Answer
$area \simeq 0.143$
Work Step by Step
From Exercise 28 in Section 12.3 we have obtained the following results:
$\begin{array}{*{20}{c}}
{Point}&\theta \\
{C = \left( {1,0} \right)}&{0,\pi }\\
{D = \left( {1,1} \right)}&{\frac{\pi }{4}}\\
{A = \left( {0,1} \right)}&{\frac{\pi }{2}}\\
{B = \left( {0,0} \right)}&{\frac{{3\pi }}{4}}
\end{array}$
So, we see that part of the curve in the fourth quadrant traces out $B$ and $C$ corresponding to the interval $\frac{{3\pi }}{4} \le \theta \le \pi $. Using Eq. (2) of Theorem 1, the area in the fourth quadrant is
$area = \frac{1}{2}\cdot\mathop \smallint \limits_{3\pi /4}^\pi {\left( {\sin \theta + \cos \theta } \right)^2}{\rm{d}}\theta $
$area = \frac{1}{2}\cdot\mathop \smallint \limits_{3\pi /4}^\pi \left( {{{\sin }^2}\theta + 2\sin \theta \cos \theta + {{\cos }^2}\theta } \right){\rm{d}}\theta $
Since ${\sin ^2}\theta + {\cos ^2}\theta = 1$ and $\sin 2\theta = 2\sin \theta \cos \theta $, the integral becomes
$area = \frac{1}{2}\cdot\mathop \smallint \limits_{3\pi /4}^\pi \left( {1 + \sin 2\theta } \right){\rm{d}}\theta $
$area = \frac{1}{2}\left( {\theta - \frac{1}{2}\cos 2\theta } \right)|_{3\pi /4}^\pi $
$area = \frac{1}{2}\left( {\frac{\pi }{4} - \frac{1}{2}} \right) = \frac{\pi }{8} - \frac{1}{4} \simeq 0.143$
