 0 1 4 − 3 cos θ 2 cos θ r 2 3 + r 2 cos θ + 3 sin θ ⅆ r ⅆ θ ∫ 0 2 π ∫ 0 1 4 − 3 cos θ 2 r 3 + r 2 cos θ + 3 sin θ ⅆ r ⅆ θ , ∫ 0 2 π ∫ 0 1 4 − 3 cos θ 2 sin θ r 2 3 + r 2 cos θ + 3 sin θ ⅆ r ⅆ θ ∫ 0 2 π ∫ 0 1 4 − 3 cos θ 2 r 3 + r 2 cos θ + 3 sin θ ⅆ r ⅆ θ
(6)
Explicit values for r &conjugate0; and θ &conjugate0;
>
Student MultivariateCalculus CenterOfMass , r = .. , θ = .. , coordinates = polar r , θ
1 48 73 , arctan 3 8
(7)
Plot:
>
Student MultivariateCalculus CenterOfMass , r = .. , θ = .. , coordinates = polar r , θ , output = plot , caption =
Table 6.5.10(b) Calculation of center of mass via task template
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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.
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The polar coordinates of the centroid are therefore r &conjugate0; , θ &conjugate0; = , which, in Cartesian coordinates would be r &conjugate0; cos θ &conjugate0; , r &conjugate0; sin θ &conjugate0; = 1 / 6 , 1 / 16 . In other words, the task template calculates the mass and first moments in Cartesian coordinates, but returns the center of mass in polar coordinates.
Maple Solution - Coded
Initialize
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Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
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Define the density function ρ .
ρ ≔ 3 + 2 x + 3 y :
Obtain m , the total mass
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Use the MultiInt command in the Student MultivariateCalculus package.
MultiInt ρ , x , y = Ellipse x 2 + 4 y 2 = 1 , output = integral ; m ≔ MultiInt ρ , x , y = Ellipse x 2 + 4 y 2 = 1
3 2 π
Obtain M x and M y , the first moments
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Use the MultiInt command in the Student MultivariateCalculus package.
MultiInt y ⋅ ρ , x , y = Ellipse x 2 + 4 y 2 = 1 , output = integral ; M x ≔ MultiInt y ⋅ ρ , x , y = Ellipse x 2 + 4 y 2 = 1
3 32 π
MultiInt x ⋅ ρ , x , y = Ellipse x 2 + 4 y 2 = 1 , output = integral ; M y ≔ MultiInt x ⋅ ρ , x , y = Ellipse x 2 + 4 y 2 = 1
1 4 π
Obtain x &conjugate0; , y &conjugate0; , the coordinates of the center of mass
M y m , M x m = 1 6 , 1 16
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