# Integral involving population density

1. Aug 18, 2008

### Andrusko

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

Hey all I've got a problem to do with population density. It asks for you to show the number of people living between arbitrary points a and b from the centre of the town is equal to:

$$P(a,b)=2\pi\int^{b}_{a}rf(r)dr$$

Where f(r) is the population density. Note that a is not the centre of the town, but a distance out from it, with a < b.

So I have stated that the number of people living in this area will be the population density multiplied by area. Easy enough. I have then partitioned up the large ring in between a and b into small rings.

Here's what I have so far:

$$Area_{i} = \pi(r^{2}_{i}-r^{2}_{i-1})$$

and

$$\Delta r_{i} = r_{i}-r_{i-1}$$

and it follows that

$$Population \approx \sum^{n}_{i=1}f(r^{i}_{*})\pi(r^{2}_{i}-r^{2}_{i-1})$$

My question is, how do I convert this to an integral if there's no $$\Delta r$$ involved?

I do realize that as the rings get really small they can be approximated as circumferences but this still doesn't help me get the term I need.

Thanks for any help.

2. Aug 18, 2008

### tiny-tim

Welcome to PF!

Hi Andrusko! Welcome to PF!

(I assume you mean the number of people living between arbitrary distances a and b from the centre of the town? )

No, using $$\text{Area}_{i} = \pi(r^{2}_{i}-r^{2}_{i-1})$$ is just a nuisance, and it doesn't incorporate d-anything.

You have correctly sliced up the area … in this case, into rings.

ok, call the thickness of each ring dr …

then its area (to first order of magnitude) is … ?

and so the total area is ∫ … dr ?

3. Aug 18, 2008

### Andrusko

Ah, so I should treat it like a rectangle and go:

$$Area_{i} = 2\pi r^{*}_{i} \Delta r_{i}$$

Then the Riemann sum becomes:

$$Population \approx \sum^{n}_{i=1} f(r^{*}_{i})2\pi r^{*}_{i} \Delta r_{i}$$

Which in the limit as $$\Delta r_{i} \rightarrow 0$$ becomes:

$$Population = \int^{b}_{a} f(r)2\pi rdr$$
$$Population = 2\pi\int^{b}_{a} rf(r)dr$$

This seems to have solved it. Thankyou for the help.