Answer
$\oint_C xy^2 dx +(x^2y +2x) \ dy=2 \times ( \ Area \ of \ the \ Square)$
Work Step by Step
The tangential form for Green Theorem is given as:
Counterclockwise Circulation: $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy $
$\oint_C xy^2 dx +(x^2y +2x) \ dy= \iint_{R} (\dfrac{\partial (x^2y+2x) }{\partial x}-\dfrac{\partial (xy^2) }{\partial y}) dx dy $
or, $= \iint_{R} 2xy+2-2xy \ dx \ dy$
or, $\oint_C xy^2 dx +(x^2y +2x) \ dy=2 \times ( \ Area \ of \ the \ Square)$
