Possibility of finding exact value of circle area

Question

I am taking a Calculus course and my current theme is calculating a circle's area from scratch, and the tutor is splitting the circle in smaller circle shapes, draws them as a rectangle and putting them on graph to get the shape of the right triangle. and as the width of the rectangle reduces graph is getting more similar to triangle. and after he uses basic trigonometry formula to calculate the triangle area (Area = 1/2 * bh) where h = 2PIr and b = r and my current concern is. since the "triangle" on graph is made by rectangles we can not get the exact triangle even if the width of the rectangle is 1^-Infinity, and so on we can never get exact Area of Circle with the formula of PIr^2. I want to know if this is true, and correct me if I am wrong because I just started learning Calculus and I do not want to go by the wrong way

Illustration:

Triangle

Answer 1

The handwavy idea of calculus is that an image of rectangles approximates the image of your area so that if you take the limit when your rectangles get thinner and you have a lot of them, the get closer and closer to superimposing the image.

Now, here's the big hand wavy part, if you make these rectangles so thin that they are just lines and that they are so thin you have an infinite number of them then they will exactly superimpose on your image.

The problem is you can't actually calculate the area of your image by adding up the areas of the line/rectangle because have $\frac 1{\infty}$ width and finite height the have $0$ area, and having infinite number of them adding an infinite number of them you have the "salami paradox" (that $\sum_{k=1}^{\infty} \frac {a_i}{\infty} = ?????$) and can have any answer you want.

But that's not the point. You can't ever actually "get to" infinity. For any finite number of $n$ the larger the $n$ the closer the the sum of the rectangles get to the area of the image. And therefore, we realize, the LIMIT of the sums of the rectangles as to values of $n$ "go to" (but NEVER "get to") infity will be the area of the image.

That's the idea. The sums of the rectangles approach an absolute limit; a limit with a real value (even though we can't actually make a collection of infinite number of infinitely thin rectangles) and that limit is the area of our image.

......

So yes. For any $n$, adding up the $n$ number of rectangles each $\frac 1n$ thin won't be the same as the triangle but the thinner the rectangles the closer the limit is to the triangle and so the LIMIT of the sums will be the same as the area of the triangle.

.......

All that said, I think this is a very confusing exercise and not well presented.

The idea is we break the circle of radius $r$ into $n$ many thin rings. The $k$the one of the rings with be $\frac rn$ thick and ill have a radius of $k*\frac rn$.

We try to figure what the area of the rings are be cutting them and "unrolling" them into rectangular strips. (Frankly, I think this is just begging the question.... we do not explain how we assume a shape in the shape of a ring can be "unrolled"). As the circumference of a circle is $2\pi radius$ and the $k$th one of these rings has a radius of $k*\frac r{n}$, the height of this $k$th ring is $2\pi k\frac r{n}$ and its width is $\frac r{n}$ so it's area is $2\pi k\frac {r^2}{n^2}$.

So the area of this rectangles is $\sum_{k=1}^n 2\pi k\frac {r^2}{n^2}$ and we know the area of the circle will be the limit of this as $n\to \infty$. ...

But we don't have to calculate that (Thank god). We know these rectangles lined up get closer and closer to a triangle with height equal to the very last ring, i.e. the circle itself. That is a height of $h = 2\pi r$.

So the area of the circle =

the LIMIT of the sum of $n$ rectangle $\frac rn$ thick and $2\pi k\frac rn$ tall =

the area of a triangle that has a base of $r$ and a height of $2\pi r$ =

$\frac 12 bh =$

$\frac 12 (r)(2\pi r =$

$\pi r^2$.

