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

To see why the "diameter" formula gets the wrong answer, first let's consider why your first formula, $$A=\int_{0}^{r_0}2r\pi\; dr=r_0^2\pi,$$ gets the correct answer. Note that in order to remove some confusion that might be caused by using the same symbol, $r$, for two different things (the upper bound of the interval of integration, and the variable of integration, which falls within that interval), I have written $r_0$ for the radius of the circle whose area you want to find, and left $r$ as the radius of one of the concentric circles within the larger circle, so that $0 \leq r \leq r_0$.

Consider the Riemann integral of $2r\pi$. In particular consider what happens when we subdivide $[0,r_0]$ (the interval of integration) into $n$ equal subintervals of width $\Delta r = \frac{r_0}{n}$.

If we construct all the circles of radius $\Delta r$, $2\Delta r$, $3\Delta r$, etc., these circles cut the larger circle into $n$ concentric rings whose total area is the area of the circle. Now look at one of these rings, let's say the ring between the circles of radius $k\Delta r$ and $(k+1)\Delta r$. Its area is a little less than we would get if we had a rectangle with length $2\pi(k+1)\Delta r$ (the outer circumference of the ring) and width $\Delta r$, but a little more than we would get if we had a rectangle with length $2\pi k\Delta r$ (the inner circumference of the ring) and width $\Delta r$. We can make the difference between these two areas as small as we want by making $\Delta r$ smaller, so either of these is a good approximation for the actual area of the ring, which is $$(\Delta r) \times (2\pi \rho)$$ (width $\times$ length) for some $k\Delta r \leq \rho \leq (k+1)\Delta r$.

Now look at the interval between $r = k\Delta r$ and $r = (k+1)\Delta r$ in the Riemann sum. The area under the function $2\pi r$ in that interval is somewhere between the areas of the upper and lower rectangles on that interval, so it is $$(\Delta r) \times (2\pi \rho')$$ for some $k\Delta r \leq \rho' \leq (k+1)\Delta r$.

As it happens, the two areas (of the ring and of the region under the function) are exactly the same, but we don't need to know that. All we need to know is that they are approximately the same, and as we reduce the size of $\Delta r$ to zero, we reduce the total difference in area (over all the rings and rectangles) to zero.

Now look what happens if you use the diameter $R$ as your variable of integration.

Suppose we start with a circle of diameter $R_0$ whose area we want to measure. We can divide the interval $[0,R_0]$ into $n$ subintervals of width $\Delta R$, and draw concentric circles with diameters $\Delta R$, $2\Delta R$, $3\Delta R$, etc., inside the original circle.

Again, the concentric circles cut the larger circle into rings. Look at the ring whose inner diameter is $k\Delta R$ and whose outer diameter is $(k+1)\Delta R$. The "width" of that ring (the gap between the two circles) is just $\frac12 \Delta R$; that is, if we look at a horizontal diameter, half of the increase in the diameter "bumps out" the circle to the right and half "bumps out" the circle to the left. So the approximate area of the ring is just $$\left(\frac12\Delta R\right) \times (2\pi \rho)$$ for some $k\Delta R \leq \rho \leq (k+1)\Delta R$. But the area under the function in the Riemann integral is $$(\Delta R) \times (2\pi \rho')$$ for some $k\Delta R \leq \rho' \leq (k+1)\Delta R$. This is twice as large as the area of the corresponding ring, so the integral gives us twice the area of the circle.

In general, the key idea in using a single-variable integral like $A=\int_{0}^{r_0}2r\pi\; dr$ to measure area is to set it up so that the area we want to measure is subdivided into thin "strips" (not necessarily rectangular) such that each strip corresponds to some value of $r$, the "length" of the strip is the function we are integrating (in this case, $2r\pi$), and the "width" of the strip is uniformly equal to the amount we have to increase $r$ to get to the next strip. When our "strips" are concentric rings, the radius increases by the width of the ring as we go from one ring to the next, so we can use the radius as the variable of integration without any adjustment. But the diameter increases by twice the width of the ring, so to turn it into a variable that increase by the same amount as the width of the ring we have to divide by two.

Answer 2

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

Answer 3

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 4

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

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