Answer
$\dfrac{51\sqrt 3}{16}$
Work Step by Step
$r=2 +cos 2 \theta$ and $r=2+sin \theta$
$Area,A=
2[\frac{1}{2}\int_{-\pi/2}^{\pi/6}(2 +cos 2 \theta)^2-( 2+sin \theta)^2]d\theta$
$=\int_{-\pi/2}^{\pi/6}\frac{cos 4\theta}{2}+4 cos 2 \theta+\frac{cos 2\theta}{2}-4 sin \theta d\theta$
$=[\frac{sin 4\theta}{8}+2 sin 2 \theta+\frac{sin 2\theta}{4}+4 cos \theta]_{-\pi/2}^{\pi/6}$
$=\dfrac{51\sqrt 3}{16}$
