$\displaystyle Let\ P=xy+y^{2\ }and\ Q=x^2 \\[2ex] [image] \\[3ex] \therefore{}\dfrac{\partial{}P}{\partial{}y}=x+2y\ and\ \dfrac{\partial{}Q}{\partial{}x}=2x\\[3ex]Given,\ y=x\ and\ y^2=x\\[2ex]Solving\ simultaneouslyx^2=x\\[2ex]\therefore{}x^2-x=0\\[2ex]\therefore{}x=0\ or\ x=1\\[3ex]But\ y=x\\[3ex]\therefore{}y=0\ or\ y=1 \\[2ex]Two\ curves\ intersect\ at\ (0,0)\ and\ (1,1)\\[3ex]$

$Part\ 1:\\[2ex]Consider \\[3ex] \displaystyle I=\int_cPdx+Qdy=\int_c\left(xy+x^2\right)dx+x^2dy \\[2ex] Along\ line\ OA\ equation:y=x\\[3ex]\therefore{}dx=dy\\[2ex]\therefore{} \displaystyle I_{OA}=\int_0^1\left(x\bullet{}x+x^2\right)dx+x^2dx\\[2ex] \displaystyle =\int_0^13x^2dx\\[2ex] \displaystyle ={\left[\frac{3x^3}{3}\right]}_0^1\\[2ex]=1-0=1\ \ \rightarrow{}(i) $

$\displaystyle Along\ AO\ i.e.\ parabola\ equation\ x=y^2 \\[2ex] \displaystyle \therefore{}dx=2ydy \\[2ex] \displaystyle \therefore{}I_{AO}=\int_1^0\left(y^2\bullet{}y+y^2\right)\bullet{}2ydy+{\left(y^2\right)}^2dy \\[2ex] \displaystyle =\int_1^0\left({2y}^4+2y^3+y^4\right)dy \\[2ex] \displaystyle =\int_1^0\left({3y}^4+2y^3\right)dy \\[2ex] \displaystyle {=\left[\frac{{3y}^5}{5}+\frac{{2y}^4}{4}\right]}_1^0 \\[2ex] \displaystyle =0-\left[\frac{3}{5}+\frac{1}{2}\right]\\[2ex] =\dfrac{-11}{10} \\[2ex] $

$\displaystyle Hence\ \\[2ex] \displaystyle \int_cPdx+Qdy=\ IAO+IOA \\[2ex] \displaystyle =1-\frac{11}{10}=\frac{-1}{10} \\[2ex] $

$\displaystyle Part\ 2: \\[2ex] \displaystyle Consider \\[2ex] \displaystyle \\[2ex] \displaystyle \iint_R\left(\frac{\partial{}Q}{\partial{}x}-\frac{\partial{}P}{\partial{}y}\right)dxdy \\[2ex] \displaystyle \\[2ex] \displaystyle =\iint_R\left[2x-\left(x+2y\right)\right]dxdy \\[2ex] \displaystyle \\[2ex] \displaystyle =\int_0^1\int_x^{\sqrt{x}}\left(x-2y\right)dxdy \\[2ex] \displaystyle \\[2ex] \displaystyle =\int_0^1{\left[xy-\frac{2y^2}{2}\right]}_x^{\sqrt{x}}dx \\[2ex] \displaystyle \\[2ex] \displaystyle =\int_0^1\left\{\left[x\sqrt{x}-x\right]-\left[x^2-x^2\right]\right\}dx \\[2ex] \displaystyle \\[2ex] \displaystyle =\int_0^1\left[x^{\frac{3}{2}}-x\right]dx \\[2ex] \displaystyle ={\left[\frac{x^{\frac{3}{2}}}{\frac{5}{2}}-\frac{x^2}{x}\right]}_0^1 \\[2ex] \displaystyle =\frac{2}{5}-\frac{1}{2}=\frac{-1}{10}\rightarrow{}\left(ii\right) \\[2ex] $

$\displaystyle \therefore{}From\ \left(i\right)and\ \left(ii\right) \\[2ex] \displaystyle \int_cPdx+Qdy=\iint_R\left(\frac{\partial{}Q}{\partial{}x}-\frac{\partial{}P}{\partial{}y}\right)dxdy \\[4ex] \displaystyle \therefore{}Green^{'}s \ Theorem\ is\ verified \\[2ex] $