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Area of a circle is addition of circumference of layers of a onion. If n is radius of a onion then area is

$$ A = 2 \pi \cdot 1 + 2 \pi \cdot 2 + 2\pi \cdot 3 + \ldots + 2 \pi \cdot n $$ which

$$ = 2 \pi \cdot \frac{n(n+1)}{2} = \pi (n^2 + n) $$

But the answer is wrong. Please tell me where am I wrong?

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  • $\begingroup$ Why area of $n$-th layer is $2\pi n$? $\endgroup$ – Michael Galuza Jun 27 '15 at 10:42
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    $\begingroup$ You're confusing circumference of a layer with its area $\endgroup$ – uranix Jun 27 '15 at 10:44
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    $\begingroup$ @DanielV why is that ? (2 *r - 1) instead of just 2 * r? $\endgroup$ – Pointer Jun 27 '15 at 11:10
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    $\begingroup$ I think that the thickness of each layer is not 1. Try reformulating after viewing this animated gif 38.media.tumblr.com/2445fc904426c259dfd9f36b41219827/… $\endgroup$ – John Joy Jun 27 '15 at 15:25
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    $\begingroup$ Just throwing this out there, if you're curious about more of this stuff (proofs in early mathematics), Journey Through Genius is a really great book for this kind of stuff, and an easy read. $\endgroup$ – Alex Mathers Jun 28 '15 at 20:57
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Well, let's take a close look at what you're doing. Right now, if you want to calculate the area of a circle of radius $4$, you're summing up the circumferences in the following picture, times the "thickness" of each skin, which is $1$:
$\hskip1in$ enter image description here
However, this method has a problem: Each circumference represents the area of everything inside it, but before the next circle. So, for instance, the circumference of the red circle is approximating the area in which the blue circle lies in the following image:
$\hskip1.1in$enter image description here
which is troublesome because the red circumference is bigger than the blue circumference - which is a problem, meaning your method is an overestimate.

More rigorously speaking, we need that the area of a circle is invariant of how we slice it - if we divide a circle into rings of half the thickness as follows:
$\hskip1.1in$enter image description here
then your method suggest less area - in particular, if the original area your method obtained was $\pi (n^2+n)$, then the new area is $\pi (n^2+\frac{n}2)$ (A quick way to get this result without summing a new series is to see that this is the same as doubling $n$, and taking a quarter of the area thus calculated). This should trouble us some, because the area of a circle shouldn't depend on how we measure it - and it should scale so that doubling the radius quadruples the area. However, if we subdivided each of the original rings into $c$ sections, we'd get an area of $\pi (n^2+\frac{n}c)$ and, as we increase $c$, we get closer to the correct area of $\pi n^2$. In a rigorous sense, we may consider the process of dividing finer and finer sections of the circles to be an example of integration - but we can avoid such such tools by refining our geometric approach.

In particular, instead of estimating area of a slice as the outer ring times its thickness, we can estimate the area of a slice by a circle through the "midpoint" of the ring times the thickness - that is, we use the dotted lines in the following diagram to estimate the area of the sections through which they pass:
$\hskip1.1in$enter image description here
which yields the sum: $$2\pi \cdot \frac{1}2 + 2\pi \cdot \frac{3}2 +2\pi \cdot \frac{5}2 +\ldots + 2\pi \cdot \frac{2n-1}2 $$ which equals $$\pi n^2.$$ This happens to be the correct answer - and we can convince ourselves of this through various methods. The most elementary is to consider that if we make finer and finer* divisions as before, this value does not change - hence, it must be the correct value as any other value would be shifted towards the actual value as we do this. Calculus also tells us that since the circumference of a circle increases linearly with its radius and we are integrating over this, the amount we over/under estimate by choosing the center balances out. In any case, this yields the correct answer and gives a plausible geometric reason for it to be so.

*We can also make more coarse divisions - we could have one division and guess $2\pi \cdot \frac{n}2 \cdot n$ as the circumference of a circle of half the radius times the "thickness" of the whole circle.

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I like this (entirely non-rigorous) approach:

Area of circle

(The image has been taken from this article on Wikipedia)

It should be "intuitively obvious" that increasing the number of sectors makes the approximation better and better.

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    $\begingroup$ I have a vivid image of myself seeing this argument in primary school... and strongly objecting to it. With no calculus knowledge, the crucial part "the error approaches zero" was described only in vague terms (and would have been beyond me, anyway). I kept insisting that no matter how large the number of slices, these were no triangles. My poor teacher probably hated me for that... $\endgroup$ – chi Jun 28 '15 at 22:01
  • $\begingroup$ How pi is calculated here? $\endgroup$ – Mr_Green Jun 29 '15 at 9:28
  • $\begingroup$ @Mr_Green This method doesn't calculate the value of pi (nor do any of the other answers). We only need to know that pi is the ratio between the circumference and diameter; the numerical value of pi doesn't play a role in the formula derivation. $\endgroup$ – Aky Jul 1 '15 at 18:02
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The problem is that your onion doesn't have enough layers. If you were to integrate the circumference function $2\pi r$, which is equivalent to making infinitely many layers, you would indeed get $$2\pi \cdot \frac{1}{2}r^2=\pi r^2$$ However, I'm not sure if you call integration "basic math".


