I'm looking at example 7 in 16.2 of Stewart. If there is a force field $F(x,y)=x^2i - xyj$ that moves a particle along the quarter-circle $r(t)=cos(t)i +sin(t)j$, $0\leq t\leq \frac{\pi}{2}$, and you take:
$$F(r(t))=cos^2(t)i-cos(t)sin(t)j$$ $$r'(t)=-sin(t)i+cos(t)j$$
the work done by the force field in moving the particle is apparently given by:
$$\int_CF\cdot dr=\int_0^{\frac{\pi}{2}}F(r(t))\cdot r'(t)\,dt = \int_0^{\frac{\pi}{2}}(-2cos^2(t)sin(t))\,dt = -\frac{2}{3}$$
The line integrals I have been doing up to this point were the type where you have a function and you integrate it along a curve, meaning you evaluate the function $f(x,y)$ along the curve, and the sum gives you the area of one side of a "wall" or some similar shape as it appears in three dimensions.
When you have a two-dimensional force field, and you evaluate a line integral along a curve in this force field, what is the result in a geometric sense, aside from work done? I'm just picturing a line on a plane with no three dimensional function to evaluate; why isn't the integral just the area of the line (zero)?