Integral depending on a path?

Question

I need to check whether differential form $\omega$ has, in the domain $G$, such property that it's integral doesn't depend on path. In my exercise: $\omega = \frac{ydx -xdy}{x^2+xy+y^2}$ and $G= R^2 \setminus {(0,0)}$. So I found the antiderivative, it's equal to $\frac {2}{\sqrt{3}}\arctan\left(\frac{2x+y}{\sqrt{3}y}\right)$. And my problem is that it isn't defined for $y=0$. So what should I do now to answer my question? Define it for $y=0$? Or maybe it will turn out that this integral depends on a path?

Answer 1

You may wish to consider the limit as $y\to 0$ of your function. This limit depends on whether $x>0, x=0,x<0$ and $y>0,y<0$. Is it possible to define a value anywhere along the line $y=0$ and keep continuity? (Remember that you can choose which arctan to take freely!) If so, how many places can you make a consistent definition?

The question is: once you've made as many consistent definitions as possible, do you have a leftover boundary with a discontinuity you can't fix? If so, you have path-dependent integrals.

Answer 2

You can write your $\omega$ $$\omega=\frac{y}{x^2+xy+y^2}dx-\frac{x}{x^2+xy+y^2}dy$$ For integral it has the form $$\int\bigg(Pdx+Qdy\bigg)$$ which you can test for beeing conservative such as $$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}$$ where $$\frac{\partial P}{\partial y}=\frac{(x-y)(x+y)}{\big(x^2+xy+y^2\big)^2}$$ $$\frac{\partial Q}{\partial x}=\frac{(x-y)(x+y)}{\big(x^2+xy+y^2\big)^2}$$ Which means that the field is conservative and the line integral is path-independent.

EDIT

I must have said in the beginning "Assuming that the path is closed..."

and at the end "This is a necessary requirement but not sufficient"

Answer 3

It is true that your differential $$\omega=Pdx+Qdy:={y\over x^2+xy+y^2}\ dx-{x\over x^2+xy+y^2}\ dy$$ satisfies the integrability condition $Q_x-P_y\equiv0$. Therefore in simply connected subsets $\Omega\subset G$ there exist potential functions $f:\ \Omega\to{\mathbb R}$ with $df =\omega$, or $\nabla f=(P,Q)$.

But there is no global $f:\ G\to{\mathbb R}$ of this sort. For a proof consider the elliptic curve $$\gamma:\quad t\mapsto\bigl(x(t),y(t)\bigr):=\left(\cos t-{1\over\sqrt{3}}\sin t,\ {2\over\sqrt{3}}\sin t\right)\qquad(0\leq t\leq2\pi)\ .$$ One computes $$x^2(t)+x(t)y(t)+y^2(t)\equiv1,\quad \dot x(t)=-\sin t-{1\over\sqrt{3}}\cos t,\quad \dot y(t)={2\over\sqrt{3}}\cos t\ .$$ It follows that $$\int_\gamma \omega=\int_0^{2\pi}\bigl(y(t)\dot x(t)-x(t)\dot y(t)\bigr)\ dt=-{4\pi\over\sqrt{3}}\ne0\ .$$