Answer
$-\frac{16}{3}$
Work Step by Step
Green's Theorem states that:
$\oint_CP\,dx+Q\,dy=\iint_{D}\left(\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right)dA$
Conversion to polar coordinates: $x=r \cos \theta; y= r \sin \theta$
note that since the Curve C is being traversed clockwise, we must multiply in a negative.
We set up the line integral and find out the integrand of the double integral as follows:
$\oint_CP\,dx+Q\,dy=-\iint_{D} y-x \sin x +\cos x -(\cos x -x \sin x) dA$
or, $= -\int_{0}^{2} \int_{0}^{-2x+4} y dy dx$
or, $= -\int_{0}^{2} [y^2/2]_{0}^{-2x+4} y dy dx$
or, $=-\frac{1}{2} \int_{0}^{2} (4x^2-16x +16) dx$
or, $=-2\left[\frac{x^3}{3}-2x^2+4x\right]_0^2$
or, $=\frac{-16}{3}$
