#### Answer

Length: $s = 8$

#### Work Step by Step

In Exercise 7, we see that the cardioid $r = 1 - \cos \theta $ traces out the entire curve for $0 \le \theta \le 2\pi $.
We have
$r = f\left( \theta \right) = 1 - \cos \theta $, ${\ \ }$ $f'\left( \theta \right) = \sin \theta $
Using Eq.(7), we get the length of the cardioid for $0 \le \theta \le 2\pi $
$s = \mathop \smallint \limits_0^{2\pi } \sqrt {{{\left( {1 - \cos \theta } \right)}^2} + {{\sin }^2}\theta } {\rm{d}}\theta $
$s = \mathop \smallint \limits_0^{2\pi } \sqrt {1 - 2\cos \theta + {{\cos }^2}\theta + {{\sin }^2}\theta } {\rm{d}}\theta $
$s = \mathop \smallint \limits_0^{2\pi } \sqrt {2 - 2\cos \theta } {\rm{d}}\theta = \sqrt 2 \mathop \smallint \limits_0^{2\pi } \sqrt {1 - \cos \theta } {\rm{d}}\theta $
Since $\cos \theta = 1 - 2{\sin ^2}\frac{\theta }{2}$, the integral becomes
$s = \sqrt 2 \mathop \smallint \limits_0^{2\pi } \sqrt {2{{\sin }^2}\frac{\theta }{2}} {\rm{d}}\theta = 2\cdot\mathop \smallint \limits_0^{2\pi } \sin \frac{\theta }{2}{\rm{d}}\theta $
$s = 4\left( { - \cos \frac{\theta }{2}} \right)|_0^{2\pi } = 8$