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This one is tough:

A straight road goes through the center of a circular city of radius $5\text{km}$. The density of the population at a distance $r$ is well represented by $D(r)=20-4r$ (in thousand people per $\text{km}^2$). Find the population of the city.

Am I correct in that this is problem involving a "centroid"? I am not sure how to set this problem up with with the information given.

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    $\begingroup$ You'll want a double integral that looks like $\int_0^{2\pi}\int_0^5 \text{(something)}r\mathrm dr\mathrm d\theta$... $\endgroup$
    – J. M. ain't a mathematician
    Dec 8, 2011 at 4:49
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    $\begingroup$ At a distance $r$ from where? The road or the centre?. If it is the centre, what does the road have to do with it? $\endgroup$
    – André Nicolas
    Dec 8, 2011 at 4:49
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    $\begingroup$ @J.M. Not really necessarily; I've seen this problem before in Calc II textbooks (that's before multiple integrals). $\endgroup$
    – Arturo Magidin
    Dec 8, 2011 at 5:00
  • $\begingroup$ @Dylan: Note, you aren't, as your title claims, using integration to find population density, you are using integration to find total population. $\endgroup$
    – Arturo Magidin
    Dec 8, 2011 at 5:05
  • $\begingroup$ @Arturo: I didn't realize, sorry. I just wrote out why I'd have done if I were given it... $\endgroup$
    – J. M. ain't a mathematician
    Dec 8, 2011 at 5:24

2 Answers 2

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Assuming this is the kind of standard problem:

Imagine the circular city. Divide it into $n$ annuli of very small width, $\Delta r$. The density of population in an annular region is almost constant. If we are $r_i$ away from the center on the inner edge of the annular region, the density is well approximated by $D(r_i) = 20-4r_i$ thousand people per square kilometer. The area of the annular region is also well-approximated by "slicing it open and stretching it out", which will give you a shape that is very close to a rectangle of height $\Delta r$, and of width $2\pi r_i$, so the area is approximately $2\pi r_i\Delta r$ square kilometers. So the population in the annular region just described can be approximated by: $$\text{Population in the }i\text{th region}\approx (20-4r_i)(2\pi r_i)\Delta r\text{ thousand people.}$$ Adding it up over all the annular regions we have that $$\text{Population of the city} \approx \sum_{i=1}^n (20-4r_i)(2\pi r_i)\Delta r.$$ If we take the limit as $n\to\infty$, the approximations gets better and better (both the area approximations and the density approximations), and the error goes to zero. So $$\text{Population of the city} = \lim_{n\to\infty}\left(\sum_{i=1}^n(20-4r_i)(2\pi r_i)\Delta r\right).$$

But these sums are Riemann sums of a particular function, and so the limit will equal an integral. Figure out what integral, and then performing the integration will give you the population.

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  • $\begingroup$ Thank You. Solving, I got $\frac {500 \pi}{3}$ thousand people per $km^2$. $\endgroup$
    – dtg
    Dec 8, 2011 at 7:02
  • $\begingroup$ @Dylan: Your units, at least, are wrong. $(20-4r_i)$ is measured in thousands of people per square km. $r_i$ is measured in km, as is $\Delta r$. That means that the units of the sum are $$\left(\frac{\text{thousands of people}}{\text{km}^2}\right)(\text{km})(\text{km}) = \text{thousands of people}.$$Your answer should be a population, not a density. $\endgroup$
    – Arturo Magidin
    Dec 8, 2011 at 15:30
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To find the population you can integrate:

Let $R=$ radius of city; Since $x^2 +y^2 = R^2$ ,then $y = \sqrt{25 - x^2}$; since you are only evaluating $\dfrac{1}{4}$ of the circle multiply the integral by $4$; let $r=x$; and you get the Integral $$\int_0^54\sqrt{25 - x^2}\cdot(20 - 4x) dx$$

I think this is correct. Let me know if I am wrong.

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    $\begingroup$ I make no remark about the correctness of the answer. But, here are some things to remember when you post in an answer: Use complete sentences. And, remember to use TeX in your answers. Equations are inserted by using $...$ while display equations are inserted by $$...$$. For some of the commaands, browse through the internet there are plenty of them. And, you may click on the time stamp above my name to see some changes that went into your original posts. Lastly, and joyfully, Welcome to MSE. $\endgroup$
    – user21436
    Mar 18, 2012 at 1:31

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