## Calculus 8th Edition

$\frac{33}{16}$
$x = \frac{y^{4}}{8} + \frac{1}{4y^{2}}$ then $\frac{dx}{dy} = \frac{1}{2} y^{3} - \frac{1}{2} y^{-3}$ $1+(dx/dy)^{2} = 1+\frac{1}{4} y^{6} - \frac{1}{2} + \frac{1}{4} y^{-6} = \frac{1}{4} y^{6} + \frac{1}{2} + \frac{1}{4}y^{-6} = (\frac{1}{2} y^{3} + \frac{1}{2} y^{-3})^{2}$ So $L = \int^{2}_{1} \sqrt{(\frac{1}{2}y^{3} + \frac{1}{2}y^{-3})^{2}}dy = \int^{2}_{1} (\frac{1}{2}y^{3} + \frac{1}{2} y^{-3}) dy = [\frac{1}{8}y^{4} - \frac{1}{4}y^{-2}]^{2}_{1} = \frac{33}{16}$