Answer
$L = \int^{4}_{1}\sqrt{1+(\frac{1}{2\sqrt{y}}-1)^{2}}dy$
Work Step by Step
$x = \sqrt{y} - y$
then $dx/dy = 1/(2\sqrt{y}) -1$
and $1+(dx/dy)^{2} = 1+(\frac{1}{2\sqrt{y}}-1)^{2}$
$L = \int^{4}_{1}\sqrt{1+(\frac{1}{2\sqrt{y}}-1)^{2}}dy$