About the Green's Theorem

Question

I don't understand something about the Green's Theorem. Is it true that any integral of a closed path is always equal to zero? For example $$\int _{C}(xy^2+x^2)dx+(x^2y+x)dy$$ where $C$ is a trapezium with vertices $A=(0,0)$, $B=(3,0)$, $C=(2,1)$, and $D=(1,1)$. Sorry for my english...

Answer 1

Your shape is a trapezium. But to apply Green's theorem, you need that the differential $u(x,y)\,dx+v(x,y)\,dy$ that you are integrating over is defined throughout the interior of $C$ and that it also satisfies $$\frac{\partial}{\partial y}u(x,y)=\frac{\partial}{\partial x}v(x,y)$$ inside $C$. In your example. $$\frac{\partial}{\partial y}u(x,y)=2xy$$ and $$\frac{\partial}{\partial x}v(x,y)=2xy+1$$ so you cannot immediately prove your integral is zero. (In fact it isn't).

Answer 2

In General, an integral over a closed path is not Zero. It is only Zero, if the Integrand (here: $(xy^2+x^2)dx+(x^2y+x)dy$) can be expressed by a total differential $df(x,y) = \partial_x f dx + \partial_y f dy$ for any function $f(x,y)$. In other words,

for the Integrand of the form $Adx+Bdy$ it must hold

$\partial_x B = \partial_y A$

that the integral over a closed path vanishes.

Answer 3

Notice that, $$\frac{\partial(x^2y+x)}{\partial x}-\frac{\partial(xy^2+x^2)}{\partial y} =2xy+1-(2xy)=1.$$ Therefore, by Green's Theorem, if $C$ is a positively oriented, piecewise smooth, simple closed plane curve then $$\int _{C}(xy^2+x^2)dx+(x^2y+x)dy=\iint_D 1 dxdy=\text{Area}(D)$$ where $D$ is the region bounded by $C$. So the line integral along the perimeter of your trapezium is not zero, it is equal to its area which is $\frac{(3+1)\cdot 1}{2}=2$.