$$\overrightarrow{F}=-y\hat{i}+x\hat{j}$$ $$C:r=a \cos{t}\hat{i}+a \sin{t}\hat{j}, o \leq t \leq 2 \pi$$ $$R:x^2+y^2 \leq a^2 $$
Green's Theorem: $$\oint_C{F}dr=\int \int_R{\bigtriangledown \times \overrightarrow{F} \cdot \hat{n}}dA$$
$$\bigtriangledown \times \overrightarrow{F}=2 \hat{k}$$ But how can I find $ \int \int_R{ \bigtriangledown \times \overrightarrow{F} \cdot \hat{n}}dA $ ??
First of all Greens Theorm implies : $$\oint_C{F}dr=\int \int_R{\nabla \times \overrightarrow{F}\cdot \hat{k} }dA$$ with $\hat{k}$ is pointing the $z$ direction and the R.H.S is just the normal double integral of $\nabla \times \overrightarrow{F}$ over region $R$ which is the area inside the circle centered at the origin and of radius $a$.
Since $\nabla \times \overrightarrow{F}=2 \hat{k}$, we get
$$\oint_C{F}dr=2\int \int_R{ }dA= 2\pi a^2 $$