Chapter 6: Applications of Double Integration
Section 6.5: First Moments
Example 6.5.2
Determine the coordinates of the center of mass of a lamina in the shape of R in Example 6.5.1 if its density is equal to the square of the distance from the point 0,1/2.
Solution
Mathematical Solution
Figure 6.5.2(a) shows the region R in red, and the surface ρ=x2+y−1/22, in blue. The green dot represents the center of mass x&conjugate0;,y&conjugate0;=0,1568−15 π9π−16. The relevant calculations are tabulated to the left of the figure.
m=∫−11∫01−x2ρ ⅆy ⅆx = 38π−23
Mx=∫−11∫01−x2ρ⋅y ⅆy ⅆx = 1730−18π
My=∫−11∫01−x2ρ⋅x ⅆy ⅆx = 0
x&conjugate0;=My/m=0
y&conjugate0;=Mx/m=1568−15 π9π−16
Figure 6.5.2(a) Region R, the CM, and ρ
Maple Solution - Interactive
Table 6.5.2(a) contains a solution via the task template that implements the CenterOfMass command from the Student MultivariateCalculus package. The density is ρ=x2+y−1/22.
Tools≻Tasks≻Browse:
Calculus - Multivariate≻Integration≻Center of Mass≻Cartesian 2-D
Center of Mass for Planar Region in Cartesian Coordinates
Density:
x2+y−1/22
x2+y−122
Region: ux≤y≤vx,a≤x≤b
ux
0
vx
1−x2
−x2+1
a
−1
b
1
Moments÷Mass:
Inert Integral - dy dx
StudentMultivariateCalculusCenterOfMass, x=..,y=..,output=integral
∫−11∫0−x2+1x2+y−122xⅆyⅆx∫−11∫0−x2+1x2+y−122ⅆyⅆx,∫−11∫0−x2+1x2+y−122yⅆyⅆx∫−11∫0−x2+1x2+y−122ⅆyⅆx
Explicit values for x&conjugate0; and y&conjugate0;
StudentMultivariateCalculusCenterOfMass,x=..,y=..
0,1730−18π38π−23
Plot:
StudentMultivariateCalculusCenterOfMass,x=..,y=..,output=plot,caption=
Table 6.5.2(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 ρx,y in blue. The center of mass is represented by the green dot.
The Cartesian coordinates of the center of mass are therefore x&conjugate0;,y&conjugate0;=0,1568−15 π9π−16.
A solution from first principles is detailed in Table 6.5.2(b).
Define the density function ρ
Context Panel: Assign Name
ρ=x2+y−1/22→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
∫−11∫01−x2ρ ⅆy ⅆx = 38π−23→assign to a namem
Obtain Mx, the total moments about the x-axis
Context Panel: Assign to a Name≻Mx
∫−11∫01−x2ρ⋅y ⅆy ⅆx = 1730−18π→assign to a nameMx
Obtain My, the total moments about the y-axis
Context Panel: Assign to a Name≻My
∫−11∫01−x2ρ⋅x ⅆy ⅆx = 0→assign to a nameMy
Obtain x&conjugate0;=My/m
My/m = 0
Obtain y&conjugate0;=Mx/m
Context Panel: Simplify≻Simplify
Context Panel: Approximate≻10 (digits)
Mx/m = 1730−18π38π−23= simplify −15−68+15π9π−16→at 10 digits0.3401587472
Table 6.5.1(b) Calculation of the center of mass from first principles
Maple Solution - Coded
Density ρx,y
Define the density function.
ρ≔x2+y−1/22:
Total mass
Use the Int and value commands.
q≔Intρ,y=0..1−x2,x=−1..1
∫−11∫0−x2+1x2+y2−y+14ⅆyⅆx
m≔valueq
38π−23
First Moments
q≔Intρ⋅y,y=0..1−x2,x=−1..1
∫−11∫0−x2+1x2+y−122yⅆyⅆx
Mx≔valueq
1730−18π
q≔Intρ⋅x,y=0..1−x2,x=−1..1
∫−11∫0−x2+1x2+y−122xⅆyⅆx
My≔valueq
Coordinates of center of mass x&conjugate0;,y&conjugate0;
Apply the simplify and evalf commands.
CM≔Mym,Mxm
simplifyCM
0,−15−68+15π9π−16
evalfCM
0.,0.3401587474
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