Why does (dr)^2 equal 0 in the differential element equation?

[Apphysicist]

This is not a homework problem in itself. In my physics homework, I wanted to write the difference of two areas (thus yielding a differential disk) as:

Pi*(r+dr)²-Pi*r²

It reduces to

2*Pi*r*dr+Pi*(dr)²

Now, I seem to recall from a prior class that a quick hand-waving made (dr)² = 0.
I would like to know why this is or is not the case. It'd be of great use to let it equal 0, but right now, I'm not seeing it (other than it's really really small).

 

[Dick]

(dr)^2 isn't zero unless dr=0. But it's vanishingly small compared with dr as dr->0. Which is the limit you do when you are integrating or differentiating. So, sure, you can ignore it.

 

[Apphysicist]

(dr)^2 isn't zero unless dr=0. But it's vanishingly small compared with dr as dr->0. Which is the limit you do when you are integrating or differentiating. So, sure, you can ignore it.

I guess it's partly the weird idea of integration with such a thing.

The actual problem involves a cylinder with circular cross-sectional area, inner radius a, outer radius b. I am working on the inside of the cylinder (a<r<b) and so that's why I'm doing this.

I want to integrate from a to r to find the total area enclosed as a function of r, so I'm thinking that I would have integral from a to r of 2*Pi*r*dr+Pi*(dr)². Since it is most definitely a single integral, that's why lim->inf (dr^2) =0 ?

 

[Dick]

Factor the 2*pi*r*dr out. So you've got 2*pi*r*dr*(1+dr/(2*r)). As dr->0 the second term becomes 1, yes?