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