This is the problem I'm trying to solve:
Given vector field $F(x,y)=x^2 \hat i - yx \hat j$ and a particle travels positively over the closed curve in the first quadrant, given by: $x=0$; $y=x^2$; $x^2+y^2=1$, find work done along the curve, both using line integrals and by Green's theorem.
I already did the Green part and got -0,312 (also seems weird to me: can a line integral be negative?)
As for the variant using line integrals, I labeled curves C1, C2 and C3 as follows (excuse me for the shaky hand when using my mouse):
I parameterized the first one (circle) and got a result (again, a negative one, which sounds weird, but that's not my main question).
As for the second (segment), I did this: parameterized segment as $<0; -t+1>$ with $t$ going from 0 to 1. Then found the derivative, as $<0;-1>$. Then did the composition with the vector field: $<0^2;(-t+1).0>$ so I got <0;0>. So now I set up my integral as $\int_0^1 <0;0> \cdot <0;-1> dt$ so I end up having 0 as my integrand. Is that correct?
Of course it can be zero, note that on C2 $\vec F= \vec 0$.