I understand the reasoning behind $\pi r^2$ for a circle area however I'd like to know what is wrong with the reasoning below:

The area of a square is like a line, the height (one dimension, length) placed several times next to each other up to the square all the way until the square length thus we have height x length for the area.

The area of a circle could be thought of a line (The radius) placed next to each other several times enough to make up a circle. Given that circumference of a circle is $2 \pi r$ we would, by the same reasoning as above, have $2 \pi r^2$. Where is the problem with this reasoning?

Lines placed next to each other would only go straight like a rectangle so you'd have to spread them apart in one of the ends to be able to make up a circle so I believe the problem is there somewhere. Could anybody explain the issue in the reasoning above?

  • 1
    $\begingroup$ Maybe the two items here will help: wku.edu/~tom.richmond/Pir2b.html, additionally, see a proof here: basic-mathematics.com/proof-of-the-area-of-a-circle.html $\endgroup$ – Amzoti Sep 20 '12 at 17:21
  • 1
    $\begingroup$ Thanks but I was just there before posting this actually. I understand how the reasoning behind the proper formula works but I'd like to know what's wrong with the above. $\endgroup$ – ByteFlinger Sep 20 '12 at 17:23
  • 2
    $\begingroup$ I think the argument closest to the above logic that gets the right answer is what Wikipedia calls the rearrangement proof. $\endgroup$ – Rahul Sep 20 '12 at 17:29
  • 4
    $\begingroup$ The segments of length $r$ are placed next to each other, yes, but their two ends don't move the same distance. The outer ends of the line cover a length of $2\pi r$, while the inner ends cover $0$. Taking the average gives the right answer. Note that this works for an annulus (donut) shape as well: the lengths are $R-r$, the outer ends go $2\pi R$ and the inner ends go $2\pi r$, so the area is $2\pi(R-r)(R+r)/2=\pi R^2 - \pi r^2$. $\endgroup$ – mjqxxxx Sep 20 '12 at 17:29
  • 3
    $\begingroup$ By the way, if you really want to know why it can't be $2\pi r^2$ in particular, note that the circle with radius $r$ fits inside the square with side length $2r$, so its area must be $\le 4r^2$. $\endgroup$ – mjqxxxx Sep 20 '12 at 17:31

The main issue is that you don't form an area by placing lines next to each other--you need to place strips next to each other. As you say, to form an $a\times a$ square, you can place $n$ strips of dimension $a\times w$ next to each other, where $w=a/n$, giving total area $naw=a^2$.

enter image description here

Your suggestion amounts to forming the area of a circle of radius $r$ by placing $n$ strips of dimension $w\times r$ next to each other radially, where $w=2\pi r/n$, giving area $nrw=2\pi r^2$. But look what happens if you do this:

enter image description here

The problem is that the strips overlap, so the total area of the $n$ strips is greater than the area of the circle. If you run the animation (by reloading the page, if necessary) you can convince yourself that in the $n\rightarrow\infty$ limit, half of each strip contributes to the final area. (Observe that as strips are added in the counterclockwise direction, roughly half of each strip gets covered by subsequent strips.)

We can fix the problem of overlapping strips more easily by using triangles of base $w$ and height $r$ instead of rectangles:

enter image description here

This gives area $\frac{1}{2}nrw=\pi r^2$.

  • 8
    $\begingroup$ More thoughts: calculus seems mysterious if one tries to imagine infinitely many lines, each of zero area, somehow adding to non-zero area, but that's not the right way to think about it. The correct picture, which is mundane but provides clarity, is that any area computation must make sense for finitely many pieces - as an approximation: you can estimate the area of a shape using finitely many rectangles or triangles, but not using finitely many lines. The purpose of taking the limit is to reduce error to zero of a construction that already gives a reasonable answer at the finite level. $\endgroup$ – Will Orrick Jan 2 '13 at 19:31
  • 4
    $\begingroup$ Spectacular animations. +1 $\endgroup$ – Emily Jul 29 '13 at 20:17

The problem with that reasoning is that when showing a rectangle's area by dividing the rectangle's length and height into units, the length and height units form squares. Rotating the radius around the circle by a set number of degrees to divide the circle does not produce squares; these slices are more like triangles. Recall that the area of a triangle is $\dfrac{bh}{2}$; the math's a bit more complex but you can draw a parallel here to the area of a circle versus a rectangle. All other things being equal, the rectangle is double the triangle's area, and so the area of a shape divided into a number of congruent triangles will be half the area of a shape divided into the same number of rectangles of the same length and height as the triangles.

The actual math to prove the area of a circle is very closely related to this, but incorporates an additional concept from calculus:

For any arbitrary $n$, draw an $n$-gon (hexagon, octagon, hectogon, etc) around a circle of radius $r$. Each side of this shape will have length $s$, and $s*n > 2\pi r$; the perimeter of the n-gon will be greater than the circumference of the circle (recall that $\pi = \dfrac{c}{d}, d=2r \therefore c=2\pi r$). However, as $n$ increases, $s$ decreases, and the perimeter of the n-gon will approach the circumference of the circle. It never quite gets there for any finite n, but it gets close enough, allowing us to define what's known in calculus as a limit: $\lim_{n\to \infty}ns = 2\pi r$.

