Find the area of the enclosed region

By Calculus:

The shaded area is given by:

$\begin{aligned} A & = \frac 12 \int_\frac \pi 6^\frac {5\pi}6 \left(r^2(\theta) - 2^2\right) \ d\theta \\ & = \frac 12 \int_\frac \pi 6^\frac {5\pi}6 \left(16 \sin^2 \theta - 4\right) \ d\theta \\ & = \int_\frac \pi 6^\frac {5\pi}6 \left(4(1 - \cos 2\theta) - 2 \right) \ d\theta \\ & = \int_\frac \pi 6^\frac {5\pi}6 \left(2 - 4\cos 2\theta \right) \ d\theta \\ & = \bigg[ 2 \theta - 2\sin 2 \theta \bigg]_\frac \pi 6^\frac {5\pi}6 \\ & = \frac {5\pi}3 - \frac \pi 3 - 2\left(-\frac {\sqrt 3}2 - -\frac {\sqrt 3}2 \right) \\ & = \frac {4\pi}3 + 2\sqrt 3 \end{aligned}$

$\implies a+b+c = 4+3+2 = \boxed{9}$

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By Geometry:

In rectangular coordinate system, we have:

$\begin{cases} x = r\cos \theta = 4\sin \theta \cos \theta = 2\sin 2\theta \\ y = r\sin \theta = 4\sin^2 \theta = 2-2\cos 2 \theta \end{cases}$

$\implies x^2 + (y-2)^2 = 4$ , which is a circle centered at $(0,2)$ and radius 2.

We note that the segment in the overlap extends an angle of $120^\circ$ with the upper circle. Therefore, the area of the shaded region

$\begin{aligned} A & = \pi R^2 - 2\left(\frac 13 \pi R^2 - 2\times \frac 12 \times \frac 12 \times \frac {\sqrt 3}2R^2 \right) & \small \color{#3D99F6} \text{where }R=2 \\ & = 4 \pi - \frac 83 \pi + 2 \sqrt 3 \\ & = \frac {4\pi}3 + 2\sqrt 3 \end{aligned}$

$\implies a+b+c = 4+3+2 = \boxed{9}$