Answer
$$s = \pi $$
Work Step by Step
$$\eqalign{
& x = \sqrt {36 - {y^2}} ,{\text{ }}0 \leqslant y \leqslant 3 \cr
& \cr
& \left( {\text{a}} \right){\text{Graph below}} \cr
& \cr
& {\text{Use the formula for arc length}} \cr
& s = \int_c^d {\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} } dy \cr
& \frac{{dx}}{{dy}} = \frac{d}{{dy}}\left[ {\sqrt {36 - {y^2}} } \right] \cr
& \frac{{dx}}{{dy}} = \frac{{ - 2y}}{{2\sqrt {36 - {y^2}} }} \cr
& \frac{{dx}}{{dy}} = - \frac{y}{{\sqrt {36 - {y^2}} }} \cr
& s = \int_0^3 {\sqrt {1 + {{\left( { - \frac{y}{{\sqrt {36 - {y^2}} }}} \right)}^2}} } dy \cr
& s = \int_0^3 {\sqrt {1 + \frac{{{y^2}}}{{36 - {y^2}}}} } dy \cr
& s = \int_0^3 {\sqrt {\frac{{36 - {y^2} + {y^2}}}{{36 - {y^2}}}} } dy \cr
& s = \int_0^3 {\sqrt {\frac{{36}}{{36 - {y^2}}}} } dy \cr
& s = 6\int_0^3 {\frac{1}{{\sqrt {36 - {y^2}} }}} dy \cr
& \cr
& \left( {\text{c}} \right){\text{Integrate }} \cr
& s = 6\left[ {{{\sin }^{ - 1}}\left( {\frac{y}{6}} \right)} \right]_0^3 \cr
& s = 6\left[ {{{\sin }^{ - 1}}\left( {\frac{3}{6}} \right) - {{\sin }^{ - 1}}\left( 0 \right)} \right] \cr
& s = 6\left[ {\frac{\pi }{6} - 0} \right] \cr
& s = \pi \cr
& \cr
& {\text{Graph}} \cr} $$
