Thomas' Calculus 13th Edition

$L= \sqrt {y^4 - y^2 +1}$
$y= \sqrt {x-3}$ $y^2= x-3$ $x = y^2 + 3$ Say a is the distance between (x,0) and (4,0), then a = x-4 Say b is the distance between (4,y) and (4,0) then b = y-0 Using Pythagorean's theorem, one can say that $L^2 = a^2 + b^2$. Thus $L = \sqrt {a^2 + b^2}$. So, $L = \sqrt {(x-4)^2 + (y-0)^2}$. Substituting $x = y^2 + 3$ gives $L = \sqrt {(y^2 + 3-4)^2 + (y-0)^2}$ $L = \sqrt {(y^2 -1)^2 + y^2}$ $L = \sqrt {y^4 -2y^2 +1 + y^2}$ $L = \sqrt {y^4 -y^2 +1}$