Using dy/dx or dx/dy? to find arc length "Set up the integral for finding the arc length of the graph of $y=y^3-x$ from $(0,-1)$ to $(6,2)$".
Why use $dx/dy$ instead of $dy/dx$?
This problem is from my teacher
 A: You can't express $y$ as a function of $x$, in this case. Draw the graph, and you'll see why. So, trying to get $dy/dx$ as a function of $x$ won't work, either.  On the other hand, expressing $x$ as a function of $y$ is easy.
Hint for drawing the graph: $x=y^3-y=y(y-1)(y+1)$, so $x=0$ when $y=0,1,-1$.
A: It's just cleaner.  
$\dfrac{\operatorname d y}{\operatorname d x}=\dfrac{1}{3y^2-1}$.  But what is $y$ in terms of $x$?
$$s = \int_{0}^{6} \sqrt{1+\left(\frac{\operatorname dy}{\operatorname d x}\right)^2}\operatorname d x = \int_0^6 \sqrt{1+\left(\frac{1}{3y^2-1}\right)^2}\;\operatorname d x = \text{huh?}$$ 

Where as $\dfrac{\operatorname d x}{\operatorname d y} = 3y^2-1$
$$s = \int_{-1}^2 \sqrt{1+\left(\frac{\operatorname d x}{\operatorname d y}\right)^2} \operatorname d y = \int_{-1}^{2} \sqrt{1+(3y^2-1)^2}\operatorname d y = \ldots$$ 
A: \begin{align}
\text{arc length} & = \int_{(x,y)=(x_0,y_0)}^{(x,y)=(x_1,y_1)} \sqrt{(dx)^2+(dy)^2} \\[10pt] & = \int_{x_0}^{x_1} \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\ dx \\[10pt]
& = \int_{y_0}^{y_1} \sqrt{\left(\frac{dx}{dy}\right)^2+ 1}\  dy.
\end{align}
Either is correct if $x$ and $y$ can be expressed as functions of each other.  In some cases, one of the integrals may be more tractable than the other.
The expression $\sqrt{(dx)^2+(dy)^2}$ comes from the Pythagorean theorem.
Sometimes one uses
$$
\int_{t_0}^{t_1} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\  dt.
$$
