I am trying to find the arc length of $y^3 = x^2$ and I am suppose to use two formulas, one for in terms of x and one for in terms of y.
At first I need to find (0,0) (1,1) and I start with in terms of x
$y\prime = \frac{2}{3}x^\frac{-1}{3}$
$\left(\frac{2}{3}x^\frac{-1}{3}\right)^2 = \frac{4}{9x^\frac{2}{3}}$
$\int_0^1 \sqrt{1 + \left( \frac{2}{3}x^\frac{-1}{3}\right)^2}\, dx$
$\int_0^1 \sqrt{1 + \frac{4}{9x^\frac{2}{3}}}\, dx$
This is undefined at 0 so it is an improper function. I do not think I can continue.
Now in terms of y.
$x = y^\frac{3}{2}$
$x \prime = \frac{3\sqrt{y}}{2}$ $ \left(\frac{3\sqrt{y}}{2}\right)^2 = \frac{9y}{4}$
$\int_0^1 \sqrt{1+ \frac{9y}{4}}\, dy$
I have no idea how to factor that out, I have tried many ways but I can not get it.