Answer
$$L = 3\pi $$
Work Step by Step
$$\eqalign{
& x = \cos 3t,{\text{ }}y = \sin 3t{\text{ on the interval }}\left( {0 \leqslant t \leqslant \pi } \right) \cr
& {\text{Calculate the derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\cos 3t} \right] = - 3\sin 3t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\sin 3t{\text{ }}} \right] = 3\cos 3t \cr
& {\text{Use the Arc Length Formula For Parametric Curves}} \cr
& L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} \cr
& L = \int_0^\pi {\sqrt {{{\left( { - 3\sin 3t} \right)}^2} + {{\left( {3\cos 3t} \right)}^2}} dt} \cr
& L = \int_0^\pi {\sqrt {9{{\sin }^2}3t + 9{{\cos }^2}3t} dt} \cr
& L = \int_0^\pi {\sqrt {9\left( {{{\sin }^2}3t + {{\cos }^2}3t} \right)} dt} \cr
& L = \int_0^\pi {3dt} \cr
& {\text{Integrating}} \cr
& L = 3\left[ t \right]_0^\pi \cr
& L = 3\left( {\pi - 0} \right) \cr
& L = 3\pi \cr} $$