About the integral of the area in a circle

Question

so i start with a circle with radius $r$, and make another one with radius $r+dr$, and the area between them is $(2\times \pi \times r)\times dr$ because that area is a rectangle, and that's where my problem is... if you make a circle with radius r and another of radius $r+dr$, then the shape would be of a trapezium rather than a rectangle!

making the math, that would get a pretty bizarre result:

$dA=dr(2\times\pi\times r+2\times \pi\times r+2\times \pi\times dr)/2 $(area of the trapezium)

$dA=(2\times \pi\times r)dr + (\pi\times dr)dr$

integral (1) $dA=\int(2\times \pi\times r)dr + \int(\pi\times dr)dr$

the second term on the right shouldn't exist, so where am i wrong?

Answer 1

The area between then should be:

$$\pi(r+\Delta r)^2 - \pi r^2 = \pi (\Delta r)^2 + 2 \pi r \Delta r.$$

The integral for this area is:

$$\int_{r'=r}^{r + \Delta r} \int_{\theta ' = 0}^{2\pi} r' dr' d \theta '. $$

Answer 2

The argument is that for sufficiently small $\Delta r$, the area increment is approximately correctly equal to that of a rectangle of side $\Delta r$ and the circumference of the circle.   That is to say, as $\Delta r$ approaches zero the imprecision of this approximation approaches zero.

You can approximate it as a trapezium, with near side $(2\pi r)$ the far side $(2\pi r+2\pi\Delta r)$ and height $(\Delta r)$.   However the additional term vanishes as the increments of the sum are taken down to the limit. $$\lim_{\Delta r\to 0}\sum_{r\in\{0,..k\Delta r,..(R-\Delta r)\}} \Bigl(2\pi r\Delta r + \pi(\Delta r)^2 \Bigr) = \int_0^R 2\pi r\operatorname d r$$