O comprimento do arco é dado pela seguinte fórmula:

\(\Longrightarrow S=\int \limits_{y_A}^{y_B} \sqrt {1 + \Big [g'(y)\Big ] ^2} dy\)

Sendo \(x=g(y)\), a expressão da derivada \(g'(y)\) é:

\(\Longrightarrow g'(y) = \Big ({1 \over 8}y^4 + {1 \over 4y^2 } \Big )'\)

\(\Longrightarrow g'(y) = \Big ({1 \over 8}y^4 + {1 \over 4 }y^{-2} \Big )'\)

\(\Longrightarrow g'(y) = {4 \over 8}y^{4-1}+ {-2 \over 4 }y^{-2-1}\)

\(\Longrightarrow g'(y) = {1 \over 2}y^3 -{1 \over 2 }y^{-3}\)

Portanto, a expressão de \(\Big [ g'(y) \Big ] ^2\) é:

 \(\Longrightarrow \Big [ g'(y) \Big ] ^2 = \Big [ {1 \over 2}(y^3 -y^{-3}) \Big ] ^2\)

\(\Longrightarrow \Big [ g'(y) \Big ] ^2 = {1 \over 4}(y^3 -y^{-3}) ^2\)

\(\Longrightarrow \Big [ g'(y) \Big ] ^2 = {y^6 - 2 -y^{-6} \over 4}\)

Portanto, a expressão da soma \(1+\Big [ g'(y) \Big ] ^2\) é:

\(\Longrightarrow 1+\Big [ g'(y) \Big ] ^2 = 1 + {y^6 - 2 -y^{-6} \over 4}\)

\(\Longrightarrow 1+\Big [ g'(y) \Big ] ^2 = {4 + y^6 - 2 -y^{-6} \over 4}\)

\(\Longrightarrow 1+\Big [ g'(y) \Big ] ^2 = { y^6 + 2 -y^{-6} \over 4}\)

\(\Longrightarrow 1 +\Big [ g'(y) \Big ] ^2 = {1 \over 4}(y^3 +y^{-3}) ^2\)

Portanto, sendo \(y_A = 1\) e \(y_B= 2\), o valor de \(S\) é:

\(\Longrightarrow S=\int \limits_{y_A}^{y_B} \sqrt {1 + \Big [g'(y)\Big ] ^2} dy\)

\(\Longrightarrow S=\int \limits_{1}^{2} \sqrt {{1 \over 4}(y^3 +y^{-3}) ^2} dy\)

\(\Longrightarrow S=\int \limits_{1}^{2} {1 \over 2}(y^3 +y^{-3}) dy\)

\(\Longrightarrow S= {1 \over 2} \Big ({1 \over 3+1}y^{3+1} +{1 \over -3 + 1}y^{-3+1} \Big ) \bigg |_{1}^{2}\)

\(\Longrightarrow S= {1 \over 2} \Big ({1 \over 4}y^{4} -{1 \over 2}y^{-2} \Big ) \bigg |_{1}^{2}\)

\(\Longrightarrow S= {1 \over 2} \Big ({1 \over 4}(2)^{4} -{1 \over 2}(2)^{-2} \Big ) - {1 \over 2} \Big ({1 \over 4}(1)^{4} -{1 \over 2}(1)^{-2} \Big ) \)

\(\Longrightarrow S= {1 \over 2} \Big (4 -{1 \over 8} \Big ) - {1 \over 2} \Big ({1 \over 4} -{1 \over 2} \Big ) \)

\(\Longrightarrow S= {1 \over 2} \Big (4 -{1 \over 8} -{1 \over 4} +{1 \over 2}\Big )\)

\(\Longrightarrow S= {1 \over 2} \Big ({33 \over 8}\Big )\)

\(\Longrightarrow \fbox {$ S= {33 \over 16} $}\)