Consider the vector field
$$F(x,y) = F_1(x,y)\mathbf{i} + F_2(x,y)\mathbf{j} = -\frac{y}{x^2 + y^2}\mathbf{i} + \frac{x}{x^2 + y^2}\mathbf{j}$$
- Show that $\frac{dF_2}{dx}=\frac{dF_1}{dy}$ for values of x,y for which F is defined.
Compute integral $\int F dr$ with:
a) C1 boundary where C1 is the unit circle centred at the origin, oriented anticlockwise.
b) C2 boundary where C2 is the circle $(x-2)^2 + y^2 = 1$, oriented anticlockwise.
c) C3 boundary where C3 is the square $\left\{(x,y)\,|\,-1<x<1, -1<y<1\right\}$
For a), b), and c) which can use Green's theorem? Explain.
3) For which closed paths C does integral $\int F dr = 0$?
My thinking:
For (1), I managed to show that $\frac{dF_2}{dx}$ = $\frac{dF_1}{dy}$. But I don't understand what it means by "for which $F$ is defined". If I am correct, $F$ should be defined everywhere except when $(x,y) = (0,0)$. If so, does that mean $\frac{dF_2}{dx}$ is possible to be not equal to $\frac{dF1}{dy}$?
Also, for (a), (b), and (c), I feel like none of them can use Green's theorem since $F_1$, $F_2$, $dF_1$, and $dF_2$ are not continuous at $(0,0)$ which is part of the region C1, C2, and C3. I tried solving the integral using parametrisation instead and found that for C1 and C2, the integral equals $2\pi$, whereas for C3, the integral equals 0. Would that answer (3)?