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Why do I get such a weird angle? (Let F = x^2 i  yx j be vector field where...)
I'm trying to solve this exercise that showed up in a calc 3 final exam a few months ago: "Let F=x^{2} i  yx j be the vector field where a particle travels positively on the closed curve C in the first quadrant: x=0; y=x^{2}; x^{2}+y^{2}=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=x^{2} and x^{2}+y^{2}=1, so
y+y^{2}=1
by completing the square: (y+1/2)^{2}1/4=1So y= (1+square_root(5))/2
By replacing this value in y=x^{2} 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 x^{2} 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 y0, 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?
Thanks
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