Calc3.8

The Line Integral

Definition

We've integrated single variable functions along lines, and two variable functions over regions of the plane, and even three variable functions over regions of space. But lines, and curves, can be contained in the plane, and surfaces can be contained in space. Why have we been restricting ourselves to integrating functions over regions which have the same dimensions as their domains?

Well, because it is simpler, and useful to have under our belts before we expand our repetoire. But there is no reason we can't integrate a function over a region in its domain which is a lower dimension than the domain itself.

We'll start with the line integral. Technically, it should be called the curve integral, but let's let that slide for now and get to the definition. We're trying to integrate over a curve, say, C. At first, we'll let this curve lie in the xy plane - it's not difficult to generalize to three dimensions. So we proceed as before: we divide this region into n pieces $C_{k},1\leq k\leq n$, and we pick a point $(x_{k},y_{k})\in C_{k}$ for each piece. Letting the length of $C_{k}$ be $L_{k}$. The definition follows as expected:

$\int _{C}f(x,y)dL=\lim _{L_{k}\to 0,n\to \infty }\sum _{k=1}^{n}f(x_{k},y_{k})L_{k}$

Computation

As before, we're not going to be using the Riemann sum definition to compute line integrals. Let $x=x(t),y=y(t)$ be a parametric description of $C$. This allows us to compute the length element, dL:

$dL={\sqrt {dx^{2}+dy^{2}}}$

But we don't have any formulas for dx or dy! The closest we've got are formulas for $dx/dt,dy/dt$; we'll have to plug these in and see if we can patch things up:

$={\sqrt {\left({\frac {dx}{dt}}\right)^{2}+\left({\frac {dy}{dt}}\right)^{2}}}{\sqrt {dt^{2}}}$

Now we can just plug in our formula for dL into the integral to get an easy-to-use formula for a line integral:

$\int _{C}f(x,y)dL=\int _{t_{0}}^{t_{1}}f(x,y){\sqrt {\left({\frac {dx}{dt}}\right)^{2}+\left({\frac {dy}{dt}}\right)^{2}}}dt$