Show that the vector field $\vec F=(xf(u),xg(u))$ is not conservative

Question

I'm trying to prove that the vector field $\vec F=(xf(u),xg(u))$ with $u=xy$ is not conservative. I suppose that there is a function $\phi$ so that $\nabla \phi= \vec F$. So I need to satisfy that: $$\frac {\partial \phi}{\partial x}= yf(u)$$ $$\frac {\partial \phi}{\partial y}= xg(u)$$ Calculating the mixed partial for each I get that: $$\frac {\partial^2 \phi}{\partial x \partial y}= f(u)+y \frac{\partial f(u)}{\partial y}$$ $$\frac {\partial^2 \phi}{\partial y \partial x}= g(u)+x \frac{\partial g(u)}{\partial x}$$ Since mixed partials are equal I have that: $$f(u)+y \frac{\partial f(u)}{\partial y}=g(u)+x \frac{\partial g(u)}{\partial x}$$ and $\frac {\partial f(u)}{\partial y}=\frac {df}{du}\frac {\partial u}{\partial y}= x\frac{df}{du}$, applying this to both sides I end up with: $$f(u)+u \frac{df}{du}=g(u)+u \frac{dg}{du}$$ This is where I'm stuck, I'm not quite sure how to get a contradiction out of this or how I can show that this statement cannot hold. Should I have taken a different approach to this problem?

Answer 1

Hint: In three dimensions you would check if the vector field has a non-zero curl.

Checking $F = (x f(u), x g(u), 0)$ we get $$ \DeclareMathOperator{curl}{curl} $$ \begin{align} \curl F &= (0, 0, \partial_x(x g(u)) - \partial_y(x f(u))) \\ &= (0, 0, g(u) + x g'(u)y - x f'(u) x) \\ &= (0, 0, g(u) + u g'(u) - x^2 f'(u)) \end{align} So we need $g$ and $f$ to check if that third component is non-zero.