To understand the area inside of a polar curve $r=f(\theta)$, we start
with the area of a slice of pie. If the slice has angle $\theta$
and radius $r$, then it is a fraction $\frac{\theta}{2\pi}$ of the entire
pie. So its area is
$$\frac{\theta}{2\pi} \pi r^2 = \frac{r^2}{2}\theta.$$
Now we can compute the area inside of polar curve $r=f(\theta)$
between angles $\theta=a$ and $\theta=b$. As with all bulk
quantities, we
Break the region into $N$ small pieces.
Estimate the contribution of each piece.
Add up the pieces.
Take a limit to get an integral.
In our case, the pieces are slices of angle $\Delta \theta =
(b-a)/N$. These aren't exactly pie slices, since the radius isn't
constant, but it's a good approximation when $N$ is large and $\Delta
\theta$ is small. The $i$-th slice has area approximately
$f(\theta_i^*)^2 \Delta \theta/2$, where $\theta_i^*$ is a
representative angle between $a+(i-1)\Delta \theta$ and $a+i \Delta
\theta$, so the whole thing has area approximately
$\displaystyle{\sum_{i=1}^N \frac{f(\theta_i^*)^2}{2}\Delta
\theta}$. Taking a limit as $N \to \infty$ gives the integral
$$\int_a^b \frac{f(\theta)^2}{2} d\theta, \qquad \hbox{or
equivalently} \qquad \frac{1}{2}\int_a^b f(\theta)^2 d\theta.$$
The following video goes over the derivation of this formula, and
uses it to compute the area inside one lobe of a cardioid.