What is the area of the circle πr²

Source: Internet
Author: User
Tags cos polyline sin

Why is the area of the circle \ (S = \pi r^2 \)? How to testify?

This formula is important because it is the basis for all the volume of the rotating body (cylinder, cone, round table, ball, etc.), it is better to be more rigorous proof, rather than approximate to calculate the end.

There may be many, but the well-known "evidence law" is somewhat problematic.

The evidence law of primary School

Cut watermelon slices? Too tight to be convincing:

The arc of others is clearly curved, why do you say that people are straight? No matter how many parts you divide, the arc is always bent, and it will never be parallelogram. The approximation also has to be justified, otherwise it happens to be right.

To say that approximations are all right, with similar logic, you can also prove \ (\pi=4\):

Is that right, too?

The evidence law of the university

Integral? This is a circular argument:

The integral must change the yuan, the change the element to return the derivative, the derivative also returns to the limit. To limit \ (\lim_\limits{x \to 0}\frac{sin{x}}{x}=1\) The value is 1, we should use inequality \ (\sin{x}<x<\tan{x}\) (\ (x\) is a sharp angle) to derive inequalities \ (\COS{X}&L T;\frac{sin{x}}{x}<1 \). So here's the question: why inequality \ (\sin{x}<x<\tan{x}\) is established within a sharp angle? The textbook said, you draw a circle, and then draw a corner, from the area relationship is known, the inequality is obvious ~ but the establishment.

The question is, what is the area of the sector \ (opa\) for \ (\frac{1}{2}x\)? According to the perfect symmetry of the circle, it is known that the center angle is a fan of X, its area is the area of the whole circle of \ (\frac{x}{2\pi}\) times, ...

Wait, what's the area of the whole circle ? That's what I'm asking now, isn't it? This doesn't go around again?

Avoid cyclic argumentation

So here's the question: a circular argument? Why is that? How can we convincingly deduce the area formula of a circle?

To solve this problem, first think about the circle of the area formula exactly reflects what. On the surface, there is nothing but two points:

The area of a circle is proportional to the square of the radius.

Second, the proportional coefficient is pi \ (\pi\)

So what is PI? Pi is the circumference of a circle with a diameter of unit length. Note that it is not that we have calculated that the length is equal to pi, but that it defines the length equal to Pi. Therefore, the Circle area formula actually reflects the circle area and circumference (arc length) and radius of the relationship, namely:

\ (s=\frac{1}{2}cr\)

Once you know this, think about the geometrical meaning of the limit \ (\lim_\limits{x \to 0}\frac{sin{x}}{x}=1\). This can be understood in two ways:

First, when the angle is enough, the area of the triangle is approaching the area of the fan.

Second, when the angle is enough, the chord length is approaching the arc length.

(The same inequality, also can have area and arc length two kinds of explanation)

The reason for the above method has a circular argument, is because the use of the first and area-related interpretation, and the area of the circle is not yet know how much. In this sense, it is appropriate to use the second interpretation.

The above two points indicate the position of arc length (it is the most basic, because \ (\pi\) is directly defined with arc length). So take a look at the arc-length angle to reconsider how to push the limit (\lim_\limits{x \to 0}\frac{sin{x}}{x}=1\).

, Make \ (\angle poa=x \in (0, \frac{\pi}{2}) \), then there \ (ma=\sin{x}\), \ (\overset{\frown}{pa}=x\), \ (ta=\tan{x}\). Extend \ (am\) turn round to \ (b\), connect \ (ob\), \ (bt\) Get a symmetrical kite shape.

If can prove that the original inequality \ (\sin{x}<x<\tan{x}\), namely \ (ma<\overset{\frown}{pa}<ta\), you can introduce the limit.

And because \ (Ma<\overset{\frown}{pa}<ta \leftrightarrow 2ma<2\overset{\frown}{pa}<2ta \),

So just certificate \ (ab<\overset{\frown}{ab}<ta+tb\).

