Answer
$0$
Work Step by Step
The tangential form for Green's Theorem -- Counterclockwise Circulation $\oint_C F \cdot T ds= \iint_{R} (\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}) dx dy $
Now, $\oint_C F \cdot n ds= \iint_{R} \dfrac{\partial (x^2)}{\partial x}-\dfrac{\partial (y^2)}{\partial y} dx dy $
or, $= \int_{0}^1 \int_0^{1-x} 2x-2y dx dy$
or, $= \int_{0}^1 \int_0^{1-x} 2x-2y dx dy$
or, $= \int_{0}^1 2x(1-x) -(1-x^2) dx$
or, $= \int_{0}^1 [4x-3x^2-1] dx$
$[2x^2-x^3-x]_0^1=0$
