Chapter 8: Applications of Triple Integration
Section 8.4: Moments of Inertia (Second Moments)
Example 8.4.2
If R is the wedge the planes z=y and z=0 cut from the cylinder x2+y2=4, and δr,θ,z=r z2cosθ/3 is the density in R, obtain the moments of inertia and the radii of gyration about the Cartesian coordinate-axes.
(See Example 8.1.5.)
Solution
Maple Solution - Interactive
Initialize
Context Panel: Assign Name
δ=r z2cosθ/3→assign
The calculations for the moments of inertia are detailed in Table 8.4.2(a) where the iterated integrals are a modification of the contents of Table 8.1.5(c).
Context Panel: Evaluate and Display Inline
Context Panel: Approximate≻5 (digits)
Ix=∫0π∫02∫0r sinθδ r2sin2θ+z2 r ⅆz ⅆr ⅆθ→assign
Ix = 2187140→at 5 digits15.621
Iy=∫0π∫02∫0r sinθδ r2cos2θ+z2 r ⅆz ⅆr ⅆθ→assign
Iy = 4617560→at 5 digits8.2446
Iz=∫0π∫02∫0r sinθδ r2 r ⅆz ⅆr ⅆθ→assign
Iz = 24320→at 5 digits12.150
Table 8.4.2(a) Calculations for the moments of inertia
The total mass m and the radii of gyration are given in Table 8.4.2(b).
m=∫0π∫02∫0r sinθδ r ⅆz ⅆr ⅆθ→assign
m = 8120
kx=Ix/m→assign
kx = 37⁢21→at 5 digits1.9640
ky=Iy/m→assign
ky = 114⁢399→at 5 digits1.4268
kz=Iz/m→assign
kz = 3→at 5 digits1.7321
Table 8.4.2(b) Radii of gyration
Maple Solution - Coded
Define the density.
δ≔r z2cosθ/3:
Obtain the moments of inertia
Qx≔Intr δ r2sin2θ+z2,z=0..r sinθ,r=0..2,θ=0.. π
∫0π∫02∫0r⁢sin⁡θr2⁢z2⁢cos⁡13⁢θ⁢r2⁢sin⁡θ2+z2ⅆzⅆrⅆθ
Ix≔valueQx
2187140
Qy≔Intr δ r2cos2θ+z2,z=0..r sinθ,r=0..2,θ=0.. π
∫0π∫02∫0r⁢sin⁡θr2⁢z2⁢cos⁡13⁢θ⁢r2⁢cos⁡θ2+z2ⅆzⅆrⅆθ
Iy≔valueQy
4617560
Qz≔Intr δ r2,z=0..r sinθ,r=0..2,θ=0.. π
∫0π∫02∫0r⁢sin⁡θr4⁢z2⁢cos⁡13⁢θⅆzⅆrⅆθ
Iz≔valueQz
24320
Obtain the total mass m
M≔Intr δ,z=0..r sinθ,r=0..2,θ=0.. π
∫0π∫02∫0r⁢sin⁡θr2⁢z2⁢cos⁡13⁢θⅆzⅆrⅆθ
m≔valueM
8120
Obtain the radii of gyration
kx≔Ix/m
37⁢21
ky≔Iy/m
114⁢399
kz≔Iz/m
3
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