Integrating the area of a disc

1. Mar 14, 2008

Hakins90

1. The problem statement, all variables and given/known data

Finding the area of a disc by integration of rings.

2. Relevant equations

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

3. The attempt at a solution

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

2. Mar 14, 2008

tiny-tim

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

What would $$\pi (dr)^2$$ represent?

3. Mar 14, 2008

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.

4. Mar 14, 2008

Hakins90

Oh dear... i posted too quickly.

I just realised 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.

$$A = \pi (r+dr)^2 - \pi r^2 = \pi (r^2 + 2rdr + (dr)^2 - r^2) = \pi (2rdr + (dr)^2)$$

I now realise that the ring's radius is measured from the centre of the width of the ring.

so
$$A = \pi (r+ \frac{1}{2} dr)^2 - \pi (r- \frac{1}{2} dr)^2 =\pi (r^2 +rdr + \frac{1}{4} (dr)^2 - r^2 + rdr - \frac{1}{4} (dr)^2) = 2 \pi rdr$$

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

5. Mar 14, 2008

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.

6. Mar 15, 2008

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.

7. Mar 15, 2008

HallsofIvy

Staff Emeritus
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?