Why is dV = 2πrh dr for cylinder moment of inertia?

[gboff21]

I have been deriving the moment of inertia of a solid cylinder and have got this far:
I=∫ r² dm
and dm=dv *density

h=height
r=radius
$$\rho$$=density

To get to the correct I=1/2mr². you need to make dv=2$$\pi$$rh dr
why isn't it dv= dv=$$\pi$$r²h
as in volume=cross-sectional area*height

 

[Delphi51]

Consider a circle on the end of the cylinder at radius r. If it goes all the way through the circle, its area is 2πrh. Let dV be this area times an incremental radius, dr:
dV = 2πrh*dr
Imagine it "unrolled" into a rectangular solid to see it clearly.

Don't forget the density.

 

[gboff21]

Thanks I've been struggling to get my head round that for a while!

 

[Delphi51]

Most welcome. Always draw a picture!