Calculus proof for the area of a circle

Question

I was looking for proofs using Calculus for the area of a circle and come across this one $$\int 2 \pi r \, dr = 2 \frac {r^2}{2} \pi = \pi r^2$$ and it struck me as being particularly easy. The only other proof I've seen was by a teacher and it involved integrating $x = \sqrt{r^2 - y^2}$ from $-1$ to $1$, using trig substitutions and then doubling the area to get $\pi r^2$ but the above proof seemed much more straight forward.

Is it a valid proof, or is it based on circular logic or some other kind of fallacy?

Answer 1

Possibly the proof that you found is what the Wikipedia article for the area of a disk calls "The Onion Proof".

Although I would probably use the following double integral instead:

$$ \text{Area of circle} = \iint_{x^2 + y^2 \leq R}1 \, dx\,dy $$

and then calculate the integral using polar coordinates to get

$$ \iint_{x^2 + y^2 \leq R}1 \, dxdy = \int_0^{2 \pi} \int_0^R r \, dr\,d\theta = \int_0^R 2\pi r \, dr = \pi R^2 $$

Answer 2

$$ \int_0^a 2 \pi r \,dr $$ The above integral seems geometrically as below figure.

Answer 3

There's a particularly simple formula using line integrals: if $\,\gamma\,$ is a simple, closed and smooth (at least by parts) path (in the positive direction), the area of the inclosed region equals $$\frac{1}{2}\oint_\gamma x\,dy-y\,dx$$

In our case, we can take the path $\,\gamma(t)=(r\cos t\,,\,r\sin t)\,\,,\,t\in [0,2\pi)\,$ , and get $$\frac{1}{2}\int_0^{2\pi}r^2(\cos^2t+\sin^2t)\,dt=\frac{r^2}{2}\int_0^{2\pi}dt=\pi r^2$$

Answer 4

Just one remark. The so-called onion proof is a special case of the co-area formula. This formula is a rigorous justification of all those computations that we learned in the first course of general physics. It is a "curvilinear" generalization of Fubini's theorem: instead of slices, you integrate over hypersurfaces like a sphere. And also the fact that "differentiating the volume gives the area" is a consequence of the same theorem.

Answer 5

What you have here:

$$ \int 2 \pi r \, dr = 2 \frac {r^2}{2} \pi = \pi r^2 $$

does not represent an area because the integration is not bounded (also, a constant is missing on the RHS). An area should be for something with bounds (limits). However, the formula you mentioned is used in what is known as Onion proof for area of the circle (please do a find on 'onion'). This proof divides the circle into rings as explained in the link.

Answer 6

Here is another 'calculus proof'. (Basically integrating along $\theta$ rather than $r$.)

Approximate the circle from the inside using a regular n-sided polygon formed from the vertices $(r \cos k\theta_n, r \sin k\theta_n)$, where $\theta_n = \frac{2 \pi}{n}$, and $k = 0,...,n-1$. Draw lines between adjacent vertices and between the origin and each vertex. This splits the polygon into $n$ triangles with sides $r,r,2r \sin \frac{\theta_n}{2}$. The area of each polygon is given by $A_n = 2 r \sin \frac{\theta_n}{2} \frac{1}{2} r \sqrt{1 - (\sin \frac{\theta_n}{2})^2} = r^2 \sin \frac{\theta_n}{2} \cos \frac{\theta_n}{2}$.

Then the area of the circle is given by $\lim_{n \to \infty} n A_n = \lim_{n \to \infty} n r^2 \sin \frac{\theta_n}{2} \cos \frac{\theta_n}{2}$. Since $\lim_{x \to 0, x \neq 0} \frac{\sin x}{x} = 1$, we have $\lim_{n \to \infty} n \sin \frac{\theta_n}{2} = \lim_{n \to \infty} n \sin \frac{\pi}{n} = \lim_{n \to \infty} \pi \frac{\sin \frac{\pi}{n}}{\frac{\pi}{n}} = \pi$. Consequently we have $\lim_{n \to \infty} n A_n = \pi r^2$.

Note: To show that the area of the polygon converges to the area of the circle, note that the area between the polygon and the circle is bounded by $n r \sin \frac{\theta_n}{2} r ( 1 - \cos \frac{\theta_n}{2})$. A calculation along the above lines shows that this converges to $0$.

Answer 7

I would like to show you another method how we can prove that the area of circle is $\pi r^2$ by using infinity part of a circle. I divided only 8 parts in my picture to demostrate how to apply that method but we need to have infinite divided parts to get an exact rectangle shape. After that We can write easily that

Area of circle = $\pi r .r =\pi r^2$

Note: I assumed that we know the circumference is $2πr$

Answer 8

You can actually convince yourself geometrically that

$$A'(r)=2 \pi r (*)$$

Intuitively, the rate of change of the area of the circle is the circumference.

Formally

$$A'(r) = \lim_{\Delta r \to 0} \frac{A(r+\Delta r) -A(r)}{\Delta r}$$

Now, geometrically it is pretty clear (but not really easy to prove mathematically) that the area of a corona between circles satisfies

$$2 \pi r_\text{in} (r_\text{out}-r_\text{in}) < \text{Area} < 2 \pi r_\text{out} (r_{out}-r_{in})$$

Using these inequalities it is easy to calculate the above limit which leads to $(*)$.

The formula you provided solves $(*)$ for $A(r)$.

Answer 9

Consider a circle of radius $r$ with $O$ as its center.

Now, consider an arc $XY$of the circle which subtends an angle of $\theta$ at the center.

let, $dA$ be the area of the segment $XOY$

if we take infinitesimally small angle $\delta\theta$ then we have, $dA=\frac{1}{2}(r^2)(\sin{\delta\theta})$

we know that $\lim_{\delta\theta ->0} \frac{\sin\delta\theta}{\delta\theta} =1$

using, the above fact, we have, $dA=\frac{1}{2}(r^2)(\frac{\sin\delta\theta}{\delta\theta})({\delta\theta})$

Now, $\lim_{\delta\theta ->0} dA = \lim_{\delta\theta ->0}\frac{1}{2}(r^2)(\frac{\sin\delta\theta}{\delta\theta})({\delta\theta})$

we, have, $dA=\frac{1}{2}(r^2){\delta\theta}$

Now,$$\int{dA}=\int_{0}^{2\pi}\frac{1}{2}(r^2){\delta\theta}$$

$$\int{dA}=\frac{1}{2}(r^2)\int_{0}^{2\pi}{\delta\theta}$$

$$A=\frac{1}{2}(r^2)(2\pi)$$ $$A={\pi}r^2$$

we, now have the area of a circle of radius $r$, and it is ${\pi}r^2$