Answer
The area of the region that lies inside one but not both of the curves is equal to the shaded area:
shaded area $ = 8\sqrt 2 $
Work Step by Step
In Exercise 21, we have obtained the intersection points between the two curves $r = 2 + \sin 2\theta $ and $r = 2 + \cos 2\theta $ for the interval $0 \le \theta \le 2\pi $, that is $\theta = \frac{\pi }{8},\frac{{5\pi }}{8},\frac{{9\pi }}{8},\frac{{13\pi }}{8}$.
Now, we find portion of the shaded area between the two curves in the interval $\frac{\pi }{8} \le \theta \le \frac{{5\pi }}{8}$. Let $A$ denote this area as is shown in the figure attached. So,
$A = \frac{1}{2}\cdot\mathop \smallint \limits_{\pi /8}^{5\pi /8} \left( {{{\left( {2 + \sin 2\theta } \right)}^2} - {{\left( {2 + \cos 2\theta } \right)}^2}} \right){\rm{d}}\theta $
$A = \frac{1}{2}\cdot\mathop \smallint \limits_{\pi /8}^{5\pi /8} \left( {\left( {4 + 4\sin 2\theta + {{\sin }^2}2\theta } \right) - \left( {4 + 4\cos 2\theta + {{\cos }^2}2\theta } \right)} \right){\rm{d}}\theta $
$A = \frac{1}{2}\cdot\mathop \smallint \limits_{\pi /8}^{5\pi /8} \left( {4\sin 2\theta + {{\sin }^2}2\theta - 4\cos 2\theta - {{\cos }^2}2\theta } \right){\rm{d}}\theta $
Since $\cos 4\theta = {\cos ^2}2\theta - {\sin ^2}2\theta $, the integral becomes
$A = \frac{1}{2}\cdot\mathop \smallint \limits_{\pi /8}^{5\pi /8} \left( {4\sin 2\theta - 4\cos 2\theta - \cos 4\theta } \right){\rm{d}}\theta $
$A = \frac{1}{2}\left( { - 2\cos 2\theta - 2\sin 2\theta - \frac{1}{4}\sin 4\theta } \right)|_{\pi /8}^{5\pi /8}$
$A = \frac{1}{2}\left( { - 2\cos \frac{{5\pi }}{4} + 2\cos \frac{\pi }{4} - 2\sin \frac{{5\pi }}{4} + 2\sin \frac{\pi }{4} - \frac{1}{4}\sin \frac{{5\pi }}{2} + \frac{1}{4}\sin \frac{\pi }{2}} \right)$
$A = 2\sqrt 2 $
The area of the region that lies inside one but not both of the curves in Figure 23 is equivalent to the shaded area. By symmetry, it is four times the area $A$. So,
shaded area $ = 4\cdot2\sqrt 2 = 8\sqrt 2 $
