I'm trying to solve this exercise that showed up in a calc 3 final exam a few months ago: "Let F=x2 i - yx j be the vector field where a particle travels positively on the closed curve C in the first quadrant: x=0; y=x2; x2+y2=1. Find work done throughout C." I need to do it using Green's theorem and then using the line integral method.
So first of all I drew the curves in 3D and in the XY projection.
3d.jpg xy.jpg

Then found the intersection by doing this:
y=x2 and x2+y2=1, so
by completing the square: (y+1/2)2-1/4=1So y= (-1+square_root(5))/2
By replacing this value in y=x2 I got that x is the square root of (-1+square_root(5))/2

Ok, so now for Green's theorem I used the area bound by:
y between x2 and (-1+square_root(5))/2
x between 0 and square_root(-1+square_root(5))/2
I got derivatives from the vector field as -y-0, so I set up my integral for the said area and finally got -0,312 as the result. Not sure if that's correct, but my biggest problem came when I tried to solve it as a line integral.

I know I need to parameterize all three curves. However, when I did the circle parameterization as cos(t) i + sin(t) j and tried to find values for t, I got that x=cos(t), so t must be cos-1(square_root(-1+square_root(5))/2) which gives me a weird 0,67.
That definitely doesn't look like the angle I'm looking at in the XY plane projection... so where did I go wrong?