Now, for each side of the n-gon, we can define an isoceles triangle between the vertices of the side and the center of the circle. The symmetrical sides of this triangle have length $l$ which is $>r$ (because the line connects to the vertex of the n-gon, outside the circle) but, similar to the way $ns$ approaches $2\pi r$, $l$ approaches $r$ as $n\to \infty$. This isoceles triangle of base $s$ and height $r$ can be split into two right triangles with base $s/2$ and height $r$. The area of a right triangle is $\dfrac{bh}{2}$ as previously stated, and so the area of the isoceles triangle is two of these, or $2*\dfrac{\dfrac{s}{2}r}{2}= \dfrac{sr}{2}$. There are $n$ of these triangles, one per side, so the area of the n-gon is $\dfrac{nsr}{2}$. Finally recall our limit; as $n\to\infty$, $ns\to 2\pi r$. The limit allows us to equate these two in the general case we are considering, where for our purposes $n=\infty$; $ns=2\pi r$. Plug that into the area of the n-gon, and behold: $\dfrac{nsr}{2} = \dfrac{(2\pi r)r}{2} = \dfrac{2\pi r^2}{2} = \pi r^2$.



OK - so the heart of your intuition is linking the area with the length of line required to fill in that area. It's like shading in a picture using a ballpoint pen.

Take your example with your square: at the end, when you've drawn all your lines, you've shaded the entire area. So if you know how much ink you've used from your pen, you know the area - you've shaded the area more or less evenly, so the amount of ink you've used is proportional to the area.

Uneven shading:

However when you draw the lines for your circle, you go over the bits in the middle quite a few times. In fact the very centre of the circle gets done a ridiculous amount, because every line you draw starts or ends there. However, at the edges of the circle, your lines are much more spaced out.

The intuition you're using relies on the idea that you're shading in the area evenly. If you're repeatedly scribbling over one area (the centre), then using the length of your lines does not give you an accurate estimate of the shape's area.

How to fix it:

So to get a similar sort of intuition for the circle, you need to sketch across the area evenly. Doing that with straight lines is more complicated, geometrically. Instead, try thinking of drawing concentric circles, from the inside out.

Now - imagine that every time you draw a circle, you draw a straight line on another piece of paper that's the same length. By the time you've drawn all your circles, your other piece of paper should have a triangle shaded on it - and you can even use the formula for the area of a triangle to get $\pi r^2$.


Here is a similar approach. Split the radius into $n$ equal parts, and form concentric circles of radius $0, \frac{r}{n}, \frac{2r}{n},...,r$. Think of the cross section of an onion. Then estimate the area by unrolling each circle, approximating its area by a rectangular strip of length given by the outer radius and width $\frac{r}{n}$ and adding the lot together. Then let $n \to \infty$ to make the approximation better.

This gives $A \approx 2 \pi r \frac{r}{n} + 2 \pi (r-\frac{r}{n}) \frac{r}{n} + \cdots + 2 \pi \frac{1}{n} \frac{r}{n} = 2 \pi \frac{r}{n}r (1+ (1-\frac{1}{n})+ \cdots + \frac{1}{n})$, which gives the estimate $A \approx 2 \pi r^2 \frac{1+ \frac{1}{n}}{2}$. Taking limits gives $A = \pi r^2$, as desired.

Note that the $\frac{1}{2}$ appears because you are summing $1+ (1-\frac{1}{n})+ \cdots + \frac{1}{n}$. If you draw lines of lengths $1$, $1-\frac{1}{n}$, etc. stacked on top of each other, you see that they approximate a triangle. The area of a triangle is half that of the 'equivalent' rectangle. This explains the 'disappearing' 2 in the formula.


For each shape like a square or a circle that for sufficiently large $n$ can be split into lines of thickness of approximately width $\frac{1}{n}$, you can find a function that assigns to each sufficiently large positive integer $n$ an ordered pair whose first component is one such splitting of part of it with zero area of overlap and whose second component is a relation from length to area where length is considered equivalent to an area $\frac{1}{n}$ times that length, such that as $n$ approaches $\infty$, the unfilled area of the splitting approaches zero. That means that as $n$ approaches $\infty$, the area corresponding to the length of such a splitting should approach the area of the circle, if the area unfilled by the splitting approaches zero and the area of overlap remains zero. If you take the splitting to $2\pi n$ times a number approaching 1 equally spaced lines going from the edge to the center of thickness $\frac{1}{n}$, then it's not the case that the area of overlapping in the splitting is zero.


Actually, you can use your method, but you have to be a bit careful. Start with a circle of radius $r$ centered at the origin. We can think of the total area of the circle as adding the "lengths" of all the circumferences of all the circles centered at the origin which have a radius $x \leq r$:

$$\text{Area of Circle} = \int_0^{r} 2 \pi x~\textrm{d}x = \pi r^2$$

In reality, we are approximating the circumference of each circle by an annulus of small width, and then letting that width tend to zero.

This method works with finding the area of the square too. Suppose you have a square centered at the origin. Then, the area of the square is the perimeter of each square centered at the origin which ha a side length $x < r$.

$$ \text{Area of Square} = \int_{-r/2}^{r/2} 4x dx = r^2. $$

The reason this doesn't work the way you originally thought was in order to add "fattened" radii all around the circle, the fattened radii overlap, and so you are over counting.

  • $\begingroup$ Just to note that this works in other dimensions too - though I've always remembered it in differential form - so volume of 3-dimensional sphere is integral of surface area from 0 to $r$ etc. $\endgroup$ – Mark Bennet Dec 31 '12 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.