As I recall, the moment of inertia is
∫∫∫ r^2 p dv
which in this case would indeed be
∫∫∫ r^2 * kr dv
and dv = r dr dθ dz
so it looks like you're good to go, except you dropped that extra r. See it hiding there in your original integrand?
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]∫[0,H]∫[0,2]∫𝑟^2(kr)rdzdrd𝜃=> (2𝜋)k(𝑅^3 /3)H = (2𝜋Hk𝑅^3)/3
I = m= [0,R]∫[0,H]∫[0,2]∫(kr)rdzdrd𝜃=> (2𝜋)k(𝑅)H = 2𝜋Hk𝑅
R=I/m=>R= ((2𝜋Hk𝑅^3)/3)/(2𝜋Hk𝑅)=answer R^2
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