Chapter 6: Applications of Double Integration
Section 6.6: Second Moments
Example 6.6.9
Find the moments of inertial Ix and Iy, the total mass m, and the radii of gyration Rx and Ry of the lamina that
has the shape of R, the region bounded by the graphs of x=0, and x=y−y3, where y∈0,1. (Take ρ=1.)
See Example 6.5.9.
Solution
Mathematical Solution
The relevant calculations are in Table 6.6.9(a).
m=∫01∫0y−y3ⅆx ⅆy = 14
Ix=∫01∫0y−y3y2 ⅆx ⅆy = 112
Iy=∫01∫0y−y3x2 ⅆx ⅆy = 1120
Rx=Iy/m=1/1201/4=130 ≐ 0.58
Ry=Ix/m=1/121/4=13 ≐ 0.18
Table 6.6.9(a) Moments of inertia and radii of gyration
Maple Solution - Interactive
A solution from first principles is detailed in Table 6.6.9(b).
Obtain m, the total mass of the lamina
Calculus palette: Iterated double-integral template
Context Panel: Evaluate and Display Inline
Context Panel: Assign to a Name≻m
∫01∫0y−y31 ⅆx ⅆy = 14→assign to a namem
Obtain Ix, the moment of inertia about the x-axis
Context Panel: Assign to a Name≻Ix
∫01∫0y−y3y2 ⅆx ⅆy = 112→assign to a nameIx
Obtain Iy, the moment of inertia about the y-axis
Context Panel: Assign to a Name≻Iy
∫01∫0y−y3x2 ⅆx ⅆy = 1120→assign to a nameIy
Obtain Rx
Context Panel: Approximate≻10 (digits)
Iy/m = 130⁢30→at 10 digits0.1825741858
Obtain Ry
Ix/m = 13⁢3→at 10 digits0.5773502693
Table 6.6.9(b) Moments of inertia and radii of gyration
Maple Solution - Coded
A solution from first principles is provided in Table 6.6.9(c).
Obtain the total mass of the lamina
Display the unevaluated integral with the Int command, and evaluate the integral with the value command.
q≔Int1,x=0..y−y3,y=0..1
∫01∫0−y3+y1ⅆxⅆy
m≔valueq
14
Obtain the moments of inertia Ix and Iy
q≔Inty2,x=0..y−y3,y=0..1
∫01∫0−y3+yy2ⅆxⅆy
Ix≔valueq
112
q≔Intx2,x=0..y−y3,y=0..1
∫01∫0−y3+yx2ⅆxⅆy
Iy≔valueq
1120
Obtain the radii of gyration
Rx=sqrtIy/m;Rx=evalfsqrtIx/m
Rx=130⁢30
Ry=sqrtIx/m;Ry=evalfsqrtIy/m
Ry=13⁢3
Table 6.6.9(c) Moments of inertia and radii of gyration
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