Line Integrals

Section 13.5 Line Integrals

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 \(r=\hbox{constant}\text{,}\) so that \(dr=0\text{.}\) Inserting this fact into the expression (4.3.2) for arclength in polar coordinates, one immediately obtains

\begin{equation} ds^2 = r^2 \, d\phi^2\tag{13.5.1} \end{equation}

and finally

\begin{equation} \Lint \, ds = \int_0^{{\pi}\over{2}} r\,d\phi = {\pi r\over{2}} .\label{Answer}\tag{13.5.2} \end{equation}

But what if you didn't remember (4.3.2)? The calculation is not much harder in rectangular coordinates: You know that \(x=r\cos\phi\) and \(y=r\sin\phi\text{,}\) with \(r=\hbox{constant}\text{,}\) so that \(dx=-r\sin\phi\,d\phi\) and \(dy=r\cos\phi\,d\phi\text{.}\) Inserting this into (4.2.3) again leads to (13.5.2).

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 \(x^2+y^2=r^2=\hbox{constant}\text{,}\) so that \(2x\,dx+2y\,dy=0\text{.}\) Solving for \(dy\) and inserting this into (4.2.3) yields

\begin{equation} ds^2 = \left(1+{{x^2}\over{y^2}}\right) dx^2 = {{r^2}\over{r^2-x^2}} \> dx^2 .\tag{13.5.3} \end{equation}

This approach leads to the (improper!) integral

\begin{equation} \int_0^r {{dx}\over\sqrt{1-{{x^2}\over{r^2}}}}\tag{13.5.4} \end{equation}

which is, of course, most easily computed via a trig substitution — or numerically — yielding the same answer.