How do you find the lengths of the curve (3y-1)^2=x^3 for 0<=x<=2?

Jun 28, 2018

Isolate $y$ then apply the arclength formula.

Explanation:

${\left(3 y - 1\right)}^{2} = {x}^{3}$, $x \in \left[0 , 2\right]$

Isolate $y$:

$y = \frac{1}{3} \left({x}^{\frac{3}{2}} + 1\right)$

Take the derivative with respect to $x$:

$y ' = \frac{1}{2} \sqrt{x}$

Arc length is given by:

$L = {\int}_{0}^{2} \sqrt{1 + \frac{x}{4}} \mathrm{dx}$

Integrate directly:

$L = \frac{8}{3} {\left[{\left(1 + \frac{x}{4}\right)}^{\frac{3}{2}}\right]}_{0}^{2}$

Insert the limits of integration:

$L = \frac{8}{3} \left({\left(\frac{3}{2}\right)}^{\frac{3}{2}} - 1\right)$