\ (Ab<\overset{\frown}{ab} \) This is very intuitive, two points between the shortest segment; \ (\OVERSET{\FROWN}{AB}<TA+TB \) This is not difficult to see, after all, relative to the chord \ (ab\), Arc \ (ab\) and the polyline \ (a-t-b\) are convex, and the polyline is outside the arc, so the polyline is longer than the arc. If you admit these two points, the limit will be proven. The limit is proven, it is not a problem to calculate the area.

Maybe the 2nd (Polyline is longer than the arc) is a bit far-fetched, but it's better to admit it than to slip into a circular argument. In fact, the ancient Greek Archimedes is to rely on the two-point strict proof of the Circle area formula:

(Quoted from Ouyang Shunxiang Xiang translation of Bill Casselman's paper "Archimedes on the circumference and area of the Circle", the following affixed to the text of the two lemma description)

Define ARC length

What if we don't recognize the two lemma about the length of the curve?

If you do not admit it, you cannot compare the length of the curve (arc length) visually on the geometry. If the length can not be compared, then what is the meaning of length?

So, if you don't admit it, you have to define who is short, in other words, the length of the curve must be strictly defined . If the length of the curve in calculus is defined (the limit of the length of the polyline), then the limit after the Radian (\lim_\limits{x \to 0}\frac{sin{x}}{x}=1\) is not self-evident, the introduction of the Circle area formula can not become a problem. And if you define the length of the curve in this way, you do not need the conclusion of \ (\lim_\limits{x \to 0}\frac{sin{x}}{x}=1\) to prove the area formula of the circle:

According to the definition of curve length in calculus, the length of the smooth curve is equal to the definite integral of the following:

\ (\int_{a}^{b}\sqrt{[x ' (t)]^2+[y ' (t)]^2}\mathrm{d}t \)

Thus, the circumference of the circle can be

\ (c = 4\int_{0}^{r}\sqrt{1+ (\frac{\mathrm{d}}{\mathrm{d} x}\sqrt{r^2-x^2}) ^2}\;\mathrm{d}x \)

\ (= 4\int_{0}^{r}\frac{r}{\sqrt{r^2-x^2}}\mathrm{d}x \)

As a result, the circumference of the circle is proportional to the radius (geometrically intuitive) and defines the perimeter of the unit circle as \ (2\pi\). The resulting \ (c = 2\pi r \)

And because the area of the circle is \ (s = 4\int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x \)

So the area formula of the circle to be proven is equivalent to

\ (2s = rc \)

\ (\leftrightarrow 2\int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x=r\int_{0}^{r}\frac{r}{\sqrt{r^2-x^2}}\;\mathrm{d}x \)

\ (\leftrightarrow 2\int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x=\int_{0}^{r}\frac{r^2}{\sqrt{r^2-x^2}}\;\mathrm{d}x \ )

\ (\leftrightarrow 2\int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x=\int_{0}^{r}\frac{(r^2-x^2) +x^2}{\sqrt{r^2-x^2}}\;\ MATHRM{D}X \)

\ (\leftrightarrow 2\int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x=\int_{0}^{r}\frac{r^2-x^2}{\sqrt{r^2-x^2}}\;\mathrm{ D}X+\INT_{0}^{R}\FRAC{X^2}{\SQRT{R^2-X^2}}\;\MATHRM{D}X \)

\ (\leftrightarrow 2\int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x=\int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x-\int_{0}^{r} X\CDOT\FRAC{-X}{\SQRT{R^2-X^2}}\;\MATHRM{D}X \)

\ (\leftrightarrow \int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x=-\int_{0}^{r}x\cdot\;\mathrm{d} (\sqrt{r^2-x^2}) \)

\ (\leftrightarrow \int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x=-\left [(x\sqrt{r^2-x^2}) \bigg\rvert _{0}^{r}-\int_{0}^ {R}\sqrt{r^2-x^2}\;\mathrm{d}x\right] \)

\ (\leftrightarrow \int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x=-\left [0-\int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x\ Right] \)

\ (\leftrightarrow \int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x=\int_{0}^{r}\sqrt{r^2-x^2}\;\mathrm{d}x \)


No points

All the talk is integral, can not use points? Can you use a tedious curve length definition?

