# How do you find the lengths of the curve 8x=2y^4+y^-2 for 1<=y<=2?

Apr 26, 2017

$8 x = 2 {y}^{4} + {y}^{- 2}$

$\implies x ' = {y}^{3} - \frac{1}{4} {y}^{- 3}$

$s = {\int}_{1}^{2} \sqrt{1 + {\left(x '\right)}^{2}} \setminus \mathrm{dy}$

$= {\int}_{1}^{2} \sqrt{1 + {\left({y}^{3} - \frac{1}{4} {y}^{- 3}\right)}^{2}} \setminus \mathrm{dy}$

$= {\int}_{1}^{2} \sqrt{1 + {y}^{6} + \frac{1}{16} {y}^{- 6} - \frac{1}{2}} \setminus \mathrm{dy}$

$= {\int}_{1}^{2} \sqrt{{y}^{6} + \frac{1}{16} {y}^{- 6} + \frac{1}{2}} \setminus \mathrm{dy}$

$= {\int}_{1}^{2} \sqrt{{\left({y}^{3} + \frac{1}{4} {y}^{- 3}\right)}^{2}} \setminus \mathrm{dy}$

$= {\int}_{1}^{2} {y}^{3} + \frac{1}{4} {y}^{- 3} \setminus \mathrm{dy}$

$= {\left[{y}^{4} / 4 - \frac{1}{8 {y}^{2}}\right]}_{1}^{2}$

$= \frac{123}{32}$