Q) Consider the vector field $F(x,y)=(y,-x)$. Show that there exists no scalar field $f:R^2 \rightarrow R$ whose gradient is $F$. Find a closed path C so that $\int_{0}^{1}F . dr \neq 0$.
A) So far I have, $f(x,y)=\int_{0}^{1} F(tx,ty)(x,y)dt = \int_{0}^{1} ty^2+tx^2dt = [\frac{t^2}{2}(y^2+x^2)]$ between $0$ and $1$ which gives $\frac{1}{2} (y^2+x^2)$.
I don't know if this method is correct or if there is a simple check that will tell you if a scalar field exists or not.
For the second part I don't know how to find such a path, would the unit circle work?
1) A necessary condition for $F(x,y) = F = \langle F^1, F^2 \rangle$ to have a potential function is that,
$$\frac{\partial F^1}{\partial y} = \frac{\partial F^2}{\partial x}$$
but $1 \not = -1$ i.e $F$ has no potential function or equivalently, $F$ is not a gradient vector field.