punniya.v

2022-01-19

### Answer & Explanation

nick1337

Green Theorem:

${\oint }_{c}M\left(x,y\right)dx+N\left(x,y\right)dy=\int {\int }_{R}\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}dxdy\right)$

$\stackrel{\to }{F}\left(x,y\right)=2x\left(x+y\right)\stackrel{\to }{i}+\left({x}^{2}+xy+{y}^{2}\right)\stackrel{\to }{j}$

$\stackrel{\to }{r}=x\stackrel{\to }{i}+y\stackrel{\to }{j}$

$d\stackrel{\to }{r}=dx\stackrel{\to }{i}+dy\stackrel{\to }{j}$

$\stackrel{\to }{F}\cdot d\stackrel{\to }{r}=2x\left(x,y\right),N\left(x,y\right)={x}^{2}+xy+{y}^{2}$

Now

$M=2{x}^{2}+2xy$

$\frac{\partial M}{\partial y}=0+2x=2x$

$\frac{\partial N}{\partial x}=2x+y+0=2x+y$

$\int {\int }_{R}\left(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}dxdy=\int {\int }_{R}\left(2x+y-2x\right)dxdy$

$=\int {\int }_{R}\left(y\right)dxdy$

Let us construct horizontal strip which moves one end on $x=0$ and other end on $x=1$ and this strip moves $y=0$ to $y=1$

$\int {\int }_{R}\left(y\right)dxdy={\int }_{y=0}^{1}\left({\int }_{y=0}^{1}ydx\right)dy$

$={\int }_{y=0}^{1}\left(xy{\right)}_{0}^{1}dy$

$={\int }_{y=0}^{1}ydy$

$=\left(\frac{{y}^{2}}{2}{\right)}_{0}^{1}$

$\frac{1}{2}\left(1-0\right)$

$=\frac{1}{2}$

Do you have a similar question?

Recalculate according to your conditions!