Chapter 6: Applications of Double Integration
Section 6.6: Second Moments
Example 6.6.2
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 in Example 6.6.1 (the region bounded by y=1−x2 and the x-axis), and whose density is equal to the square of the distance from the point 0,1/2. See Example 6.5.2.
Solution
Mathematical Solution
Table 6.6.2(a) summarizes the requisite calculations.
m=∫−11∫01−x2ρ ⅆy ⅆx = 38⁢π−23
Ix=∫−11∫01−x2ρ⋅y2 ⅆy ⅆx = 1196⁢π−415
Iy=∫−11∫01−x2ρ⋅x2 ⅆy ⅆx = 1196⁢π−215
Rx=Iym=1196⁢π−21538⁢π−23 = 55⁢π−64259⁢π−16 ≐ 0.067
Ry=Ixm=1196⁢π−41538⁢π−23 = 55⁢π−128259⁢π−16 ≐0.43
Table 6.6.2(a) Moments of inertia and radii of gyration
Maple Solution - Interactive
Define the density ρ
Context Panel: Assign Name
ρ=x2+y−1/22→assign
Obtain the total mass m
Calculus palette: Iterated double-integral template
Context Panel: Evaluate and Display Inline
Context Panel: Assign to a Name≻m
∫−11∫01−x2ρ ⅆy ⅆx = 38⁢π−23→assign to a namem
Obtain the second moments
Context Panel: Assign to a Name≻Ix (or Iy, as appropriate)
∫−11∫01−x2ρ⋅y2 ⅆy ⅆx = 1196⁢π−415→assign to a nameIx
∫−11∫01−x2ρ⋅x2 ⅆy ⅆx = 1196⁢π−215→assign to a nameIy
Obtain the radii of gyration
Expression palette: Square-root template Press the Enter key.
Context Panel: Simplify≻Simplify
Context Panel: Approximate≻10 (digits)
Rx=Iy/m
Rx=1196⁢π−21538⁢π−23
= simplify
Rx=110⁢5⁢55⁢π−649⁢π−16
→at 10 digits
Rx=0.6656956599
Ry=Ix/m
Ry=1196⁢π−41538⁢π−23
Ry=110⁢5⁢55⁢π−1289⁢π−16
Ry=0.4271347563
Maple Solution - Coded
Initialize
Define the density ρ.
ρ≔x2+y−1/22:
Use the top-level int command.
m≔intρ,y=0..1−x2,x=−1..1
38⁢π−23
Ix≔intρ⋅y2,y=0..1−x2,x=−1..1
1196⁢π−415
Iy≔intρ⋅x2,y=0..1−x2,x=−1..1
1196⁢π−215
Rx=simplifysqrtIym;Rx=evalfsqrtIym
Ry=simplifysqrtIxm;Ry=evalfsqrtIxm
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