Alternatively, you can see a circle as a regular $n$-gon with $n \rightarrow \infty$ sides and side length $k$. This polygon is approximated by its circumscribed circle with radius $r$. Of course, we can divide the polygon into $n$ triangles. We can approximate $k$ by $\frac{2\pi r}{n}$. The height is $r$. Therefore the total area of such a triangle is $\frac{1}{2}\frac{2\pi r}{n}r=\frac{\pi r^2}{n}$ therefore the total area of the polygon and thus the circle is $\pi r^2$. Of course, to write it down rigorously, we also have to consider the inscribed circle and write down bounds etcetera, but this is the idea.

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    $\begingroup$ Yes, I don't call integration basic math. 2nd answer is cool but i would like to know the area using addition of layers of onion. Thanks. $\endgroup$ – Pointer Jun 27 '15 at 11:17
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    $\begingroup$ @Pointer You can't get to the area of the circle by summing the circunference of a finite number of layers of onion, as you say. You have to use "infinitely many" layers and compute the resulting integral. No way around that :P $\endgroup$ – Ant Jun 27 '15 at 12:30
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You need to add areas, not lengths. So you approximate the area of the layers by multiplying the circumference times a small radial lenght: $$ \left (2\pi\,\frac rn\right)\,\frac rn,\ \ \left (2\pi\,\frac {2r}n\right)\,\frac rn,\ \ \ldots \ \ ,\left (2\pi\,\frac {nr}n\right)\,\frac rn. $$ The sum is $$ \frac {2\pi r^2}{n^2}\, (1+2+\cdots+n)=\pi r^2\,\left (1+\frac1n\right). $$

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The error is in neglecting that the radius at the outside is larger than the radius at the inside; yet you just use the radius at the outside for calculating the onion layer's area. The mistake is most obviously seen in the innermost layer, which is just the circle of radius $1$. The inner border of that layer is the central point of the circle, which has length zero! Consequently, the area you calculate for the innermost layer is $2\pi\cdot 1 = 2\pi$ while its actual area is $\pi\cdot 1^2 = \pi$, so you overstate the area of that layer by a factor of two!

OK, so if you take the outer circumference of your layer, you obviously get a result that's too large. On the other hand, if you take the inner circumference of your layer, you get a result that's too small. So what circumference to take? Well, if the innermost circle is too small and the outermost circle is too large, then obviously you need the circumference of some circle in between.

So let's just take the circle in the middle. For the $n$-th onion layer that has the radius $n-\frac12$, and therefore you now get $$A = 2\pi\cdot\left(1-\frac12\right) + 2\pi\cdot\left(2-\frac12\right) + \ldots = 2\pi \frac{n(n+1)}{2} - n\cdot 2\pi\cdot\frac{1}{2} = \pi n^2$$

OK, but how do we now know that this is the correct value? After all, we could have chosen any other value?

Well, it makes sense that the radius of the circle to take is always at some fixed fraction between inner and outer radius, which means that for your thickness $1$ layers you'll always have a radius of $n-x$, with $x$ a number between $0$ and $1$.

Now let's look at the two innermost layers. If we take just one layer, we see that the layer is itself just the circle of radius 1. Its area is then, accoirding to out formula, $A_1=2\pi(1-x)$ If, instead, we take the first two layers, we see that we get the circle of radius two. Now the circle of radious two is just the circle of radius $1$ dilated to double size. Elementary geometry tells us that it has thus $2^2=4$ times the area of the original circle, thus its area is $$A_2=4A_1=8\pi(1-x)$$ On the other hand, the "onion formula" tells us that its area is $$A_2=2\pi(1-x) + 2\pi(2-x) = 2\pi(3-2x)$$ The only way both formulas can be true at the same time is if $$8\pi(1-x) = 2\pi(3-2x)$$ which means $$x=\frac12$$

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You are not so far: the correct surface is $\pi n^2$. The ratio of your formula to the expected area is

$$\frac{n^2+n}{n^2}=1+\frac1n.$$

As you must take $n$ as large as possible, the ratio is $1$.


Note that your computation is approximate as your layers have a thickness of $1$ and their area is that of a ring, $\pi((n^2-(n-1)^2)=\pi(2n-1)$ and not $2\pi$, but as you increase $n$, the difference becomes negligible.

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  • $\begingroup$ +1, though it's not quite correct to say "you must take $n$ as large as possible". The issue is that the OP is using $n$ both for the radius and for the number of layers; the fix is to separate these, and increase the number of layers while holding the radius fixed. $\endgroup$ – ruakh Jun 28 '15 at 0:16
  • $\begingroup$ @ruakh: no, the radius can be increased, giving $\pi n^2$ for the circle area. I avoided saying take the limit for $n\to\infty$ on purpose to avoid technicality. $\endgroup$ – Yves Daoust Jun 28 '15 at 20:41

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