# Integrating the area of a disc

1. Homework Statement

Finding the area of a disc by integration of rings.

2. Homework 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 $$\pi (2rdr + (dr)^2)$$?

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tiny-tim
Homework Helper
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
Homework Helper
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.

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?

Dick
Homework Helper
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.

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.

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