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Let $\vec{F}=(xy,x^2+y^2)$ be a vector field. Is there exist a function $f(x,y)$ such that $\vec{\nabla}f=\vec{F}$?

My attempt: if $f_x=xy$ and $f_y=x^2+y^2$, then $f(x,y)=\frac{x^2y}{2}+g(y)$. Therefore $f_y=\frac{x^2}{2}+g'(y)$. Hence $\frac{x^2}{2}+g'(y)=x^2+y^2$, so $g'(y)=\frac{x^2}{2}+y^2$. But $g(y)$ is a function of $y$, a contradiction.

On the other hand, if you take any circle $C(t)=(R\cos t,R\sin t)$, $t\in[0,2\pi]$ then $$ \oint_{C}f\cdot d\vec{l}=\int_{0}^{2\pi}(R^2\cos t\sin t,R^2)\cdot(-R\sin t,R\cos t)\,dt=R^3\int_{0}^{2\pi}\cos^3t\,dt=0 $$ Why this does not contradict the first claim? Thanks

boaz
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    That's right. Your contradiction proves, that there isn't such a function $f$. – Mundron Schmidt Jun 07 '17 at 19:37
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    Thanks, but if you take any circle around $(0,0)$, the line integral over this circle is 0. See my edit. – boaz Jun 07 '17 at 19:45
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    The theorem is that if there exist F(x, y) is such that $f(x,y)= \nabla F$ then the integral around any closed path of f(x,y) is 0. It does not say if f is not an "exact differential" that the integral around some closed path may not be 0. It is the difference between "if A then B" and "if B then A". One does not imply the other. – user247327 Jun 07 '17 at 19:50

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You can get the line integral being zero even if the field is non-conservative. It works in your case since the circular path is symmetric about the y-axis and the curl of the field is odd with respect to x. Pick a path that does not have that symmetric, for example the unit square in the first quadrant, and you will get a non-zero answer. By Stoke Theorem, you get the work as $$W=\int \nabla \times F dA$$ $$=\int_0^1 \int_0^1 x dydx$$ $$=1/2$$

Paul
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