find the area of the region lying inside the circle $r=6$ and inside the cardioid $r=4-3\sin \theta$. Well, I drew a graph to visualise it and I found the interceptions $\theta=\arcsin \left(-\frac{2}{3}\right)$. From the graph, by symmetry, I found that the area of region from $\theta$ to $\pi/2$ and from $\pi/2$ to $\pi-\theta$.
So How would I apply the formula now?

You want to compute the area integral
$$\int_A {1} dA$$
where $A$ is the intersection of the interior of the circle and the cardioid i.e.
\begin{align*}
A &= \{(x, y) : r = \sqrt{x^2 + y^2} \leq 6\} \cap \{(x, y) : r = \sqrt{x^2 + y^2} \leq 4 - 3 \sin \theta \}\\
&= \{(x, y) : r \leq \min(6, 4 - 3 \sin \theta) \}
\end{align*}
So let
$$f(\theta) = \min(6, 4 - 3 \sin \theta)$$
and you can rewrite
$$A = \{(x, y) : r \leq f(\theta)\}$$
and the problem is equivalent to computing area of the region bounded by the closed curve $r = f(\theta)$ and so is the same as computing
$$A = \int_0^{2\pi} \frac{1}{2} f(\theta)^2 d\theta$$
[See this link for details.]
Now you have to split the integration range to the region where $f$ takes value 6 and where $f$ take value $4 - 3 \sin \theta$. It is easily solved: for $\theta \in [0,2\pi)$
$$6 \leq 4 - 3 \sin \theta \iff 2 \leq -3 \sin \theta \iff -\frac{2}{3} \geq \sin\theta \iff \theta \in [\theta_1, \theta_2]$$
where
$$\theta_1 = \pi - \arcsin\left(-\frac{2}{3}\right); \theta_2 = 2 \pi + \arcsin\left(-\frac{2}{3}\right); 0 < \theta_1 < \theta_2 < 2\pi.$$
This means
$$f(\theta) = \begin{cases}
6 &\text{if }\theta_1 \leq \theta \leq \theta_2\\
4 - 3 \sin \theta &\text{otherwise}
\end{cases}$$
Going back to the integral
\begin{align*}
A &= \int_0^{2\pi} \frac{1}{2} f(\theta)^2 d\theta\\
&= \int_0^{\theta_1} \frac{1}{2} (4 - 3 \sin)^2 d\theta + \int_{\theta_1}^{\theta_2} \frac{1}{2} 6^2 d\theta + \int_{\theta_2}^{2\pi} \frac{1}{2} (4 - 3 \sin)^2 d\theta
\end{align*}
(You should be able to complete the computation.)
