[tex]y=\left(4-x^{2/3}\right)^{3/2}\implies y'=\dfrac32\left(4-x^{2/3}\right)^{1/2}\left(-\dfrac23x^{-1/3}\right)=-\dfrac{\left(4-x^{2/3}\right)^{1/2}}{x^{1/3}}[/tex]
The arc length is given by
[tex]\displaystyle\int_1^8\sqrt{1+\left(-\dfrac{\left(4-x^{2/3}\right)^{1/2}}{x^{1/3}}\right)^2}\,\mathrm dx=\int_1^8\sqrt{\dfrac4{x^{2/3}}}\,\mathrm dx=\int_1^82x^{-1/3}\,\mathrm dx=9[/tex]
