The following quotes are from this book(A First Course in Calculus by Serge Lang)
Suppose we are given a continuous function $$r=f(\theta)\tag{1}$$ which is defined in some interval$~a\le\theta\le b~$. We assume that$~f(\theta)\ge0~$and$~b\le a+2\pi~$ We wish to find an integral expression for the are encompassed by the curve$~r=f(\theta)~$between the two bounds$~a~$and$~b~$. Let us take a partition of$~[a,b]~$, say $$a=\theta_0\le\theta_1\le\cdots\le\theta_n=b\tag{2}$$ The picture between$~\theta_i~$and$~\theta_{i+1}~$might look like this:
We let$~s_i~$be a number between$~\theta_i~$and$~\theta_{i+1}~$such that$~f(s_{i})~$is a maximum in that interval, and we let$~t_i~$be a number such that$~f(t_i)~$is a minimum in that interval. In the figure, we have drawn the circles(or rather the sectors)of the radius$~f(s_i)~$and$~f(t_i)~$, respectively. Let $$A_i=\text{area between}~\theta=\theta_i,~\theta=\theta_{i+1},~~\text{and bounded by the curve}\\=\text{area of the set of points}~(r,\theta)~\text{in polar coordinates such that}\\\theta_i\le\theta\le\theta_{i+1}~~\text{and}~~~0\le r\le f(\theta)~~~~. \tag{3}$$ The area of a sector having angle$~\theta_{i+1}-\theta_{i}~$and radius$~R~$is equal to the fraction $${\theta_{i+1}-\theta_{i}\over 2\pi}\tag{4}$$ of the total area of the circle of radius$~R~$, namely$~\pi R^2~$
Currently I can't understand
$$ \color{fuchsia}{{\theta_{i+1}-\theta_{i}\over 2\pi}} \tag{5}$$
I think an area of a sector of circle is given by following, which seems quite different from the above formula.
$${\text{radius}^2\cdot\text{angle}\over 2}\tag{6}$$
In the first place, eqn4 doesn't contain$~R~$in anywhere.
But it says “of the total area” meaning that you have to multiply that fraction by $\pi R^2$. If you do that, you get your formula.