Answer
$$s = \frac{{76}}{3}$$
Work Step by Step
$$\eqalign{
& x = \frac{1}{3}{\left( {{y^2} + 2} \right)^{3/2}},{\text{ }}0 \leqslant y \leqslant 4 \cr
& {\text{Differentiate}} \cr
& \frac{{dx}}{{dy}} = \frac{d}{{dy}}\left[ {\frac{1}{3}{{\left( {{y^2} + 2} \right)}^{3/2}}} \right] \cr
& \frac{{dx}}{{dy}} = \frac{1}{3}\left( {\frac{3}{2}} \right){\left( {{y^2} + 2} \right)^{1/2}}\left( {2y} \right) \cr
& \frac{{dx}}{{dy}} = y\sqrt {{y^2} + 2} \cr
& {\text{Use the formula for arc length}} \cr
& s = \int_c^d {\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} } dy \cr
& s = \int_0^4 {\sqrt {1 + {{\left( {y\sqrt {{y^2} + 2} } \right)}^2}} } dy \cr
& s = \int_0^4 {\sqrt {1 + {y^2}\left( {{y^2} + 2} \right)} } dy \cr
& s = \int_0^4 {\sqrt {1 + {y^4} + 2{y^2}} } dy \cr
& s = \int_0^4 {\sqrt {{{\left( {{y^2} + 1} \right)}^2}} } dy \cr
& s = \int_0^4 {\left( {{y^2} + 1} \right)} dy \cr
& {\text{Integrate}} \cr
& s = \left[ {\frac{1}{3}{y^3} + y} \right]_0^4 \cr
& s = \left[ {\frac{1}{3}{{\left( 4 \right)}^3} + \left( 4 \right)} \right] - \left[ {\frac{1}{3}{{\left( 0 \right)}^3} + \left( 0 \right)} \right] \cr
& s = \frac{{76}}{3} \cr} $$
