Proving the moment of inertia of a thinwalled hollow sphere

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Discussion Overview

The discussion revolves around proving the moment of inertia of a thin-walled hollow sphere using Cartesian coordinates, specifically through the equation y² + x² = r², rather than the conventional angle method. The scope includes mathematical reasoning and technical explanation related to surface integrals.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant questions the feasibility of using y² + x² = r² to prove the moment of inertia, suggesting a preference for Cartesian coordinates over polar coordinates.
  • Another participant corrects the first by stating that y² + x² = r² describes a circle, not a sphere, and proposes a surface integral to calculate the inertia.
  • A participant requests clarification on how the proposed integral was derived, seeking details on the setup process.
  • In response, a participant explains that the definition of inertia involves integrating distance squared times mass, indicating they calculated for one octant and multiplied by eight due to symmetry. They also mention a correction regarding a typo in the integral.

Areas of Agreement / Disagreement

The discussion contains disagreements regarding the appropriateness of using Cartesian coordinates and the formulation of the integral. There is no consensus on the method or the correctness of the proposed integral.

Contextual Notes

Participants express uncertainty about the setup of the integral and the definition of surface integrals, indicating potential limitations in understanding or applying these concepts.

parsa418
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I was wondering if there is any way to prove the moment of inertia of a thin walled hollow sphere by using y^2 + x^2 = r^2 instead of using the angle method. I want to use a Cartesian coordinate system not a polar coordinate system.
 
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First of all y^2 + x^2 = r^2 describes a circle not a sphere.
And yes, you can calculate inertia in Cartesian coordinates by the definition of inertia, you just have to evaluate this surface integral:

8\rho \int_{0}^{r}\int_{0}^{\sqrt{r^2-x^2}}\sqrt{\frac{x^2 r^2}{r^2 - x^2 - y^2}}dy dx

Tell me what you get.
 
Hi thank you for replying
could you please tell me how you got that integral exactly? as in how did you set it up?
 
use the definition of inertia, distance squared times mass. So integrate x^2 times each surface element. I did one octant and multiplied by 8 due to symmetry.

To integrate you need to know what a surface integral is, otherwise I can't help you much.
http://en.wikipedia.org/wiki/Surface_integral

Also I made a typo, the x^2 should be x^4 if its going to be inside the square root
 

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