calculating the area of a circle by integrating over straight lines

Question

The Problem

Imagine having a circle of radius $R$. We all know the area of this circle is $A=\pi R^2$ and the circumference is $2\pi R$. I would like to know why this approach for calculating the area is wrong, and why it gives me the circumference instead:

If I have a line of length $R$, I can rotate it around its edge a full rotation ($2\pi$ rad) and it would cover the whole circle, meaning it should give me the area, so we have the area be equal to $A=2\pi R$ or alternatively $A=\int\limits _{0}^{2\pi}Rd\theta =2\pi R$.

Note: By rotating the line I mean taking the original line (purple in the image) and then placing its end (which isn't in the center of the circle) on some other point on the circumference of the circle, as seen in the picture (the red lines), doing so for every possible point on the circle would look like the entire area of the circle is colored red, since its all covered by the rotations of this one purple line.

Note: I tried to look at other similar SE questions such as this one, but they didn't answer my question well.

Context

The reason I thought of this is because a few days ago I tried to calculate the electric field of a ring centered at $\left(0,0,0\right)$ with some inner radius $R_1$ and outer radius $R_2$ and charge density $\sigma = \frac{\lambda}{r^{3}}$, at the point $\left(0,0,z\right)$ for some $z$. since the ring is symmetrical in the $x,y$ directions the field has affect in the $z$ direction at that point. I thought about finding the field one "line" in the ring causes, and multiply that by $2 \pi$, but I found this approach to be incorrect.

Answer 1

Using that reasoning, the length of a line segment is always $0$. Because it is made of individual points, which have length $0$, so that

$$\int_0^L 0\,dx=0.$$

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For the circle, the correct approach is to decompose the disk in tiny triangles of long sides $r$ and short side $\delta r$. Every triangle has the area

$$\frac{\delta }2\sqrt{r^2-\dfrac{(\delta r)^2}4}$$

Doing this approximation, you drop the areas of the circular segments along the short sides.

One can show that as $\delta r$ tends to $0$,

• $\delta r$ represents both the short side and the corresponding circular arc,

• the area of the triangles is well approximated by $\dfrac{r\delta r}2$,

• the total area of the segments is neglectable.

Finally, summing over a full turn,

$$A=\frac r22\pi r.$$

Answer 2

Consider in your illustration, an outer ring of some width $h$. The "spokes" covering it have an area of $9hw$, where $w$ is the width of each spoke. But near the center, spokes covering another ring of width $h$ would also have an area of $9hw$, even though the area of that ring is less. So it ends up weighing the area near the center disproportionately more than the area close to the edge of the circle.

You can do what Yves Daoust suggested of adding up triangles, or you could add up rings. For the latter, you can add up the areas of small rings (which is the circumference of the ring times $dr$) as $$\int_0^R 2\pi r dr$$