Answer
$area = 16$
Work Step by Step
In polar coordinates, the triangle in Figure 13 is the region bounded by the line $r = 4\sec \left( {\theta - \frac{\pi }{4}} \right)$ and the rays $\theta=0$ and $\theta = \pi /2$. Using Eq. (2) of Theorem 1, the area is
$area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} {\left( {4\sec \left( {\theta - \frac{\pi }{4}} \right)} \right)^2}{\rm{d}}\theta $
$area = 8\cdot\mathop \smallint \limits_0^{\pi /2} {\sec ^2}\left( {\theta - \frac{\pi }{4}} \right){\rm{d}}\theta $
Write $\phi = \theta - \frac{\pi }{4}$. So, $d\phi = d\theta $. The integral becomes
$area = 8\cdot\mathop \smallint \limits_{ - \pi /4}^{\pi /4} {\sec ^2}\phi {\rm{d}}\phi $
From Eq. 14 of Section 8.2 (page 402) we have
$\smallint {\sec ^m}x{\rm{d}}x = \frac{{\tan x{{\sec }^{m - 2}}x}}{{m - 1}} + \frac{{m - 2}}{{m - 1}}\smallint {\sec ^{m - 2}}{\rm{d}}x$
So,
$area = 8\cdot\mathop \smallint \limits_{ - \pi /4}^{\pi /4} {\sec ^2}\phi {\rm{d}}\phi = 8\tan \phi |_{ - \pi /4}^{\pi /4} = 16$
The length of $OP$ can be obtained by substituting $\frac{\pi }{2}$ in $r = 4\sec \left( {\theta - \frac{\pi }{4}} \right)$, so ${r_P} = 4\sqrt 2 $.
The rectangular coordinates of $P$ is $P = \left( {0,4\sqrt 2 } \right)$.
Likewise, the length of $OQ$ can be obtained by substituting $0$ in $r = 4\sec \left( {\theta - \frac{\pi }{4}} \right)$, so ${r_Q} = 4\sqrt 2 $.
The rectangular coordinates of $Q$ is $Q = \left( {4\sqrt 2 ,0} \right)$.
The area of the triangle is
$area = \frac{1}{2}\cdot\left( {4\sqrt 2 } \right)\left( {4\sqrt 2 } \right) = 16$.
