Answer
area = $4\pi $
Work Step by Step
Comparing Figure 22(A) and Figure 22(B), we see that the outer curve $r = 2 + \sin 2\theta $ is obtained by rotating the curve $r = 2 + \cos 2\theta $ by $\pi /4$ counterclockwise. This is verified below:
$2 + \cos \left( {2\left( {\theta - \frac{\pi }{4}} \right)} \right) = 2 + \cos \left( {2\theta - \frac{\pi }{2}} \right) = 2 + \cos 2\theta \cos \frac{\pi }{2} + \sin 2\theta \sin \frac{\pi }{2}$
Thus,
$2 + \cos \left( {2\left( {\theta - \frac{\pi }{4}} \right)} \right) = 2 + \sin 2\theta $
Since rotation does not change the area of a curve, so the curves $r = 2 + \sin 2\theta $ and $r = 2 + \cos 2\theta $ have the same area. Therefore, we can use the result in Exercise 19. So, the area between the two curves in Figure 22(B) is $4\pi $.
