Answer
The length of the curve: $s \simeq 6.11$
Work Step by Step
Since $r=\theta$, the rectangular coordinates of a point on the curve corresponding to $\theta$ is given by
$\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right)$,
$\left( {x,y} \right) = \left( {\theta \cos \theta ,\theta \sin \theta } \right)$
So,
$x' = \cos \theta - \theta \sin \theta $, ${\ \ }$ $y' = \sin \theta + \theta \cos \theta $
By Theorem 1 of Section 12.2, the length of the curve for $0 \le \theta \le \pi $ is
$s = \mathop \smallint \limits_0^\pi \sqrt {{{\left( {\cos \theta - \theta \sin \theta } \right)}^2} + {{\left( {\sin \theta + \theta \cos \theta } \right)}^2}} {\rm{d}}\theta $
Since
${\left( {\cos \theta - \theta \sin \theta } \right)^2} + {\left( {\sin \theta + \theta \cos \theta } \right)^2}$
$ = {\cos ^2}\theta - 2\theta \cos \theta \sin \theta + {\theta ^2}{\sin ^2}\theta + {\sin ^2}\theta + 2\theta \sin \theta \cos \theta + {\theta ^2}{\cos ^2}\theta $
$ = 1 + {\theta ^2}$
So,
$s = \mathop \smallint \limits_0^\pi \sqrt {1 + {\theta ^2}} {\rm{d}}\theta $
Let $\theta = \tan v$. So, $d{\rm{\theta }} = {\sec ^2}vd{\rm{v}}$. The integral becomes
$s = \mathop \smallint \limits_0^{{{\tan }^{ - 1}}\pi } {\sec ^2}v\sqrt {1 + {{\tan }^2}v} {\rm{d}}v$
Since ${\sec ^2}v - {\tan ^2}v = 1$, we get
$s = \mathop \smallint \limits_0^{{{\tan }^{ - 1}}\pi } {\sec ^3}v{\rm{d}}v$
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,
$s = \frac{{\tan v\sec v}}{2}|_0^{{{\tan }^{ - 1}}\pi } + \frac{1}{2}\mathop \smallint \limits_0^{{{\tan }^{ - 1}}\pi } \sec v{\rm{d}}v$
From Eq. 13 of Section 8.2 (page 402) we have
$\smallint \sec x{\rm{d}}x = \ln \left| {\sec x + \tan x} \right| + C$
So,
$s = \frac{{\tan v\sec v}}{2}|_0^{{{\tan }^{ - 1}}\pi } + \frac{1}{2}\ln \left| {\sec v + \tan v} \right||_0^{{{\tan }^{ - 1}}\pi }$
$s \simeq 6.11$
