I'm really trying to understand this, but I have been struggling to understand the concept for this problem applying Green's Theorem all day. Unfortunately I cannot find any material with some solved examples of this type so I can understand it better with the explanation.
I have the vector field
$$F(x, y) = (\frac{-y}{(x-1)^2+y^2} + \frac{y}{(x+1)^2+y^2}, \frac{x-1}{(x-1)^2+y^2} + \frac{-x-1}{(x+1)^2+y^2})$$
And I want to calculate the work done by $F$ along, let's say
- a circle of the form $(x+1)^2 + y^2 = 1$
- a circle of the form $(x-1)^2 + y^2 = 1$
- an ellipse $\frac{x^2}{4} + y^2 = 1$
I wonder a few things here:
Can I write $F(x, y) = (\frac{-y}{(x-1)^2+y^2}, \frac{x-1}{(x-1)^2+y^2}) + (\frac{y}{(x+1)^2+y^2}, \frac{-x-1}{(x+1)^2+y^2}) = \Delta(x, y) + \Gamma(x, y)$ and check separately that $\Gamma(x, y)$ is conservative and therefore the net work done by "that part" of the field $F$ to $\Gamma$ is zero? Why?
In order to prove that the field is a closed field does it suffice that I show that $\frac{\partial F_2}{\partial x} = \frac{\partial F_1}{\partial y}$? What is the reasoning for this to be true?
As an example, the field is not defined at $(1, 0)$ and (-1, 0). How should I apply Green's Theorem if the field has a "hole" within my circle/ellipse? How would I proceed with that?
Let's say the field is closed. If the domain of the Field was defined for $R^2$, would the work done by the field along all three curves be zero?
What are the differences between a closed field, a gradient field and a conservative field?
Thank you in advance! I'm really trying to understand this. Even if you are not able to answer all questions, every help would be very appreciated.