Answer
$$s = \int_0^2 {\sqrt {1 + 4{x^2}} } dx \approx 4.646783$$
Work Step by Step
$$\eqalign{
& y = 4 - {x^2},{\text{ }}0 \leqslant x \leqslant 2 \cr
& \cr
& \left( {\text{a}} \right){\text{Graph below}} \cr
& \cr
& {\text{Use the formula for arc length}} \cr
& s = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } dx \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {4 - {x^2}} \right] \cr
& \frac{{dy}}{{dx}} = - 2x \cr
& s = \int_0^2 {\sqrt {1 + {{\left( { - 2x} \right)}^2}} } dx \cr
& s = \int_0^2 {\sqrt {1 + 4{x^2}} } dx \cr
& \cr
& \left( {\text{c}} \right){\text{Integrate by using a CAS or graphing utility}} \cr
& s = \int_0^2 {\sqrt {1 + 4{x^2}} } dx \approx 4.646783 \cr
& \cr
& {\text{Graph}} \cr} $$
