Inertia Products: Solve Ixy, Iyz, Izx

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SUMMARY

The discussion focuses on calculating the moments of inertia Ixy, Iyz, and Izx for a solid object using triple integrals. Participants clarify that the correct differential mass element is dM = k dV = k dx dy dz, where dV represents the volume element. The integration ranges for dx, dy, and dz are confirmed as valid, with specific limits provided for each variable. Participants emphasize the importance of correctly applying the limits of integration to ensure accurate results.

PREREQUISITES
  • Understanding of triple integrals in calculus
  • Familiarity with moments of inertia concepts
  • Knowledge of differential mass elements in physics
  • Basic geometry of hemispherical coordinates
NEXT STEPS
  • Study the derivation of moments of inertia for various geometries
  • Learn about spherical coordinates and their applications in integration
  • Explore the use of kinematic variables in mass distribution calculations
  • Review examples of triple integrals in physics problems
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Students and professionals in physics and engineering, particularly those involved in mechanics and materials science, will benefit from this discussion on calculating moments of inertia.

unscientific
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Homework Statement


http://i45.tinypic.com/hwcsy0.png

The Attempt at a Solution



I'm not sure how to find the rest; Ixy, Iyz and Izx...
Usually for integrals such as moments of inertia you will be able to reduce it to only one variable. However, there are 2 variables here; xy, yz and zx. How do i reduce it to only one?
 
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You don't. You evaluate the triple integrals.
 
vela said:
You don't. You evaluate the triple integrals.
Are my Ixx, Iyy and Izz are correct?

Oh, so it's dM = dx dy dz,

then the range for dx is from -√(a2-y2-z2) to √(a2-y2-z2)

for dy it's from -√(a2-z2) to √(a2-z2)

for dz it's from 0 to a

Are my ranges of integration right?
 
unscientific said:
Are my Ixx, Iyy and Izz are correct?
No, they're not. You can't say ##y^2+z^2=a^2-x^2## because that's true only on the spherical part of the surface. Inside the hemisphere, it doesn't hold.

Oh, so it's dM = dx dy dz,
That's dV, not dM. You should write dM = k dV = k dx dy dz.

then the range for dx is from -√(a2-y2-z2) to √(a2-y2-z2)

for dy it's from -√(a2-z2) to √(a2-z2)

for dz it's from 0 to a

Are my ranges of integration right?
Yes.
 

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