## Calculus 8th Edition

$\frac{1}{27} (80\sqrt{10} - 13\sqrt{13})$
$x^{2} = (y-4)^{3}, x = (y-4)^{3/2}$(for $x>0$) then $x' = \frac{3}{2} (y-4)^{1/2}$ and $1+(dx/dy)^{2} = 1+ \frac{9}{4} (y-4) = \frac{9}{4}y - 8$ So $L = \int^{8}_{5} \sqrt{\frac{9}{4}y-8} dy = \int^{10}_{13/4} \sqrt{u} (\frac{4}{9} du) = \frac{4}{9}[\frac{2}{3} u^{3/2}]^{10}_{13/4} = \frac{1}{27} (80\sqrt{10} - 13\sqrt{13})$