Answer
area = $4\pi $
Work Step by Step
Using the relations: $x = r\cos \theta $ and $y = r\sin \theta $ we plot several points for each curve to determine the limits of integration.
Curve 1. Outer curve $r = 2 + \cos 2\theta $
$\begin{array}{*{20}{c}}
\theta &{\left( {x,y} \right)}\\
0&{\left( {3,0} \right)}\\
{\frac{\pi }{4}}&{\left( {\sqrt 2 ,\sqrt 2 } \right)}\\
{\frac{\pi }{2}}&{\left( {0,1} \right)}
\end{array}$
From the figure we see that part of the curve in the first quadrant corresponds to the interval $0 \le \theta \le \pi /2$. Using symmetry, the area of the entire curve is four times the area in the first quadrant. Let $A$ denote the area of the entire curve. So,
$A = 4\cdot\frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} {\left( {2 + \cos 2\theta } \right)^2}{\rm{d}}\theta $
$A = 2\cdot\mathop \smallint \limits_0^{\pi /2} \left( {4 + 4\cos 2\theta + {{\cos }^2}2\theta } \right){\rm{d}}\theta $
Since ${\cos ^2}2\theta = \frac{1}{2}\left( {1 + \cos 4\theta } \right)$, the integral becomes
$A = 2\cdot\mathop \smallint \limits_0^{\pi /2} \left( {4 + 4\cos 2\theta + \frac{1}{2}\left( {1 + \cos 4\theta } \right)} \right){\rm{d}}\theta $
$A = 2\left( {\frac{9}{2}\theta + 2\sin 2\theta + \frac{1}{8}\sin 4\theta } \right)|_0^{\pi /2}$
$A = 2\left( {\frac{{9\pi }}{4}} \right) = \frac{{9\pi }}{2}$
Curve 2. Inner curve $r = \sin 2\theta $
$\begin{array}{*{20}{c}}
\theta &{\left( {x,y} \right)}\\
0&{\left( {0,0} \right)}\\
{\frac{\pi }{4}}&{\left( {\frac{1}{2}\sqrt 2 ,\frac{1}{2}\sqrt 2 } \right)}\\
{\frac{\pi }{2}}&{\left( {0,0} \right)}
\end{array}$
From the figure we see that part of the curve in the first quadrant corresponds to the interval $0 \le \theta \le \pi /2$. Using symmetry, the area of the entire curve is four times the area in the first quadrant. Let $B$ denote the area of the entire curve. So,
$B = 4\cdot\frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} {\sin ^2}2\theta {\rm{d}}\theta $
Since ${\sin ^2}2\theta = \frac{1}{2}\left( {1 - \cos 4\theta } \right)$, the integral becomes
$B = \mathop \smallint \limits_0^{\pi /2} \left( {1 - \cos 4\theta } \right){\rm{d}}\theta $
$B = \left( {\theta - \frac{1}{4}\sin 4\theta } \right)|_0^{\pi /2} = \frac{\pi }{2}$
Thus, the area between the two curves is
$A - B = \frac{{9\pi }}{2} - \frac{\pi }{2} = 4\pi $
