Why is work around a closed path zero only in some cases? I've got two examples and the apparent arbitrarity is very confusing to me.
First, we've got the field and path that follow (I got the parametrization of the path from the given equation $16x^2 + y^2 = 4$)
$F(x,y) = \left(\dfrac{y}{(x-1)^2 + y^2 },\dfrac{1-x}{(x-1)^2 + y^2 } \right)$ and $r(t) = \left(\dfrac{cos^2(t)}{2}, sin^2(t) \right) $
As you can see the field is conservative so the work done is 0. The other example is the problem though:
$F(x,y) = \left(6xy^2 - y^3, 6x^2y - 3xy^2 \right)$ and the path from (0,1) to (2,4) composed by the line segment from (0,1) to (1,1) and the segment from (1,1) to (2,4).
This one is easy to solve since the field is conservative, so every line segment that connects (0,1) to (2,4) is the same (64). This is a contradiction with the above though, as there seem to be TWO different "theorems" around line integrals in this context: A) In a conservative field every line integral is the same and can be calculated using the initial and final segments or B) in a conservative field when two points are connected in a curve the work/ line integral is ZERO.
Which of these is true? Is this because the top line is a circle (if so, why?). I can understand that a circular path returns to its origin, but so does the triangle so that doesn't make sense.
Conservative does not mean the line integral $\int_\gamma\mathbf F\cdot d\mathbf r $ between any two points is zero. It means it is given by $\phi(\mathbf b)-\phi(\mathbf a),$ where $\mathbf a$ and $\mathbf b$ are the starting and ending point of the path $\gamma,$ and $\phi$ is the scalar potential such that $\nabla \phi = \bf F.$ So it is dependent on the endpoints, but independent of the path between them.
When the path is closed, i.e. when $\mathbf b = \mathbf a,$ then it is zero. So the fact that the elliptical path starts and ends at the same point, along with the fact that the field is conservative means that the integral is zero. The second path is not closed, so that integral doesn't need to be zero.