Answer 2

As I understood your question, you are concerned if we cannot sum infinitely many rectangles to get a triangle. The thing is, to understand what infinite sum of those rectangles means, you first need to get a concept that we can't just sum and count the total area of all infinitely many rectangles but we can watch what total area equals to, as we draw many thin rectangles. For instance, if we count the total area of 100 of rectangles, we get $area = \pi r^2 - 0.004$, then we count the total area of 1000 of rectangles, we get $area = \pi r^2 - 0.00003$, then we count the total area of 100000 of rectangles, we get $area = \pi r^2 - 0.0000000002$. So we can see that the sum approaches a value $ \pi r^2$; but as we sum all this, geometrically it means that the area of the triangle gets better and better approximated by this sum. But we could also do it with "bigger rectangles" that are slightly bigger than needed (their height is right-side value) so that the sum is getting like $ \pi r^2 + 0.00003, \pi r^2 + 0.000000001, \pi r^2 + 0.000000000000001$. So the area is guaranteed to be less than $ \pi r^2 + 0.000000000000001 $ but bigger than $ \pi r^2 - 0.0000000002$. So wouldn't it be reasonable to call the "area" of the triangle precisely equal to $ \pi r^2$?
Think about it. All the axioms by which the "area" is defined hold. Moreover, it is the only reasonable value to assign to that area: all the other numbers are bigger or less than it. That value is called the limit of the sum. Nice thing is that that limit is often easy to find, for example, we can see that $\{..., \pi r^2 + 0.00003, \pi r^2 + 0.000000001, \pi r^2 + 0.000000000000001,...\} $converges to $ \pi r^2$. So as we have all this: $ \pi r^2$ - is only reasonable value, all axioms hold, easy to find, so we define the area of a shape as this limit. So all areas which we encounter could be thought as these limits: even the area of a 1 by 1 square: the limit of a sequence $\{..., 1, 1, 1, 1, 1, ....\} \rightarrow 1$. So this new way to define the area is an extension of our thoughts of an area, and this could be thought of as a real pure area of a shape. So the real pure area of the triangle could be only $ \pi r^2$, as all other values are clearly not exact.

What's interesting, with a similar method we define all areas under curves, length of a curly line and volumes and many many more. The limit - the thing that we can never approach by an approximation but we can easily evaluate numerically, is the thing to define all non-linearly (curvy) defined things. And it's the only reasonable technique to evaluate those, on first glance impossible to get, areas, volumes, etc. The infinite sums are defined to be equal to the limit (in a sense, but we can set axioms so that there are no contradictions when doing "infinite summing"), and only after all this, it's perfectly reasonable to think about summing infinite number of things, however, it's quite intuitive in the first place.

P.S. For some time I also wasn't sure if number $ \pi $ as many irrational numbers are precise in a sense that we can show them in decimal only as an infinite number like $ \pi = 3.1415926...$ .But, even though we can't show it as a finite thing in decimal, it could be thought of, as a precise point on a number line, as we can define the set of real numbers in such a way. Moreover, when we show a number in a form like $3.1415926...$ we mean the limit (namely the certain point on a number line to which the sequence $\{3.14,3.141,3.1415,3.14159,...\} $of points on a number line approaches, and we know that there is one and only one number (the concrete point on a number line) to which this sequence approaches and we define it to be equal to $3.1415926...$ as it's the only reasonable value to assign to this infinite representation of a number).

All these thoughts are only what I've been thinking of because I am a calculus student just like you and for a long time I was struggling to get rid of that sense of incorrectness when talking about limits. So, in some points I can be wrong.

I hope this has helped you!

Answer 3

As I see all of your answers seems correct and totally fits with what calculus idea is(as far as i know it) but I tried to overcome with exact value, by treating every rectangle individually, maybe I am wrong but I would be really glad if you correct me. so the concept is that when we draw many rectangles and visualize it as a triangle we got right sided triangle left for every rectangle which is equal sided triangle also. and the main concept is that we should calculate all of the triangle's area and substract it to the triangle's area we visualized (PIr^2) and since it would take long time to calculate each one and it can not be used in everyday formulas we should see that the equal sided right triangle left from the rectangles h and b equals to dA and hence every left triangle is equal to each other we can assume that their area is eqaul to 1/2 dA^2 and to get the number we should multiply it to get every triangles area we must divide radius r of the circle by dA and finally we should substract it to PIr^2 to get the actual area of all the rectangles hence the circle. So the formula I figured out to calculate the area of circle is

πr^2 - (1/2 dA^2 * (r /dA))

Answer 4

This explanation by your tutor is a little weird, as you could compute the area of the large triangle straight away, without dissecting in rectangles. It is indeed pretty obvious that the area of the triangle is half that of the enclosing rectangle, by symmetry. Hence

$$A=\frac{2\pi r\cdot r}2.$$

To answer your question, the area under the rectangles is indeed always smaller than that under the large triangle, but getting closer and closer as you reduce the rectangle width. Indeed, if you halve the rectangle width, observe that the area of the "holes" is also halved, so it decreases proportionally to the width.

Answer 5

The area of a circle is $\pi \times r^2$
Your question is: Can we get the exact area of a circle with the equation?
My answer is yes, my test is kind of dumb but here it is:
Let $r=\frac{1}{\sqrt\pi}$ $$A = \pi\times r$$ $$A=\pi\times(\frac{1}{\sqrt\pi})^2$$ $$A=\pi\times\frac 1\pi$$ $$A = 1$$ If a circle had a radius of $\frac 1{\sqrt\pi}$ the area of the circle would be exactly $1$
When solving for the area of a circle leave $\pi$ as the symbol; your solution will be exact