parametrics and vector valued functions

[StephenPrivitera]

Considering the problem of finding the area under a parametric curve, I thought,
y=y(t), x=x(t)
A=[inte]ydx=[inte]yx'(t)dt
That result seems straighforward.

I also thought, what if I let the VVF v=<x(t),y(t)> represent the same curve. To find the area, under the curve (I have in the back of my mind the concept of velocity and position), I would solve the integral, r=[inte]vdt.

Should these two results be related? I think they should, but the math shows they aren't. Should the magnitidue of the latter equal the absolute value of the former? Looks like no. Why not?

[HallsofIvy]

Thinking of t as time, integrating the velocity vector give the position as a vector with tail at the original point. r= &int;v dt is a vector not a number. Since ||v|| is the speed, &int; ||v||dt gives arc-length, not area under the curve.