Answer
$s = 4\pi $
Work Step by Step
$$\eqalign{
& r = 4\sin \theta ,{\text{ 0}} \leqslant \theta \leqslant \pi \cr
& {\text{Find the Arc Length of a Polar curve}}{\text{, use the Theorem 10}}{\text{.14}} \cr
& s = \int_\alpha ^\beta {\sqrt {{{\left[ {f\left( \theta \right)} \right]}^2} + {{\left[ {f'\left( \theta \right)} \right]}^2}} } d\theta \cr
& {\text{Let }}r = f\left( \theta \right) = 4\sin \theta ,{\text{ }}\underbrace {{\text{0}} \leqslant \theta \leqslant \pi }_{\alpha \leqslant \theta \leqslant \beta } \cr
& f\left( \theta \right) = 4\sin \theta \cr
& f'\left( \theta \right) = 4\cos \theta \cr
& {\text{Therefore}}{\text{,}} \cr
& s = \int_0^\pi {\sqrt {{{\left[ {4\sin \theta } \right]}^2} + {{\left[ {4\cos \theta } \right]}^2}} } d\theta \cr
& s = \int_0^\pi {\sqrt {16{{\sin }^2}\theta + 16{{\cos }^2}\theta } } d\theta \cr
& s = \int_0^\pi {\sqrt {16\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right)} } d\theta \cr
& s = 4\int_0^\pi {d\theta } \cr
& {\text{Integrate}} \cr
& s = 4\left[ \theta \right]_0^\pi \cr
& s = 4\left( {\pi - 0} \right) \cr
& s = 4\pi \cr
& \cr} $$
