Finding moment of inertia of cone

[Vitani11]

Homework Statement

Why is it that when you integrate to find the moment of inertia of a cone standing on its vertex (like a spinning top) with height h mass M and radius R do you integrate the R limits as 0 to (R/h)z in the triple integral (cylindrical coordinates) below?

Homework Equations

I = moment of inertia
D = density (M/πR²h)
ρ = R distance from rotation axis (limits from 0 to (R\h)z)
φ = 2π the angle swept (limits 0 to 2π)
z = h the height of the cone (limits 0 to h)

The Attempt at a Solution

I = ∫ρ²dm = D∫∫∫ρ³dρdφdz

 

[haruspex]

I assume you mean the r limits, not R limits.
It depends on the order of integration. If the integration wrt r is the last step then the range is 0 to R. If it is an earlier step then the maximum value of r is constrained by the current value of z in the outer integral.

 

[Vitani11]

If it is an earlier step then the maximum value of r is constrained by the current value of z in the outer integral.

Yeah this is the case but I don't understand where it came from. Why doesn't it involve sines? How do you see it is (R/h)z from the picture? For the problem is was given as capital R for radius

 

[haruspex]

Yeah this is the case but I don't understand where it came from. Why doesn't it involve sines? How do you see it is (R/h)z from the picture? For the problem is was given as capital R for radius

R is the maximum radius, i.e. the radius at height h. For the integral, you need a variable for the radius at height z. r seems a reasonable choice.
If the angle of the cone (slope to vertical) is θ then tan(θ) = R/h = r/z.

 

[Vitani11]

Okay thanks