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Find the moment of inertia of a right circular cylinder of radius of base R and height H, about the axis of the cylinder, if the density at the point P is proportional to the distance from P to the axis of the cylinder. Write down the resultin terms of the mass of the cylinder.

did I do this right and do you think my answer correct?

formulais I/m=R, denisty=kr
I = I= [0,R]โˆซ[0,H]โˆซ[0,2]โˆซ๐‘Ÿ^2(kr)rdzdrd๐œƒ=> (2๐œ‹)k(๐‘…^3 /3)H = (2๐œ‹Hk๐‘…^3)/3
I = m= [0,R]โˆซ[0,H]โˆซ[0,2]โˆซ(kr)rdzdrd๐œƒ=> (2๐œ‹)k(๐‘…)H = 2๐œ‹Hk๐‘…
R=I/m=>R= ((2๐œ‹Hk๐‘…^3)/3)/(2๐œ‹Hk๐‘…)=answer R^2

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1 answer
  1. As I recall, the moment of inertia is
    โˆซโˆซโˆซ r^2 p dv
    which in this case would indeed be
    โˆซโˆซโˆซ r^2 * kr dv
    and dv = r dr dฮธ dz
    so it looks like you're good to go, except you dropped that extra r. See it hiding there in your original integrand?

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