Different ways to prove that area of circle with radius $r$ is $A=r^2\pi$

Question

I want to compute the area of a circle in different ways. I know that any circle with radius $r$ have area $A=2\int_{-r}^r\sqrt{r^2-x^2}dx=r^2\pi$, but I want to prove it in other ways.
My first way is to split a circle into infinitely many concentric circles. The smallest circle have radius near to $0$ and the largest circle have radius $r$, so area will be $$A=\int_{0}^{r}2r\pi dr=r^2\pi$$ Then I tried to do it with diameter. let $R=2r$, then smallest circle will have diameter near to $0$ and the largest will have diameter $R$. Circumference of a circle with diamater $R$ is $R\pi$, so area should be $$A=\int_0^RR\pi dR=\frac12R^2\pi$$ but it is actually $\frac14R^2\pi$. I am sure there must be an easy explanation why my second way is incorrect. I will get correct result if I divide it by $2$, but I don't understand why we can integrate it w.r.t. radius, but cannot integrate it w.r.t diameter. What is the difference?

Answer 1

With an increase in diameter $dR$, the increase in area is $\dfrac{\pi RdR}{ 2}$, and not $\pi RdR$

Answer 2

You have the following $\int_0^{r*} 2r\pi \ dr$ and want to substitute $R = 2r$, $dR = 2dr$ so you get $$\int_0^{r*} 2\pi r\ dr = \int_0^{\frac{R^*}{2}}\pi R\ \frac{dR}{2} = \pi\big(\frac{R^*}{2}\big)^2 = \frac\pi 4{R^*}^2 = \pi {r^*}^2$$

Answer 3

There are lots of different ways of calculating the area bounded by a circle with integrals. Here are a few that I could come up with off the top of my head:

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It's especially simple with a double integral (using Fubini's theorem to make it into an iterated integral):

$$\iint_{\text{disk}}dA = \int_0^{2\pi}\int_0^R rdrd\theta = \pi R^2$$

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Or use Green's theorem (with the vector field $\mathbf F(x,y) = \frac 12(y,x)$) to get

$\frac 12 \oint _{\text{circle}} xdy-ydx$ with the parametrization $(x,y) = (R\cos(\theta), R\sin(\theta))$, $0 \le \theta \le 2\pi$.

Then

$$\frac 12 \oint _{\text{circle}} xdy-ydx = \cdots = \pi R^2$$

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Or we could sum over infinitesimal sectors -- which at the infinitesimal scale are really just triangles of base $R$, and height $dt$ -- over the entire arc length. Then we have

$$\int_0^{2\pi R} \frac 12 Rdt = \pi R^2$$