I'm trying to understand the application for Green's Theorem for a field with singularities and taken from 3D to 2D point of view. So lets say I have this field $F(x, y, z) = (\frac{z}{x^2+y^2} + x, y, \frac{-x}{x^2+y^2} + z)$. I need the work done by the field along the ellipse $2(x-1)^2+\frac{y^2}{4} = 2; z = 0$.
Because $z = 0$ can I consider the field that does work really just $F^z(x, y) = (x, y)$? And if I were, for example, at the point $(0, 10, 0)$, could I just ignore the second term, being $F^y(x, z) = (\frac{z}{x^2+10}, \frac{x}{x^2+10} + z)$?
On the plane $z=0$ . The vector field is $F(x,y,z)=x\hat{i}+y\hat{j}-\frac{x}{x^{2}+y^{2}}\hat{k}$ .
The ellipse in the plane is parametrized by $\bigg\{\big(1+\cos(t),2\sqrt{2}\sin(t)\big):t\in[0,2\pi]\bigg\}$ .
Thus we have $\vec{dr}=-\sin(t)dt\hat{i}+2\sqrt{2}\cos(t)dt\hat{j}$
Thus $$\oint F\cdot \vec{dr}=\int_{0}^{2\pi}\bigg(-\big(1+\cos(t)\big)\sin(t)+8\cos(t)\sin(t)\bigg)dt = 0$$ .
You cannot directly apply Green's Theorem because of the singularity(Although it would lead to the correct answer). You have to show why and how the work done by the $\hat{k}$ component is $0$ . And this is done by showing that the vector $\vec{dr}$ . is orthogonal to $\hat{k}$ .
To answer your question about the "point" (0,10,0). Well ofcourse this is a function so it would have a value at each point. So at $(0,10,0)$ F would be just $10\hat{j}$ .
If you were on the plane $y=10$ . Then you can consider that the vector field is just $(\frac{z}{x^2+10}, \frac{x}{x^2+10} + z)$ .