Calculus – Valid Proof for Area of a Circle

calculuscirclesdefinite integralsgeometryintegration

My teacher challenged my class to prove that the area is

$$A=\pi r^2.$$

We recently learned about Riemann sums, so I thought it would be possible to apply them to them to deriving the formula for the area of the circle. I know similar proofs exist, but this is one that I genuinely came up with on my own and I am wondering if it is valid. Please tell me if there is anything invalid about this proof or how it can be improved.

Imagine splitting a circle into an infinite number of isosceles triangles, where two legs extend from a vertex at the center of the circle to the edge of the circle. The central angle that each triangle makes can be represented as $\frac{2\pi}{n}$, where $n$ is the number of triangles in the circle.

The area of a triangle is $A=\frac{1}{2}ab\sin{C}$. Since the legs of each of the triangles extends from the center of the circle to the edge, that means that $a=b=r$, the radius of the circle. Therefore, the summation of every triangle in the circle as the number of triangles approaches infinity can be represented as:

$\lim_{ n\to\infty} \sum_{i=1}^{n} \frac{1}{2}r^2\sin(\frac{2\pi}{n})$

which can be rewritten as:

$\lim_{ n\to\infty} \frac{1}{2}r^2n\sin(\frac{2\pi}{n}) = 2\pi(\frac{1}{2})r^2 = \pi r^2$

Best Answer

Nice proof! The only property of $\pi$ and the circle you seem to be relying on is that a central angle of $\alpha$ radians corresponds to an arclength of $\alpha 2\pi r$, that is, you rely on the formula for the circumference of a circle, but not on the formula for its area, so I don't see a circularity here. Note though that you've given an argument which does not use the standard mathematical definition of area given by definite integrals, since you've used specific triangles instead of arbitrary rectangles, but I wouldn't count that against your proof. I like your trick as it makes things simple computationally.

You can also use the standard approach via Riemann sums, where we fit rectangles inside. Let $f(x)=\sqrt{r^2-x^2}$. The graph of this function is the upper part of a circle of radius $r$ centered at the origin. When we fit an infinite number of rectangles we get $\int_{-r}^r \sqrt{r^2-x^2}\,dx$ which can be calculated using the fundamental theorem of calculus and the substitution $x=r\sin t$:

\begin{align*} \int_{-r}^r \sqrt{r^2-x^2}\,dx &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{r^2-r^2\sin^2t}\cdot r\cos t \, dt = r^2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^2 t\,dt\\ &= \frac{r^2}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1+\cos(2t)\,dt=\frac{r^2}{2}\left[t+\frac{\sin(2t)}{2}\right]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\\ &= \frac{\pi r^2}{2} \end{align*} so that the area of the whole circle is $2\cdot\frac{\pi r^2}{2}=\pi r^2$.

Your idea of taking isosceles triangles, instead of the more standard rectangles used in integration, utilizes the symmetry of the circle and definitely makes the calculations easier in this case!

Related Solutions

[Math] Maximal area covered by two triangles in unit circle

I whipped up a quick Matlab program that computes area stochastically (generate 20000 points; count how many are inside either triangle), and then ran it on 10,000 randomly generated triangle pairs (i.e., 6 random points on the unit circle; the first 3 define one triangle; the next three define the other. WLOG, I made the first angle be 0). Each time a new "largest area" was found, I printed out the triangles. It instantly found an area of ~1.25; then steadily improved this to 1.87 over the next couple of minutes. A previous instance of the program reached 1.97 or so...and this one just jumped to ~1.92, with a solution that's very nearly the two 45-90-45 triangles erected on a diameter.

That suggests to me that the correct answer is indeed this pair of triangles, with area 2 as the optimum.

>> testCircles
area = 1.2587; angles are [0 4.13265878886387 1.69578766384379 2.78030964099446 4.4988452659303 2.81878425776269]
area = 1.4833; angles are [0 1.03997694447012 4.97128628805005 5.53558413650189 4.31034051247746 2.03741390656017]
area = 1.5065; angles are [0 2.583714405482 3.95028117360417 3.41185810372066 4.81492678793123 1.47312279640955]
area = 1.8173; angles are [0 4.99055083947817 1.93427156493409 1.86260140259647 3.04043013930809 5.0276959720494]
area = 1.8517; angles are [0 3.01875721392096 1.30210531664363 4.12564514448203 2.38462146306761 5.8391459560257]
area = 1.87; angles are [0 1.78060248058588 4.66075033844476 2.08372315689924 5.4697587833186 3.76689571284346]
area = 1.9184; angles are [0 4.43108946197308 1.4092884007527 4.31113371053254 2.84429076671212 1.32017294630991]

Here's a picture of the most recent success: a near-optimal pair of triangles

The associated angles/area are

area = 1.9428; angles are [0 1.51926494165392 3.23093144636552 0.0773804130776986 3.34779844425289 4.92681116635565]

[Math] Use calculus to derive area of circle using n triangles

Let $r$ be the radius of the circle. The area of the triangle is $\frac12 (base \times height)$, if you split each of the triangles down the middle (to give a right angled triangle) then the height will be given by $$height = r \cos(\tfrac\pi n),$$ notice how we have halfed the angle as we split the triangle in 2. Similarly the base will be given by $$ base = 2 r \sin(\tfrac\pi n), $$ where we have remembered to multiply by two as the base of the right angled triangle is only half that of the isoceles.

Putting these quantities into the formula for the area of the isoceles triangle we get \begin{align*} A_{triangle} &= \frac 12 \times 2 r \sin(\tfrac\pi n) \times r \cos(\tfrac\pi n) \\ &= r^2 \sin(\tfrac\pi n)\cos(\tfrac\pi n) \end{align*}

Now there are n triangles so the total area of all the triangles would be given by $$ A_{total} = r^2 n\sin(\tfrac\pi n)\cos(\tfrac\pi n) , $$ which can be rewritten using the double angle formula as $$ A_{total} = r^2 n \frac 1 2 \sin(\tfrac{2\pi} {n}). $$ The area of the circle can then be found if we take the limit as $n\rightarrow\infty$. $$ A_{circ} = \lim \limits_{n \rightarrow \infty} r^2 n \frac 1 2 \sin(\tfrac{2\pi} {n}), $$ and using the fact that for very small $x$, $\sin(x) \approx x$ this can be seen to be \begin{align*} A_{circ} &= \lim \limits_{n \rightarrow \infty} r^2\frac{2\pi n} {2n} \\ &=\pi r^2, \end{align*} which is precisely what we want.

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[Math] Maximal area covered by two triangles in unit circle

[Math] Use calculus to derive area of circle using n triangles

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