Exercise about Green theorem and line integral

Exercise:

Evaluate $\oint\limits_{\left( {{C}} \right)}{xydx+{{x}^{2}}dy}$  using Green theorem then line Integral

Solution: $\oint\limits_{\left( {{C}
} \right)}{xydx+{{x}^{2}}dy}=\iint\limits_{D}{\left( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right)dA}$

\[=\iint\limits_{D}{\left( 2x-x \right)dA=\iint\limits_{D}{xdA}}=\iint\limits_{D}{x\,dxdy}=\int_{1}^{0}{\left( \int_{0}^{x-1}{xdy} \right)dx}\]

$=\int_{0}^{1}{x\left( x-1 \right)dx}=-\int_{0}^{1}{\left( {{x}^{2}}-x \right)dx}=-\left( \frac{{{x}^{3}}}{3}-\frac{{{x}^{2}}}{2} \right)_{0}^{1}=-\left( \frac{1}{3}-\frac{1}{2} \right)=\frac{1}{6}$

2nd by Method by Line  integral :

Consider ${{C}_{1}}:\,\left( 1,0 \right)\to \left( 0,0 \right)$ , ${{C}_{2}}:\left( 0,0 \right)\to \left( 0,-1 \right)$ and ${{C}_{3}}:\left( 0,-1 \right)\to \left( 1,0 \right)$

So ${{C}_{1}}:\left( x=1-t,y=0 \right)\,\,\,,\,\,{{C}_{2}}:\left( x=0,y=-t \right)\,\,\And \,\,{{C}_{3}}\left( x=t,y=t-1 \right)$    $0\le t\le 1$

So $\int\limits_{C}{xydx+{{x}^{2}}dy}=\int\limits_{{{C}_{1}}}{xydx}+\int\limits_{{{C}_{2}}}{{{x}^{2}}dy}+\int\limits_{{{C}_{3}}}{\left( xydx+{{x}^{2}}dy \right)}$

$=\int_{0}^{1}{\left( 1-t \right)0\,dt+\int_{0}^{1}{0\,dt}+\int_{0}^{1}{t\left( t-1 \right)dt+{{t}^{2}}dt=\int_{0}^{1}{\left( 2{{t}^{2}}-t \right)dt=\frac{2}{3}-\frac{1}{2}=\frac{4-3}{6}=\frac{1}{6}}}}$