making up your F (Green's theorem)

Question

Use Green's Theorem to find the areas enclosed by curve $$C: 2x^2+3y^2=2y$$ in $\mathbb{R}^2$.

Can someone tell me how to make up $F.\,\,\,$

Answer 1

First you want to complete the square. This should be the "obvious" thing to do.

Write $2x^2 + 3y^2 - 2y + 1/3 = 2x^2 + 3(y-1/3)^2 = 1/3$. Apply a change of variables $x \rightarrow u= \sqrt{2}x$ and $y\rightarrow v = \sqrt{3}(y-1/3)$. Note that the determinant of the Jacobian is $1/\sqrt{6}$ and proceed by integration to get $\text{Area} = 1/\sqrt{6}\cdot (\pi\cdot (1/3)) = \frac{\pi}{3\sqrt{6}}$.

Note that Green is not necessary here. If you must, $\oint_{C}\; -y/2\; dx + x/2\; dy$ shall do the trick (at least in theory), (why?), where we must "pullback" $dx$ and $dy$. But this will give you an elliptic integral, which is computationally far uglier than computing the area directly.