<s>[INST] Arc length of $y^3 = x^2$ 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.
 [/INST] Neater is to do it parametrically, setting $x=t^2$ and $y=t^3$, and using the formula
$$
s = \int_0^1\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}dt\\
=\int_0^1t\sqrt{9t^2+4}dt\,.
$$
Now you set $u=9t^2+4$, getting $\int_4^{13}u^{1/2}du/18$, evaluating to $(13^{3/2}-8)/27$.
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