Show that the ellipse defined by the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ in $\mathbb{R}^2$ is a submanifold and orient it counterclockwise. Compute $\int_M \omega$ for $\omega := (x+y) \ dx + (x-y) \ dy$, where $x$ and $y$ refer the standard coordinates in $\mathbb{R}^2$.
Here is what I got so far:
We consider $f: \mathbb{R}^2 \rightarrow \mathbb{R}, (x,y) \mapsto \frac{x^2}{a^2} + \frac{y^2}{b^2}$ and note that $f \in \mathcal{C}^1$ and $f^{-1}(1)$ is equal to our ellipse. Since $Df = \nabla f$ has rank $1$ the Regular Value Theorem implies that our ellipse $E$ is a $1$-dimensional submanifold of $\mathbb{R}^2$.
However, I am not sure how to evaluate the integral $\int_M \omega$. I would be tempted to use Green's Theorem $\oint_{\partial M} P(x,y) \ dx + Q(x,y) \ dy = \int_M \big( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \big) \ dxdy$.
But when I use Green's Theorem I get
$$P(x,y) := x+y, \quad\frac{\partial P}{\partial y} = 1$$ $$Q(x,y) := x-y, \quad \frac{\partial Q}{\partial x} = 1$$ $$\oint_M P(x,y) \ dx + Q(x,y) \ dy = \int_M \bigg( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \bigg) \ dxdy = \int_M 0 \ dxdy = 0.$$
Could you please tell me if this is right?