Chapter 10 - Conics, Parametric Equations, and Polar Coordinates - 10.5 Exercises - Page 732: 55

$s = 8$

$$\eqalign{ & r = 1 + \sin \theta ,{\text{ }}0 \leqslant \theta \leqslant 2\pi \cr & {\text{Find the Arc Length of a Polar curve}}{\text{, use the Theorem 10}}{\text{.14}} \cr & s = \int_\alpha ^\beta {\sqrt {{{\left[ {f\left( \theta \right)} \right]}^2} + {{\left[ {f'\left( \theta \right)} \right]}^2}} } d\theta \cr & {\text{Let }}r = f\left( \theta \right) = 4\sin \theta ,{\text{ }}\underbrace {0 \leqslant \theta \leqslant 2\pi }_{\alpha \leqslant \theta \leqslant \beta } \cr & f\left( \theta \right) = 1 + \sin \theta \cr & f'\left( \theta \right) = \cos \theta \cr & {\text{Therefore}}{\text{,}} \cr & s = \int_0^{2\pi } {\sqrt {{{\left[ {1 + \sin \theta } \right]}^2} + {{\left[ {\cos \theta } \right]}^2}} } d\theta \cr & s = \int_0^{2\pi } {\sqrt {1 + 2\sin \theta + {{\sin }^2}\theta + {{\cos }^2}\theta } } d\theta \cr & s = \int_0^{2\pi } {\sqrt {1 + 2\sin \theta + 1} } d\theta \cr & s = \int_0^{2\pi } {\sqrt {2 + 2\sin \theta } } d\theta \cr & s = \int_0^{2\pi } {\sqrt {2\left( {1 + \sin \theta } \right)} } d\theta \cr & s = \sqrt 2 \int_0^{2\pi } {\sqrt {1 + \sin \theta } } d\theta \cr & {\text{Using the symmetry}} \cr & s = 2\sqrt 2 \int_{\pi /2}^{3\pi /2} {\sqrt {1 + \sin \theta } } d\theta \cr & {\text{Rationalizing}} \cr & s = 2\sqrt 2 \int_{\pi /2}^{3\pi /2} {\sqrt {1 + \sin \theta } } \left( {\frac{{\sqrt {1 - \sin \theta } }}{{\sqrt {1 - \sin \theta } }}} \right)d\theta \cr & s = 2\sqrt 2 \int_{\pi /2}^{3\pi /2} {\frac{{\sqrt {1 - \sin \theta } }}{{\sqrt {1 - \sin \theta } }}} d\theta \cr & s = 2\sqrt 2 \int_{\pi /2}^{3\pi /2} {\frac{{\sqrt {{{\cos }^2}\theta } }}{{\sqrt {1 - \sin \theta } }}} d\theta \cr & {\text{For the interval }}\frac{\pi }{2} \leqslant \theta \leqslant \frac{{3\pi }}{2}{\text{ }}\sqrt {{{\cos }^2}\theta } = - \cos \theta \cr & s = 2\sqrt 2 \int_{\pi /2}^{3\pi /2} {\frac{{ - \cos \theta }}{{\sqrt {1 - \sin \theta } }}} d\theta \cr & {\text{Integrate}} \cr & s = 4\sqrt 2 \left[ {\sqrt {1 - \sin \theta } } \right]_{\pi /2}^{3\pi /2} \cr & s = 4\sqrt 2 \left[ {\sqrt {1 - \sin \left( {\frac{{3\pi }}{2}} \right)} - \sqrt {1 - \sin \left( {\frac{\pi }{2}} \right)} } \right] \cr & s = 4\sqrt 2 \left[ {\sqrt {1 - 1} - \sqrt {1 - \left( { - 1} \right)} } \right] \cr & s = 4\sqrt 2 \left( {\sqrt 2 } \right) \cr & s = 8 \cr} $$