The area of a sector of a circle with radius r and central angle $\mathrm{\θ}$ is given by $A\=\frac{1}{2}{r}^2\mathrm{\theta }$.

We can approximate the area bounded by the polar curve $r \= r\left(\mathrm{\θ}\right)$ and the rays $\mathrm{\θ} \= a$ and $\mathrm{\θ} \= b$ by using sectors of circles.

First, divide $\left[a\, b\right]$ up into n subintervals with endpoints $a \= {\mathrm{\θ}}_0\, {\mathrm{\θ}}_1\, {\mathrm{\θ}}_2\, ..\. \, {\mathrm{\θ}}_{n-1}\, b \= {\mathrm{\θ}}_n$ and equal width $\mathrm{\Δ\θ}$.

Consider the $i^{ \mathrm{th}}$ subinterval $\left[{\mathrm{\θ}}_{i-1}\, {\mathrm{\θ}}_i\right]$ and choose some ${\mathrm{\θ}}_i^{\ast } \in \left[{\mathrm{\θ}}_{i-1}\, {\mathrm{\θ}}_i\right]$.

The area on this interval is approximately the area of a sector of a circle with central angle $\mathrm{\Delta \theta }$ and radius $r\left({\mathrm{\θ}}_i^{\ast }\right)$. So,  $A_i\=\frac{1}{2}{\left(r\left({\mathrm{\θ}}_i^{\ast }\right)\right)}^2\mathrm{\Delta \theta }$.

Therefore, an approximation of the entire area is: $A \approx \sum _{i\=1}^n{A}_i \= \sum _{i\=1}^n\frac{1}{2}{\left(r\left({\mathrm{\theta }}_i^{\ast }\right)\right)}^2\mathrm{\Δ\θ}$.

Allowing the width of each subinterval to become infinitely small by letting n approach infinity, we obtain $A \= \underset{n\to \infty }{lim}\sum _{i\=1}^n\frac{1}{2}{\left(r\left({\mathrm{\theta }}_i^{\ast }\right)\right)}^2\mathrm{\Delta \theta } \= {\int }_a^b\frac{1}{2}{\left(r\left({\mathrm{\theta }}_i^{\ast }\right)\right)}^2 \ⅆ\mathrm{\theta }$.

Thus, $A\=\frac{1}{2}{\int }_a^b{r\left(\mathrm{\theta }\right)}^2\ⅆ\mathrm{\θ}$ is the area of the region bounded by $r\left(\mathrm{\θ}\right)$ on $a \le \mathrm{\θ} \le b$.