Answer
$$\dfrac{-16}{3}$$
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 $
Conversion to polar coordinates is as follows: $x=r \cos \theta; y= r \sin \theta$
We need to set up the line integral and compute the integrand of the double integral as follows:
$$\oint_CP\,dx+Q\,dy=\iint_{D} y-x \sin x +\cos x -(\cos x -x \sin x) dA \\= \int_{0}^{2} \int_{0}^{-2x+4} y \ dy \ dx \\= \int_{0}^{2} [\dfrac{y^2}{2}]_{0}^{-2x+4} y dy \ dx \\=\dfrac{1}{2}\times \int_{0}^{2} (4x^2-16x +16) dx \\=2[\times \dfrac{x^3}{3}-2x^2+4x]_0^2 \\=\dfrac{-16}{3}$$
