Trying to understand why circle area is not $2 \pi r^2$

Question

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?

Answer 1

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 $naw=a^2$.

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:

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:

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

Answer 2

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 an equal 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: $\dfrac{nsr}{2} = \dfrac{(2\pi r)r}{2} = \dfrac{2\pi r^2}{2} = \pi r^2$.

QED.

Answer 3

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 $\pi r^2$.

Answer 4

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.

Answer 5

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.

Answer 6

I understand how the reasoning behind the proper formula works but I'd like to know what's wrong with the above.

The answers given, including one that I posted, do not resolve this point, they only provide derivations by other methods that happen to give the correct answer. Here is an analysis of the reasoning on its own terms.

The principle in question, understood in its natural generality, is that in a plane figure constructed from a "base" (which can be an arbitrary piecewise-smooth curve, not necessarily made of straight line segments and arcs of circles) of length $B$, and with a perpendicular height of $h$ above any point of the base, and the perpendicular segments have disjoint interiors, then the area should be "base x height", $Bh$. The argument for this is that the perpendicular line segments are like disjoint miniature rectangles and these can be integrated into a solid area as in calculus.

This leads to the wrong answer in particular cases, such as a sector of an annulus, where there are two sides either of which can serve as the base, and there is no reason to pick one length over the other as the value of $B$. This indicates that a problem exists, not the origin of the problem.

A closer analysis shows that, for any such generalized rectangle, there are two base lengths (the paths of the endpoints of the perpendicular segment of length $h$ as it sweeps out the figure), and the correct formula is $\frac{b_1 + b_2}{2} \times h$. That tells us what is true about the area, and justifies the comment under the question relating the trouble to the fact that the endpoints of the segment move at different speeds. Still, it does not directly indicate what is wrong with the picture of infinitesimal rectangles all of height $h$.

The best answer is to view the sides of the bent rectangle as paths of the endpoints of a moving segment of length $h$, where the segment covers the figure in one unit of time. If for a large $N$ one slices the time interval into $N$ equal pieces this gives a slicing of the pseudo-rectangle into $N$ small areas. Each small area will be (up to some bounded factor) of order $1/N$, and is another rectanguloid with two long opposite sides of length $h$ (generally not parallel) and two tiny curved sides of length comparable to $1/N$, but these two short lengths are generally not the same. If the "base" curves, that have lengths $b_1$ and $b_2$, are smooth, then the curved sides of these mini-rectangles can be replaced by straight line segments between the same points. The error in calculating the area and length in each quadrilateral is of order $1/{N^2}$ and therefore (when adding $N$ such) will not affect the overall area calculations when the limit of large $N$ is taken, and it also does not affect the results if the vertices of the quadrilaterals are moved by amounts comparable to $1/{N^2}$. The crucial point is:

( up to an error of order $1/{N^2}$ in the lengths of the sides and/or the positions of the points, an error that does not affect the end result )

• the small quadrilaterals have the shape of isosceles trapezoids with long sides equal to $h$ and unequal bases.

The inequality of the small bases, meaning that the ratio of their lengths tends to a limit different from $1$ (and varying from point to point on the base) as the sides of length $h$ approach each other, is the same as the remark about length being accumulated at different rates on the two bases of the rectangle.

The figure being a trapezoid, that is, the parallelism of the small but unequal segments, is a more subtle property and reflects the fact (that I will not prove here, but is true) that the osculating circles to the two "base" curves at the endpoints of the moving segment of length $h$, always are concentric. This is part of the theory of evolutes, involutes and parallel curves, all of which are covered in Wikipedia.

The upshot of all this is that any curved rectangle figure to which the "base x height" argument applies, is the limit not of a collection of small rectangles, but of a strip of thin isosceles trapezoids of side $h$ (and with varying and unequal base lengths) chained together along the $h$ sides. The construction of a circle as a limit of triangular pieces is a special case.