If you want to add up something along a curve, you need to compute a line integral. Common examples are determining the length of a curve, the mass of a wire, or how much work is done when moving a mass along a particular path.

Consider the problem of trying to find the length of a quarter of a circle. What do you know? In polar coordinates, a circle is given by $r=\hbox{constant}$, so that $dr=0$. Inserting this fact into the expression (2) of § {Other Coordinate Systems} for arclength in polar coordinates, one immediately obtains \begin{equation} ds^2 = r^2 \, d\phi^2 \end{equation} and finally \begin{equation} \Lint \, ds = \int_0^{{\pi}\over{2}} r\,d\phi = {\pi r\over{2}} \label{Answer} \end{equation}

But what if you didn't remember (2) of § {Other Coordinate Systems}? The calculation is not much harder in rectangular coordinates: You know that $x=r\cos\phi$ and $y=r\sin\phi$, with $r=\hbox{constant}$, so that $dx=-r\sin\phi\,d\phi$ and $dy=r\cos\phi\,d\phi$. Inserting this into (3) of § {The Vector Differential} again leads to ($\ref{Answer}$).

But what if you didn't even remember how to parameterize a circle, or, equivalently, how to use polar coordinates? Well, you still know that $x^2+y^2=r^2=\hbox{constant}$, so that $2x\,dx+2y\,dy=0$. Solving for $dy$ and inserting this into (3) of § {The Vector Differential} yields \begin{equation} ds^2 = \left(1+{{x^2}\over{y^2}}\right) dx^2 = {{r^2}\over{r^2-x^2}} \> dx^2 \end{equation} This leads to the (improper!) integral \begin{equation} \int_0^r {{dx}\over\sqrt{1-{{x^2}\over{r^2}}}} \end{equation} which is, of course, most easily computed via a trig substitution — or numerically — yielding the same answer.