Why isn't the area of a ring \pi (2rdr + (dr)^2)?

[Hakins90]

Homework Statement

Finding the area of a disc by integration of rings.

Homework Equations

A ring of radius r and thickness dr has an area of 2 \pi rdr.

The Attempt at a Solution

Why isn't it \pi (2rdr + (dr)^2)?

 

[tiny-tim]

Why isn't it \pi (2rdr + (dr)^2)?

I don't understand - why do you think it might be?

What would \pi (dr)^2 represent?

 

[Dick]

The exact area of the ring is pi*((r+dr)^2-r^2)=pi(2*r*dr+dr^2), indeed. But in the integration we are taking the limit of a sum of these where dr->0. The limit of the terms involving dr^2 will go to zero. You can ignore corrections involving higher powers of 'infinitesimals' like dr.

 

[Hakins90]

Oh dear... i posted too quickly.

I just realized why it isn't.

My old reasoning was that the ring's area = the area enclosed by the outside circumference - the area enclosed by the inside circumference.

A = \pi (r+dr)^2 - \pi r^2 <br /> = \pi (r^2 + 2rdr + (dr)^2 - r^2)<br /> = \pi (2rdr + (dr)^2)I now realize that the ring's radius is measured from the centre of the width of the ring.

so
A = \pi (r+ \frac{1}{2} dr)^2 - \pi (r- \frac{1}{2} dr)^2<br /> =\pi (r^2 +rdr + \frac{1}{4} (dr)^2 - r^2 + rdr - \frac{1}{4} (dr)^2)<br /> = 2 \pi rdr

EDIT: Oh you posted before i saw... We both have different reasons. Who is right?

 

[Dick]

If r and dr are the same, then those are the areas of two different rings. If dr is a finite number then you'd better pay attention to which is which. But,either calculation works for integration once you ignore higher powers of dr.

 

[Google_Spider]

Oh dear... i posted too quickly.

I just realized why it isn't.

My old reasoning was that the ring's area = the area enclosed by the outside circumference - the area enclosed by the inside circumference.

A = \pi (r+dr)^2 - \pi r^2 <br /> = \pi (r^2 + 2rdr + (dr)^2 - r^2)<br /> = \pi (2rdr + (dr)^2)

This reasoning is correct, in my opinion.
When I was in school, the explanation I got from my teacher was that while doing integration we add infinitesimally small quantities. The quantity dr is very small and so dr^2 will be even smaller...like 0.000001^2=0.000000000001 , so we neglect the dr^2 thing from that expression.

 

[HallsofIvy]

The simplest way to do the original problem is to say that the area of a small ring, of inner radius r and thickness dr is approximately 2\pi r, the length of the ring, times dr, the thickness. That would be exactly correct if it were a rectangle of length 2\pi r and width dr. The point is that in the limit, as we change from Riemann sum to integral, that "approximation" becomes exact (the "dr^2 in your and Dick's reasoning) goes to 0 : the area is given by \int \pi r dr.

Can someone explain to my why this is a physics problem and not a mathematics problem?