Try This One on for Size!

We first set the curves equal to each other and we see that they intersect at $\theta = \frac{\pi}{3}$ . Next we need to define our two regions $D_1$ and $D_2$ .

$D_1= \left \{ (r, \; \theta) : 0 \le r \le 1+ \cos \theta, \: \frac{\pi}{3} \le \theta \le \pi \right \}$

$D_2= \left \{ (r, \; \theta) : 0 \le r \le 3 \cos \theta, \: \frac{\pi}{3} \le \theta \le \frac{\pi}{2} \right \}$

And now we need to find the area between the two curves.

$\textrm{Area}= 2 \left [ \iint_{D_1} dA - \iint_{D_2} dA \right ]$

Being more forthcoming we see

$2 \left [ \int_{\pi / 3}^{\pi} \int_{0}^{1+\cos \theta} r \; dr d \theta - \int_{\pi / 3 }^{\pi / 2} \int_{0}^{2 \cos \theta} r \; dr d \theta \right ] = \frac{\pi}{4}$

$a=1$ and $b=2$ and so $1 \times 2 = \boxed{2}$