Arc length integration Find the length of the arc formed by
$x^2=10y^3$
from point A to point B, where
A=(0,0) and B=(100,10).
My attempt: 
$\int_0^{100} \! \sqrt{1+(\frac{2}{3x})^2} \, \mathrm{d}x. $ However this integral does not converge.
 A: $y = \dfrac{x^{2/3}}{\sqrt[3]{10}} \to y' = \dfrac{2}{3\sqrt[3]{10}}\cdot x^{-1/3} \to (y')^2 = \dfrac{4}{9\sqrt[3]{100}}\cdot x^{-2/3}$. You can now continue...
A: There is no issue with convergence. You wanted to integrate with respect to $x$. It is more pleasant to integrate with respect to $y$. But we follow your path. 
If we are integrating with respect to $x$, the arclength is given by
$$\int_0^{100}\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx.$$ 
We have $y=x^{2/3}/\sqrt[3]{10}$. So $\frac{dy}{dx}=ax^{-1/3}$, where $a$ is a constant I don't want to worry about. Thus $1+\left(\frac{dy}{dx}\right)^2=1+a^2 x^{-2/3}$. Simplify, and take the square root. We get $x^{-1/3}\sqrt{x^{2/3}+a^2}$. This function behaves like a constant times $x^{-1/3}$ near $0$, so we have a convergent improper integral. Make the substitution $u=x^{2/3}+a^2$.
Remark: It is much more pleasant to note that the arclength is
$$\int_0^{10}\sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy.$$ 
We have $x=\sqrt{10}y^{3/2}$. Then $\frac{dx}{dy}$ is a constant times $y^{1/2}$, and we end up integrating something of shape $\sqrt{1+by}$.
