Moment of inertia about the origin for the lamina

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SUMMARY

The moment of inertia about the origin for a lamina defined by the surface of the sphere \(x^2 + y^2 + z^2 = 9\) with the condition \(z > 2\) is calculated to be \(16\pi k\), where \(k\) represents the area density. The discussion highlights errors in the initial calculations, particularly the omission of a square root in the expression for \(z\) and unnecessary substitutions that complicate the integration process. The correct approach involves integrating over circular bands of constant \(z\), leading to a straightforward integral that confirms the result of \(16\pi k\).

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fonseh
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Homework Statement


find the moment of inertia about the origin for the lamina which the surface of sphere (x^2) + (y^2) +(z^2) = 9 . z>2 . Given that density is a constant . Here's my wroking

The ans is 16pi (k) , but my ans is different , is my ans wrong ?
If so , which part is wrong ?
I have checked thru the working many times, yet , still can't find the error

Homework Equations

The Attempt at a Solution

 

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Note that a moment of inertia is about an axis, not a point. In this case I assume you meant the z axis.

I haven't worked right through your answer, and it contains some errors, for example the expression for ##z## is missing a square root, which however seems to have been assumed below. You then appear to have done a substitution which doesn't seem necessary and makes it more complicated.

I would do this question by integrating over circular bands of constant ##z##, where the mass of each band is ##2 \pi r## times its width which is ##\sqrt{dz^2 + dr^2}## where obviously you need to express ##dr## in terms of ##dz## before you integrate it. The resulting integral is simple. I make the answer ##16 \pi k## where ##k## is the area density.
 

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