Answer
$area = \frac{1}{2}\sqrt 3 $
Work Step by Step
The line $r = \sec \theta $ can be written as $r\cos \theta = 1$. Since $x = r\cos \theta $, it is a vertical line at $x=1$.
We draw the vertical line at $x=1$. Then draw two rays that make the angles $\theta=0$ and $\theta = \frac{\pi }{3}$, respectively, with the $x$-axis. So, we obtain the region bounded by the line and the rays.
Using Eq. (2) of Theorem 1, the area is
$area = \frac{1}{2}\mathop \smallint \limits_0^{\pi /3} {\sec ^2}\theta {\rm{d}}\theta $
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 = \frac{1}{2}\mathop \smallint \limits_0^{\pi /3} {\sec ^2}\theta {\rm{d}}\theta = \frac{1}{2}\tan \theta |_0^{\pi /3} = \frac{1}{2}\sqrt 3 $
We notice that the region is a right triangle $OAB$. From the figure we see that $\left| {OA} \right| = 1$.
Since $\tan \frac{\pi }{3} = \frac{{\left| {AB} \right|}}{{\left| {OA} \right|}} = \sqrt 3 $. It follows that $\left| {AB} \right| = \sqrt 3 $.
So, the area of the triangle is
$area = \frac{1}{2}\cdot\left| {OA} \right|\cdot\left| {AB} \right| = \frac{1}{2}\sqrt 3 $.
The two results agree.
