Proving moment of inertia of hollow sphere with y^2+x^2=R^2 method

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The discussion focuses on proving the moment of inertia of a thin-walled hollow sphere using the equation y^2 + x^2 = R^2. The original poster seeks a solution that avoids multivariable integrals and polar coordinates, except for trigonometric substitution. A participant notes the necessity of including the z^2 term to account for the third dimension in the calculation. The moment of inertia is expressed as I = ∫ r^2 dm, suggesting that dm can be defined in terms of the area of a ring and density. The integration should be performed from y = -R to +R to find the moment of inertia.
parsa418
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Could anyone prove the moment of inertia of a thin walled hollow sphere using the y^2 + x^2 = r^2. I have only studied up to single variable calculus. I can take regular integrals but not multivariable integrals. I don't want to use the angle method or any polar coordinate systems except in the part where I might use it for trigonometric substitution to evaluate the integral.
Thanks
 
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parsa418 said:
Could anyone prove the moment of inertia of a thin walled hollow sphere using the y^2 + x^2 = r^2. I have only studied up to single variable calculus. I can take regular integrals but not multivariable integrals. I don't want to use the angle method or any polar coordinate systems except in the part where I might use it for trigonometric substitution to evaluate the integral.
Thanks
You are missing a z^2 there for the third dimension.

The moment of inertia is simply:

I = \int r^2dm

Let dm be a slice of the sphere of thickness dy. Can you express dm in terms of area of that ring and density? Then integrate from y = -R to +R

AM
 
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