Answer
$-\dfrac{\pi}{2}+2$
Work Step by Step
Here, we have $f(x, y)=x^2+y^2$ and $g (x, y)=4x+y^3$
This gives: $\dfrac{\partial g}{\partial x}=4$ and $\dfrac{\partial f}{\partial y}=2y$
Next, we will compute the circulation integral.
$\oint_C F \ dr=\int_R (\dfrac{\partial g}{\partial x}-\dfrac{\partial f}{\partial y}) \ dA$
or, $=\int_{0}^{\pi} \int_0^{\sin x} (-2y+4) \ dy \ dx$
or, $=\int_{0}^{\pi} (-\sin^2 x+4 \sin x) \ dx$
or, $=[-(\dfrac{1-\cos 2x}{2})-4 \cos x]_{0}^{\pi} $
or, $=-\dfrac{\pi}{2}+2$
