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My professor had an interesting statement at the beginning of first year integral calculus.

What does area really mean?

How do we know that the area of a circle is $\pi r^2$? Archimedes used the method of exhaustion.

Is there a good generalisation for the meaning of area with respect to how it concerns integral calculus? I apologise if the question is either too general or just stupid. But it's interesting, and I would appreciate more interesting examples on the nature of area in general.

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Isn't that what the field called measure theory is all about? – Arthur Jan 7 '14 at 11:43
It turns out to be not that easy, if you define some map to measure the area of any subset of $\mathbb{R}^n$ and let it have some reasonable properties (for example that the sum of two disjoint sets have the sum of the area), you will easily run into problems. Like Arthur suggested, if you are interested in those things you should get a book about measure theory. – Listing Jan 7 '14 at 11:48
Try – lhf Jan 7 '14 at 11:48
The way I've always interpreted it is as follows: we define a 1x1 square to have an area of 1. We also say that if region A is made up of nonoverlapping regions B and C glued together, then $\operatorname{Area}(A)=\operatorname{Area}(B)+\operatorname{Area}(C)$. And, last but not least, if region A is completely inside inside region B (and they are not identical), then $\operatorname{Area}(A)<\operatorname{Area}(B)$. The hard part is trying to show that these are self-consistent, but I'm sure you can do it by using limits. – Akiva Weinberger Sep 16 '14 at 2:05

The traditional way that these terms gain meaning is as follows. People start with a consensus on the area of a very basic object, like a square. The trick is then to extend the definition to more general shapes in a way that (1) agrees with the old definition for basic objects, (2) satisfies nice properties. It is often the case that these two conditions entirely determine the generalized definition. For example, suppose you cut a shape into pieces: you want the sum of the area of the pieces to be the sum of the whole shape. Well, a square can be cut into triangles, so this already determines the area of a triangle. Since a polygon can always be cut into triangles, we now have defined the area for all polygons. Since circles can be approximated by polygons, this determines the area of a circle (so long as there is a property that allows us to make sense of "approximation").

In this way, a definition of area arises from an understanding of the way area should behave, and the knowledge of the area of a very basic object. For the purposes of integral calculus, you want to generalize the definition to very strange sets, so you need to think more carefully about the sort of properties you expect area to have. This leads to a subject called measure theory.

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What does area really mean?

An intuitive meaning would be the amount of paint that would be needed to paint the surface in question.

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Yeah - though what does amount mean here? Volume, you say? Well, then you have just defined area in terms of volume, which is a bit circular. Now, maybe you're keeping your paint in cubic containers whose footprint are unit squares, so that the amount equals the height of the container. You pour that over your surface, and again measure the height of the paint. That'll work - if you believe that the volume of the paint hasnt changed by pouring it - which, again, brings you back to what a volume really is... – fgp Jan 7 '14 at 12:06

First, we talk about area of shape on Euclidean plane:

Back in ancient time, Greek (such as Euclid) do not truly assign numerical real value as area of shape, unlike us today. In fact, object have same area if we can cut them up into finite piece and use isometry (reflect, rotate, translate) to turn one into another. That is minimally good enough, because all we care about is that what ever algebraic structure that contains all possible value for area to have the operation of addition and a total ordering. So strictly, we do not truly need measure theory (as it restrict you to just real value), and of course, measure theorem have COUNTABLE additivity, which can be considered too strong a requirement. A modern theorem (Bolyai-Gerwien theorem) told us that given any 2 polygonal shape of the same conventional area, you can indeed always cut them into finite piece and use isometry to convert one into another. This validate Euclidean approach to defining area.

In modern time, we think of "shape" as more like set of points, and we want to assign numerical value as area to all subset, satisfying certain condition such as being invariant under some transformation, being finite additive, monotone, and a some normalization. This is already good enough for Archimedes to approximate $\pi$, without having ever deal with an infinite sum.

In 2-dimension, von Neumann showed that a square can be decomposed, apply area-preserving affine transformation to, and get back 2 square of the same size instead. However, luckily, such paradox cannot happen on polygonal shape and circular disk if only rigid motion are allowed, thanks to a theorem due to Banach. In fact, we can construct an assignment of area to all subset that is invariant under rigid motion and satisfy all other reasonable definition of area.

Now, to the question of area in higher dimension:

Hausdorff paradoxical decomposition of spherical shell (very related to the infamous Banach-Tarski paradox) means that we cannot hope to assign surface area to all possible subset that is invariant under rotation. Though of course, it's intuitively clear before that, because if a subset is too badly looking, there is no sense to talk about area because we don't even know what is its "surface".

Attempt to define it based on analogy with curve turn out to be a failure. A standard definition of length of curve is by using approximation with piecewise linear curve, and find the supremum of such approximation. It's a good approach for curve, as it can measure all sort of curve, unless of course if the curve behave so badly that we would intuitively assign a length of $\infty$ to it anyway. That approach turn out to be a failure for area on surface with curvature: you would get infinite area for a simple surface such as a cylinder (cylinder area paradox).

However, for nicest kind of area, the one on a differentiable manifold, we could in fact define area using analogy from curve, but this time analogy with differentiable curve. Beyond that, things get fuzzy, with various definition disagree with each other.

In conclusion, we don't really know once it comes to higher dimension. But we do have a consistent definition for area on 2 dimensional Euclidean plane.

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do you have an aversion to pluralisation? – Dan Rust Jan 7 '14 at 14:17
@DanielRust:oh I am sorry. I typed this in just a few minute right after the question was posted, and I am using a computer without those fancy spell checking thingie. I just do not find it worthwhile to do spell check manually since I considered this site to be about relaying knowledge rather than proper formal presentation. Beside, it's all abstract mathematical objects right? We can easily imagine them to be uncountable (in the layman sense of "uncountable"). – Gina Jan 8 '14 at 4:44
unfortunately it made it very hard to read. For instance Euclid was one of the Greeks not one of the Greek. – Dan Rust Jan 8 '14 at 11:39
th thin abut englsh s tht its vry rdundnt. u can leev out most of th lttrs n itl stil b readabl. (I may or may not have just given myself cancer.) – Akiva Weinberger Sep 16 '14 at 2:13

A naive approach: Pick a plane region and fill it with as many circles of radius $1$ as you can (without overlapping), and then count them all. Let the result be $a_0$. After that, pick circles of radius $r/2$, count them all and divide the result by 2. Let this result be $a_1$. Iterate the process, so every time you fill with circles of radius $2^{-n}$, count them all and divide the result by $2^n$, letting this value be $a_n$. Then, the area of the region can be thought as the limit when $n \xrightarrow[]{} \infty$ of the sequence $\left\{ a_n \right\}$.

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Why not use squares? – littleO Jan 7 '14 at 12:18
I did not say you cannot use squares. ^^ You know, any Riemann sum sequence converges to the integral. – busman Jan 7 '14 at 12:20

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