I'm at a loss on what to do with the following question from a previous differential topology exam:
Let $\gamma \rightarrow \mathbb{R}^{2}$ be the curve shown below with self-intersecting points $(0,0), (2,1)$ and $(4,2)$:
I'd imagine since this is in the context of differential topology, we are supposed to use Stoke's theorem, where we get:
$\int_{\gamma} ydx = \int_{R} dydx$ where $R$ is the region bounded by $\gamma$, but I am honestly having trouble doing the calculus on the right hand side.
Note that for a simply closed curve $\gamma=\partial R$ going around $R$ counterclockwise we have $$\int_\gamma y\>dx=-\int_R 1\> {\rm d}(x,y)=-{\rm area}(R)\ .$$ Split your $\gamma$ into a closed lower part $\gamma_1$ and a closed upper part $\gamma_2$ by choosing $(0,0)$ as dividing point. Then $$\int_\gamma y\>dx=\int_{\gamma_1} y\>dx+\int_{\gamma_2} y\>dx$$ is the area of the topmost part (above the $\infty$) of your figure minus the area of the lowest part (below the $\infty$), which is $=0$ if the figure is symmetric with respect to the line $(0,0)\vee(4,2)$.