Answer
$$L = \frac{4}{3}$$
Work Step by Step
$$\eqalign{
& x = \frac{1}{3}{\left( {{y^2} + 2} \right)^{3/2}}{\text{ from }}y = 0{\text{ to }}y = 1 \cr
& {\text{Calculate }}\frac{{dx}}{{dy}} \cr
& \frac{{dx}}{{dy}} = \frac{d}{{dy}}\left[ {\frac{1}{3}{{\left( {{y^2} + 2} \right)}^{3/2}}} \right] \cr
& \frac{{dx}}{{dy}} = \frac{3}{2}\left( {\frac{1}{3}} \right){\left( {{y^2} + 2} \right)^{1/2}}\left( {2y} \right) \cr
& \frac{{dx}}{{dy}} = y{\left( {{y^2} + 2} \right)^{1/2}} \cr
& {\text{Use }}L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} dy} \cr
& L = \int_0^1 {\sqrt {1 + {{\left( {y{{\left( {{y^2} + 2} \right)}^{1/2}}} \right)}^2}} dy} \cr
& L = \int_0^1 {\sqrt {1 + {y^2}\left( {{y^2} + 2} \right)} dy} \cr
& L = \int_0^1 {\sqrt {1 + {y^4} + 2{y^2}} dy} \cr
& L = \int_0^1 {\sqrt {{{\left( {{y^2} + 1} \right)}^2}} dy} \cr
& L = \int_0^1 {\left( {{y^2} + 1} \right)dy} \cr
& {\text{Integrating}} \cr
& L = \left[ {\frac{1}{3}{y^3} + y} \right]_0^1 \cr
& L = \left[ {\frac{1}{3}{{\left( 1 \right)}^3} + \left( 1 \right)} \right] - \left[ {\frac{1}{3}{{\left( 0 \right)}^3} + \left( 0 \right)} \right] \cr
& L = \frac{1}{3} + 1 \cr
& L = \frac{4}{3} \cr} $$
