Why does $\int_0^R 2 \pi r \,\mathrm d r$ give the area of a circle?

Question

There's a method of computing the area of a circle by dividing it in concentric rings with infinitesimal width. Let $R$ be the radius of the circle and $r$ be the radius of the rings. The area of the circle is

$$\int_0^R 2 \pi r \,\mathrm d r$$

My questions are:

1. I do not understand, though, how to justify the $2 \pi r \,\mathrm d r$ approximation for the area of each ring. Its actual area would be $$\pi (r + \mathrm d r)^2 - \pi r^2 = 2 \pi r \,\mathrm d r + \pi \left( \mathrm d r \right)^2$$ right? Could I use this more precise formula if I wanted to? How?

2. The area on the integral above looks more like the lateral area of a cylinder of height $\mathrm d r$, which is different from the actual area between two concentric circles. So why does that work?

3. Would the lateral area of a truncated cone (which seems to be the intermediate between the ring area and the cylinder lateral area) also work as an approximation?

4. Also, how do you come up with such an idea for an approximation that makes the calculation so beautifully simple (i.e. adopting the lateral area of a cylinder as the area of the rings)? It is considered a trivial integral, but there is a huge and mostly ignored step to be taken there.

Answer 1

The idea is that the approximation gets better and better as you make the rings finer and finer, so that you eventually end up with the approximate-style result even if you use the more exact version.

That the approximation is justified becomes clear if you think in terms of differentials instead, so that the width of the rings is just a difference; because as $\Delta r\to 0,$ the quadratic term in $\Delta r$ approaches $0$ more quickly, or, as we say, is of smaller order than $\Delta r.$ This is simply because $x^2=o(x),$ in general, as $x\to 0.$

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To answer the other questions, which you've numbered and reorganized:

The key idea to have in mind is that any initial approximation will do, provided the error vanishes in the limit sufficiently fast.

Interpretation in mathematics is multidimensional. The same expression can be interpreted differently depending on context or ease. Thus, for example, while Euclid would have thought of $xy$ (both factors positive reals) as a rectangular area, Descartes simply thought of it as a line segment. There are infinitely many other ways of thinking of this product; e.g., one group would be $\underbrace{1×1×1×\cdots×1}_{n} × xy,$ where $n$ is a positive integer.

Thus, you could well think of $2πr\Delta r$ as the curved area of a cylinder. Unrolling this gives a rectangular strip of dimensions $2πr × \Delta r.$ The annulus with median radius $r$ and width $\Delta r$ can also be thought of as being represented (approximately) by that expression. Now as $\Delta r\to 0,$ the strip approaches a string of length $2πr,$ which is the same as the circumference of the circle that the annulus approaches too. The surface area of a conical frustum should work similarly.