Inertia Products: Solve Ixy, Iyz, Izx

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Homework Help Overview

The discussion revolves around calculating the moments of inertia (Ixy, Iyz, Izx) for a given shape, as indicated in the problem statement. The subject area is primarily focused on the application of triple integrals in the context of physics and engineering mechanics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the method of evaluating triple integrals for moments of inertia, with some questioning how to reduce the problem to a single variable. There are also inquiries about the correctness of integration ranges and the definitions of differential mass elements.

Discussion Status

Some participants have provided guidance on evaluating the integrals and confirmed the ranges of integration. However, there is a lack of explicit consensus on the correctness of the initial moments of inertia calculations, with some corrections being suggested.

Contextual Notes

There is a mention of specific integration ranges and the distinction between differential volume and mass elements, indicating a focus on the correct setup for the integrals. The discussion reflects the complexity of the problem due to the geometric constraints of the shape involved.

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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