Chapter 8: Applications of Triple Integration
Section 8.4: Moments of Inertia (Second Moments)
Example 8.4.1
If R is the region interior to the cylinder x2+y2=9 that is bounded below by the xy-plane, and above by the paraboloid z=x2+y2, and δr,θ,z=z r2sinθ/6 is the density in R, obtain the moments of inertia and the radii of gyration about the Cartesian coordinate-axes.
(See Example 8.1.3.)
Solution
Maple Solution - Interactive
Initialize
Context Panel: Assign Name
δ=z r2sinθ/6→assign
The calculations for the moments of inertia are detailed in Table 8.4.1(a) where the iterated integrals are a modification of the contents of Table 8.1.3(c).
Context Panel: Evaluate and Display Inline
Context Panel: Approximate≻5 (digits)
Ix=∫02 π∫03∫0r2δ r2sin2θ+z2 r ⅆz ⅆr ⅆθ→assign
Ix = 43099865111440→at 5 digits37675.
Iy=∫02 π∫03∫0r2δ r2cos2θ+z2 r ⅆz ⅆr ⅆθ→assign
Iy = 43029006311440→at 5 digits37613.
Iz=∫02 π∫03∫0r2δ r2 r ⅆz ⅆr ⅆθ→assign
Iz = 17714720→at 5 digits8857.4
Table 8.4.1(a) Calculations for the moments of inertia
The total mass m and the radii of gyration are given in Table 8.4.1(b).
m=∫02 π∫03∫0r2δ r ⅆz ⅆr ⅆθ→assign
m = 1968316
kx=Ix/m→assign
kx = 3715⁢1739595→at 5 digits5.5338
ky=Iy/m→assign
ky = 3715⁢1736735→at 5 digits5.5292
kz=Iz/m→assign
kz = 65⁢5→at 5 digits2.6833
Table 8.4.1(b) Radii of gyration
Maple Solution - Coded
Define the density.
δ≔z r2sinθ/6:
Obtain the moments of inertia
Qx≔Intr δ r2sin2θ+z2,z=0..r2,r=0..3,θ=0..2 π
∫02⁢π∫03∫0r2r3⁢z⁢sin⁡16⁢θ⁢r2⁢sin⁡θ2+z2ⅆzⅆrⅆθ
Ix≔valueQx
43099865111440
Qy≔Intr δ r2cos2θ+z2,z=0..r2,r=0..3,θ=0..2 π
∫02⁢π∫03∫0r2r3⁢z⁢sin⁡16⁢θ⁢r2⁢cos⁡θ2+z2ⅆzⅆrⅆθ
Iy≔valueQy
43029006311440
Qz≔Intr δ r2,z=0..r2,r=0..3,θ=0..2 π
∫02⁢π∫03∫0r2r5⁢z⁢sin⁡16⁢θⅆzⅆrⅆθ
Iz≔valueQz
17714720
Obtain the total mass m
M≔Intr δ,z=0..r2,r=0..3,θ=0..2 π
∫02⁢π∫03∫0r2r3⁢z⁢sin⁡16⁢θⅆzⅆrⅆθ
m≔valueM
1968316
Obtain the radii of gyration
kx≔Ix/m
3715⁢1739595
ky≔Iy/m
3715⁢1736735
kz≔Iz/m
65⁢5
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