Answer
$=\frac{4}{3}$
Work Step by Step
The formula for the arc length is
$s=\int_{a}^b \sqrt{1 + (x')^2} dx$
Differentiate $x =\frac{1}{3}\sqrt y(y-3)$ with respect to y.
$x' = \frac{1}{2}(\sqrt y-\frac{1}{\sqrt y})$
Substitute the value of y' in $1 + (x')^2$
$1+(x')^2=1+[\frac{1}{2}(\sqrt y -\frac{1} {\sqrt {y}})]^2$
$=[\frac{1}{2}(\sqrt y +\frac{1} {\sqrt {y}})]^2$
Substitute the value of $1 + (x')^2$ in the formula for the arc length $s=\int_{a}^b \sqrt{1 + (x')^2} dx$
Solve for the distance s
$s=\int_{1}^4\sqrt{[\frac{1}{2}(\sqrt y +\frac{1} {\sqrt {y}})]^2} dy$
$=\int_{1}^4 [\frac{1}{2}(\sqrt y +\frac{1} {\sqrt {y}})]dy$
$=[\frac{(y-3)\sqrt y}{3}]_1^4\ $
$=\frac{4}{3}$
