Answer
$4 \pi$
Work Step by Step
Green's Theorem states that:
$\oint_C M \,dx+ N \,dy=\iint_{D}(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y})dA$
We will use polar coordinates as follows: $x=r \cos \theta; y= r \sin \theta$
We need to work out with the line integral and evaluate the integrand of the double integral as follows:
Consider $I=- \iint_{D} [\sin y -1-\sin y] dA =- \iint_{D} (-1) dA$
The double integral of $1$ is the area of D. Thus, the surface above is constant and the circle of radius $2$ can just be moved to the center and then converted to polar co-ordinates.
Now, $$I=- \int_{0}^{2\pi} \int_{0}^{2} -r \ dr \ d \theta = - (-2) \int_{0}^{2\pi} d \theta = 2[ 2\pi-0] = 4 \pi $$
