= Polar area formula
{title2=$A=\frac12\int_\alpha^\beta(r_2^2-r_1^2)\,d\theta$}
Suppose $0\le r_1(\theta)\le r_2(\theta)$ and the region is described without overlap by $\alpha\le\theta\le\beta$ and $r_1(\theta)\le r\le r_2(\theta)$. Integrating the <polar area element> gives its <area>:
$$
A=\int_\alpha^\beta\int_{r_1(\theta)}^{r_2(\theta)}r\,dr\,d\theta
=\frac12\int_\alpha^\beta\left(r_2(\theta)^2-r_1(\theta)^2\right)\,d\theta.
$$
For a region bounded by a single radial graph and two rays, take $r_1=0$. If the two bounding graphs cross, split the angular interval at their intersections so that the larger radius is used consistently.
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