# 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).

Falak Kinney 2020-10-18 Answered
Use Green's Theorems to evaluate ${\oint }_{C}xydx+{x}^{2}{y}^{3}dy$, where C is the triangle with vertices(0,0),(1,0)and (1,2).
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## Expert Answer

diskusje5
Answered 2020-10-19 Author has 82 answers

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

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