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

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Homework Help Overview

The discussion revolves around the mathematical treatment of differential elements in calculus, specifically addressing the expression (dr)^2 in the context of area calculations involving a cylinder. The original poster seeks clarification on why (dr)^2 can be considered equal to 0 in certain limits.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the implications of setting (dr)^2 to zero in their area differential equation. Some participants question the conditions under which (dr)^2 can be ignored, while others explore the relationship between dr and its square in the context of integration.

Discussion Status

Participants are engaging in a productive exploration of the mathematical concepts involved. Some guidance has been offered regarding the treatment of (dr)^2 as negligible in the limit as dr approaches zero, but no consensus has been reached on the broader implications of this treatment.

Contextual Notes

The discussion includes considerations of the limits involved in integration and the specific setup of the problem related to a cylinder with varying radii. There is an acknowledgment of the need to integrate over a defined range to find the total area.

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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).
 
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(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.
 
Dick said:
(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 ?
 
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?
 

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