Use Green's Theorems to evaluate oint_C xydx + x^2y^3dy, where C is the triangle with vertices(0,0),(1,0)and (1,2).

Step 1
Consider the provided question,
${\displaystyle {\oint }_{C}xydx+x^{2}y63dy}$, where c is the triangle with vertices (0,0),(1,0) and (1,2)
We know that, Green's theorem,
${\displaystyle {\oint }_{C}P(x,y)dx+Q(x,y)dy=\int {\int }_D(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dxdy}$
here, ${\displaystyle P(x,y)=xy,Q(x,y)=x^2{y}^3}$
and D is the region inside the triangle,
Now,
${\displaystyle \frac{\partial P}{\partial y}=x,\frac{\partial Q}{\partial x}=2x{y}^3}$
${\displaystyle (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})=2x{y}^3-x=x(2y^3-1)}$
Step 2
We can describe the triangle D by, ${\displaystyle 0\le x\le 1,0\le y<+2x}$
${\displaystyle {\oint }_{C}xydx+x^2{y}^{3}dy={\int }_{x=0}^1{\int }_{y=0}^{x}x(2y^3-1)dxdy}$
${\displaystyle ={\int }_{x=0}^{1}xdx{\int }_{y=0}^x(2y^3-1)dy}$
${\displaystyle ={\int }_{x=0}^{1}xdx{[\frac{2y^4}{4}-y]}_0^x}$
${\displaystyle ={\int }_{x=0}^1[\frac{x^5}{5}-x^2]dx}$
${\displaystyle ={[\frac{x^6}{12}-\frac{x^3}{3}]}_0^1}$
${\displaystyle =\frac{1}{12}-\frac{1}{3}}$
${\displaystyle =-\frac{1}{4}}$