This is a question from an old qualifying exam at my university. I expect there's some trick but I couldn't spot it.
The question is to find the supremum over all simple closed curves $C$ of $$\int_C (y^3 - y) \ dx - 2x^3 dy$$ and then determine if the supremum is actually obtained or not.
I thought that perhaps if this integral was actually independent of path, then it would follow that over simple closed curves, this integral would always be $0$ and that would make the supremum easy - just $0$. But this isn't true, as just integrating around the regular unit circle gives $-5\pi/4$.
My next thought was maybe Green's Theorem. I can write this as $$\int_{S_C} (-6x^2 - 3y^2 + 1) \ dx \ dy$$ where $S_C$ is the interior of the simple closed curve $C$ Then by linearity of the integral it appears that there is a term which just keeps track of the area of the region, which I can make as large as I like. I don't know how the other terms might effect things in this situation though, and if the supremum isn't infinity, I don't know if I'm any closer to working it out like this than I was the first way.
All the problems on the exam are straight forward, and none require any particularly high powered weaponry. So probably there is an easy way to do it. Can anyone point me in the right direction?
If you take, say, a rectangle oriented clockwise with vertices $[0,0]$, $[0,1]$, $[X,1]$, $[X,0]$ you can make the integral arbitrarily large. Or was there an assumption on the orientation of the curve?