Answer
$ 4 \pi$
Work Step by Step
Green's Theorem states that:
$\oint_C A\,dx+B \,dy=\iint_{D}(\dfrac{\partial B}{\partial x}-\dfrac{\partial A}{\partial y}) dA $
W need to set up the line integral and compute the integrand of the double integral as follows:
$\oint_C A\,dx+B \,dy=\iint_{D}(\dfrac{\partial B}{\partial x}-\dfrac{\partial A}{\partial y}) dA $
Consider $I=- \iint_{D} [\sin y -1-\sin y] dA$
or, $=- \iint_{D} (-1) dA$
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, $I=- \int_{0}^{2\pi} \int_{0}^{2} -r \ dr \ d \theta$
$= - \int_{0}^{2\pi} (-2) d \theta$
$= 2[ 2\pi-0]$
$=4 \pi$
