Moment of Inertia of a cylinder with varying density

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SUMMARY

The discussion focuses on calculating the moment of inertia of a cylinder with a linearly varying density, defined as ρ = αr, where α is a constant. The user begins with the integral I = ∫r²dm and substitutes dm using the relationship between mass and volume. The integration process involves breaking the cylinder into thin rings and applying the volume formula, leading to an expression for I in terms of M and R. The final expression derived is I = α(Lr²)/2 + α(r³)/2, with integration bounds from R1 to R2.

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A cylinder with radius R and mass M has density that increases linearly with distance r from the cylinder axis, ρ = αr, where α is a positive constant.
Calculate the moment of inertia of the cylinder about a longitudinal axis through its center in terms of M and R.

I have I = ∫r^2dm
and ρ = αr = m/V, I solved for m and took the derivative to get dm = α(dV = dr).
Putting this into the Inertia equation, I get I = α∫r^2 (dV + dr). Using the equation for volume, i took the derivative and got dV = L * 2∏r dr. Plugging this into the original equation, and integrating, I get I = α(Lr^2)/2 + α(r^3)/2 with bounds of R2 and R1. Not sure where to go from here...
 
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dm=2πr L ρ dr - Because you can break the disk into thin rings distance r from center, each of which has volume 2πr L dr

M=∫dm

Try using these.
 

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