$\overline {F}= (x^2 - y^2)\widehat {i} + (x+y)\widehat{j} \ dr = dx \widehat {i}+dy \widehat {j} $ $\ F.dr = (x^2-y^2)dx +(x+y)dy \ Let, \: P=x^2-y^2 , \: Q =x+y $ $\ \dfrac{\partial P}{\partial y} = -2y, \quad \dfrac{\partial Q}{\partial x} = 1 \ \text{Green's Theorem,} $ $\int_c Pdx+Qdy =\iint_R\left (\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right )dxdy \Now, \iint_R\left (\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right )dxdy $  $ \ =\iint_R (1+2y)dxdy \ \text{The region : x =2y to x=y, y=0 to y =1} $ $\ = \int \limits_0^1\int \limits_{2y}^{y}(1+2y)dxdy

\ = \int \limits_0^1 (1+2y) \int \limits_{2y}^{y}dxdy $ $\ =\int \limits_0^1 (1+2y) [x]_{2y}^{y}dy \ =\int \limits_0^1 [(1+2y) \times (y-2y) ]dy $ $\ =-\int \limits_0^1 (y+2y^2)d \=- [\dfrac{y^2}{2}+2\dfrac{y^3}{3}]_0^1 $ $\ =-\dfrac{1}{2}-\dfrac{2}{3} -0 \ = -\dfrac{7}{6} ...(A)$ $\text{1. Along path x=y, dx =dy, y =0 to 1} \ \int \limits_0^1Pdx + Qdy $ $\ = \int \limits_0^1(x^2-y^2)dx +(x+y) dy \ =\int \limits_0^1(y^2-y^2)dy +(y+y)dy ....as \:x =y $ $\ = \int\limits_0^1 2ydy \ = [y^2]_0^1 = [1-0] \= 1$ $\text{2. Along path y=1, dy = 0, x =1 to 2} \ \int \limits_1^2 Pdx $ $\ =\int \limits_1^2(x^2-y^2)dx

\ =\int \limits_1^2(x^2-1)dx ...as \: y =1 $ $= [\dfrac{x^3}{3}-x]_1^2 = [\dfrac{8}{3}-2-\dfrac{1}{3}+1] = [\dfrac{7}{3}-1] \ = \dfrac{4}{3}$ $\text{3. Along path x=2y, dx =2dy, y =1 to 0} \ \int \limits_1^0Pdx + Qdy $ $\ = \int \limits_1^0(x^2-y^2)dx +(x+y) dy \ =\int \limits_1^0(4y^2-y^2)2dy +(2y+y)dy ....as \:x =2y $ $\ =\int \limits_1^0[6y^2 +3y ]dy \= [6\dfrac{y^3}{3}+3\dfrac{y^2}{2}]_1^0 $ $\ =[0-2-\dfrac{3}{2}]

\ = \dfrac{-7}{2}$ $\text{Adding the result from all 3 paths,} $ $\int_c Pdx+Qdy \ = 1+\dfrac{4}{3}-\dfrac{7}{2} \ = \dfrac{-7}{6} ....(B)$ $\text{From (A) and (B),} \ \bbox[3pt, yellow]{\int_c Pdx+Qdy =\iint_R\left (\dfrac{\partial Q}{\partial x}-\dfrac{\partial P}{\partial y}\right )dxdy }\ \text{hence, Green's Theorem is verified.}$