## Calculus (3rd Edition)

Length: $s = \frac{1}{4}\sqrt 2 \pi$
We have $r = f\left( \theta \right) = \sqrt {1 + \sin 2\theta }$, ${\ \ }$ $f'\left( \theta \right) = \frac{{\cos 2\theta }}{{\sqrt {1 + \sin 2\theta } }}$ Using Eq.(7), we get the arc length for $0 \le \theta \le \pi /4$ $s = \mathop \smallint \limits_0^{\pi /4} \sqrt {1 + \sin 2\theta + \frac{{{{\cos }^2}2\theta }}{{1 + \sin 2\theta }}} {\rm{d}}\theta$ $s = \mathop \smallint \limits_0^{\pi /4} \sqrt {\frac{{{{\left( {1 + \sin 2\theta } \right)}^2} + {{\cos }^2}2\theta }}{{1 + \sin 2\theta }}} {\rm{d}}\theta$ $s = \mathop \smallint \limits_0^{\pi /4} \sqrt {\frac{{1 + 2\sin 2\theta + {{\sin }^2}2\theta + {{\cos }^2}2\theta }}{{1 + \sin 2\theta }}} {\rm{d}}\theta$ $s = \mathop \smallint \limits_0^{\pi /4} \sqrt {\frac{{2 + 2\sin 2\theta }}{{1 + \sin 2\theta }}} {\rm{d}}\theta = \mathop \smallint \limits_0^{\pi /4} \sqrt 2 {\rm{d}}\theta$ $s = \frac{1}{4}\sqrt 2 \pi$