I $\heartsuit$ You!... well, almost

First, we must determine for which values of $\theta$ does the limaçon cross the origin. To do that, we set $r = 0$ and solve:

$\begin{aligned} 3 - 6 \sin \theta &= 0 \\ \sin \theta &= \dfrac{1}{2} \\ \theta &= \dfrac{\pi}{6}, \dfrac{5\pi}{6}. \end{aligned}$

Now, we focus on one part of the shaded area:

The region between $\theta = \frac{5\pi}{6}$ and $\frac{3\pi}{2}$

This region starts at $\theta = \dfrac{5\pi}{6}$ and ends at $\theta = \dfrac{3\pi}{2}$ . Notice that doubling this region gives the region that Little Albert colored in, so we want to find twice the area of that region.

Recall that the area of a polar curve represented by $r(\theta)$ between $\theta = \alpha$ and $\beta$ is $\dfrac{1}{2} \displaystyle \int_\alpha^\beta r(\theta)^2 \, \mathrm{d} \theta$ . Let $A$ be the area of the region Little Albert colored in. We have

$\begin{aligned} A &= 2 \cdot \dfrac{1}{2} \displaystyle \int_\frac{5\pi}{6}^\frac{3\pi}{2} (3 - 6 \sin \theta)^2 \, \mathrm{d} \theta \\ &= 9 \displaystyle \int_\frac{5\pi}{6}^\frac{3\pi}{2} \left(1 - 4 \sin \theta + 4 \sin^2 \theta \right) \, \mathrm{d} \theta \\ &= 9 \displaystyle \int_\frac{5\pi}{6}^\frac{3\pi}{2} \left[1 - 4 \sin \theta + 4 \left(\dfrac{1 - \cos 2\theta}{2} \right)\right] \, \mathrm{d} \theta \\ &= 9 \displaystyle \int_\frac{5\pi}{6}^\frac{3\pi}{2} (3 - 4 \sin \theta - 2 \cos 2\theta) \, \mathrm{d} \theta \\ &= 9 \left [3 \theta + 4 \cos \theta - \sin 2 \theta \right]_\frac{5\pi}{6}^\frac{3\pi}{2} \\ &= 9 \left [ \left (\dfrac{9\pi}{2} + 4 \cos \dfrac{3\pi}{2} - \sin 3\pi \right) - \left (\dfrac{5\pi}{2} + 4 \cos \dfrac{5\pi}{6} - \sin \dfrac{5\pi}{3} \right) \right] \\ &= 18\pi + \dfrac{27 \sqrt{3}}{2}. \end{aligned}$

Thus, $a = 18$ , $b = 27$ , and $c = 2$ , so $a + b + c = \boxed{47}$ .

Notice in the graph, there is a loop inside of the shaded region where r is negative, or when $\sin(\theta) > \frac{1}{2}$ . Therefore, we exclude all values in the interval $(\frac{\pi}{6}, \frac{5\pi}{6})$ since that region of the polar expression would've been redundant. Now, in order to find the area of the region bounded by the polar expression between 2 points, we can form infinitesimal triangles. Using the triangle formula, $\frac{1}{2} \cdot base \cdot height$ , we can obtain the series $\sum_{n=0}^{\infty} \frac{r_{n} + r_{n+1}}{2} \approx \int_{a}^{b} \frac{(3-6\sin(\theta))^2}{2} d\theta$ . Now we solve for the integral $\int_{a}^{b} \frac{(3-6\sin(\theta))^2}{2} d\theta \\ \int_{a}^{b} \frac{9(1-2\sin(\theta))^2}{2} d\theta \\ \frac{9}{2} \cdot \int_{a}^{b} 4\sin^2(\theta) - 4\sin(\theta) +1 d\theta \\ \frac{9}{2}(4\int_{a}^{b}\sin^2(\theta)d\theta-4\int_{a}^{b}\sin(\theta)d\theta+\int_{a}^{b}1d\theta \\ \frac{9}{2}(2\theta-2\cos(\theta)\sin(\theta)+4\cos(\theta)+\theta) \\ \frac{27\theta}{2}-9\cos(\theta)\sin(\theta)+18\cos(\theta) \\ \frac{27\theta}{2}-9\cos(\theta)(\sin(\theta)-2)$ Plugging this into our equation, we get $[\frac{27\theta}{2}-9\cos(\theta)(\sin(\theta)-2)]_{0}^{\frac{\pi}{6}} + [\frac{27\theta}{2}-9\cos(\theta)(\sin(\theta)-2)]_{\frac{5\pi}{6}}^{2\pi} \\ 27\pi+\frac{9\pi}{4}-\frac{45\pi}{4}+18-18+2(9\frac{\sqrt{3} \cdot 3}{4}) \\ 18\pi+\frac{27\sqrt{3}}{2}$ Therefore, $a=18, b=27, c=2, a+b+c = 18+27+2=\boxed{47}$

Area $A = \dfrac{9}{2} (\int_{0}^{2\pi} (1 - 2 sin\theta) d\theta - \int_{\dfrac{\pi}{6}}^{\dfrac{5\pi}{6}} (1 - 2 sin\theta) d\theta) = \dfrac{9}{2} (6\pi - (2\pi - 3 \sqrt{3})) = \dfrac{9}{2} (4\pi + 3 \sqrt{3}) =$

$18 \pi + \dfrac{27 \sqrt{3}}{2} = a\pi + \dfrac{b\sqrt{3}}{2} \implies a + b + c = 47.$