Find the Work $\oint{Pdx+Qdy}$ of the $F(x,y)= (-y^2,x^2)$ on the rectangle (pic) $x-y=0, x-y=1, x+y=1, x+y=2$ (positive direction)
From Green's theorem: $$\oint{Pdx+Qdy}= \iint{\left(\frac{\partial Q}{\partial x} -\frac{\partial P}{\partial x}\right) dxdy}$$ Where, $P=-y^2 , Q=x^2 $
Normal to $x$:
$$ \int_{0}^{0.5}{} \int_{1+y}^{1-y}{(2x+2y)dxdy+\int_{0.5}^{1}{} }\int_{y}^{2-y}{(2x+2y)dxdy} $$
My question is: Do I have to add or subtract the second integral (because of the directions of $a$ and $w$ (shown in above pic))?
Notice, divide the region bounded by the lines $x-y=0, x-y=1, x+y=1$ & $x+y=2$ into two parts and consider two horizontal rectangular slabs in these parts and apply proper limits of $x$ & $y$ (as shown in above figure). Add these two double integrals as follows
The first integral must have proper limits of $x$ from $x=1-y$ to $x=1+y$ and $y$ from $y=0$ to $y=0.5$ as follows $$ \int_{0}^{0.5}{} \int_{\color{blue}{1-y}}^{\color{blue}{1+y}}{(2x+2y)dxdy+\int_{0.5}^{1}{} }\int_{y}^{2-y}{(2x+2y)dxdy} $$
$$=\color{blue}{-}\int_{0}^{0.5}{} \int_{\color{blue}{1+y}}^{\color{blue}{1-y}}{(2x+2y)dxdy+\int_{0.5}^{1}{} }\int_{y}^{2-y}{(2x+2y)dxdy} $$ $$=\color{red}{\frac32}$$