Of course. Second, we can define only the length of the arc, and make the definition fit within the framework of the normal curve length definition. And as mentioned above, Archimedes has no points (there is no such thing as points at that time). But his method is a bit troublesome (see the paper mentioned above). Here, a simpler proof is given by combining the Liu Hui method. Although no integrals are used, the concept and nature of the limit are also available, as well as the fundamental theorem of real numbers.

(Pictures from Wikipedia, really lazy to draw)

The length of the arc is defined first, and the perimeter is obtained.

Make a circular inner positive \ (6\cdot 2^n \) Edge shape. It is easy to know that the side length of the polygon is strictly monocytogenes with \ (n\) (the sum of the two waists of the red and blue triangles is greater than the bottom length). Because the edge length of the Nes must be less than the circumference of the circle (admitting the shortest segment between two points), it is known by the monotone bounded convergence theorem, and the n\ is approaching infinity, and the edge length will converge. Defines the limit of the edge length for the circumference of the circle. This is proportional to the circumference and radius of the circle. Defines the circumference of a circle with a radius of \ (\frac{1}{2}\) as \ (\pi\).

The area of the regular polygon is convergent to the area of circle, then the area of circle is calculated by the uniqueness of the perimeter and limit.

The area of the circle with a radius of \ (r\) is \ (s\), and the side length of the inner positive \ (6\cdot 2^n \) edge is \ (l_n\), and the area is \ (s_n\). Known by definition \ (\lim_\limits{n\to\infty}l_n=2\pi r\). Set \ (\delta_n = s_n-s\). Make the inner positive \ (6\cdot 2^n \) edge of the circle and the inner positive \ (6\cdot 2^{n+1} \) Edge shape. Investigate the relationship between \ (\delta_{n+1} \) and \ (\delta_n \). You may wish to set the green part of the figure to be a circle inner positive \ (6\cdot 2^n \) Edge shape. In the observation diagram, the rectangle \ (abcd\) indicates

\ (6 \cdot 2^n \cdot S_{ABCD} > \delta_n \)

\ (6 \cdot 2^n \cdot \FRAC{1}{2}S_{ABCD} > \frac{1}{2}\delta_n \)

\ (S_{n+1}-s_n > \frac{1}{2}\delta_n \)

\ (\delta_n-\delta_{n+1} > \frac{1}{2}\delta_n \)

\ (\delta_{n+1} < \frac{1}{2}\delta_n \)

This leads to

\ (\delta_{m+p} < \frac{1}{2^p}\delta_m \)

\ (\delta_{n+1} < \frac{1}{2^{n}}\delta_1 \)

\ (\delta_n < \frac{1}{2^{n-1}}\delta_1 \;\; (n \geqslant 2) \)

Because \ (\frac{1}{2^{n-1}}\delta_1 \) when \ (n\) is large enough to be arbitrarily small, so \ (\delta_n \) can be arbitrarily small, so \ (\lim_\limits{n\to\infty}s_n = S \).

Also because \ (S_n=\frac{1}{2}h\cdot l_n = \frac{1}{2}r (\cos{\frac{180^{\circ}}{2^n}}) \cdot l_n \)

So \ (\lim_\limits{n->\infty}s_n = \frac{1}{2}r\cdot 2\pi r = \pi r^2 \)

Also by the limit of the uniqueness of the knowledge \ (S = \pi r^2 \).

The certificate is completed.


The Circle area formula reflects the relationship between the area of the circle and the perimeter, so the arc length on the circle cannot be a vague concept.

The area formula of a circle can be strictly proved by meeting one of the following two conditions:

First, the recognition of geometrical intuitive comparison of arc length and line and line length of two conclusions

Second, strictly define the arc length

Last tear.

I was shocked when I first found the cycle argument in the textbook and got it confirmed. How can the math textbook make such a big loophole? Although the geometry is involved in the intuitive place, because the geometry itself is not rigorous, so it is not strict and harmless, but such a big problem there, the textbook does not say that the default is no problem, this is not tolerated. At least add a note to explain it.

What is the area of the circle πr²

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