What is the maximum value of this line integral?

[Ackbach]

Here is this week's POTW:

-----

Find the positively oriented (counterclockwise) simple closed curve $C$ in the $xy$ plane for which the value of the line integral $\displaystyle\int_{C}(y^3-y) \, dx-2x^3 \, dy$ is a maximum.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

Congratulations to Opalg for his correct solution, which I give you below:

Spoiler

By Green's theorem, $$\displaystyle\oint_C (y^3-y)\,dx - 2x^3dy = \iint_D\Bigl(\tfrac{\partial}{\partial x}(-2x^3) - \tfrac{\partial}{\partial y}(y^3-y)\Bigr)\,dx\,dy = \iint_D(-6x^2 - 3y^2 + 1)\,dx\,dy,$$ where $D$ is the region bounded by $C$. To maximise the integral, we want $D$ to comprise the entire region where the integrand is positive, namely where $6x^2 + 3y^2 \leqslant1.$ So the contour $C$ should go round the ellipse $6x^2 + 3y^2 = 1$. Thus $C$ can be described by the path $t\mapsto \frac1{\sqrt6}(\cos t, \sqrt2\sin t)$ $(0\leqslant t \leqslant 2\pi).$