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The question is as follows:

Given vector field

$$V = \left(\frac{1-y}{x^2 + (y-1)^2}, \frac{x}{x^2+(y-1)^2}\right)$$

Evaluate

$$\int_{l_1}V \bullet dr\text{, }\int_{l_2}V \bullet dr$$

Where $l_1$ and $l_2$ are given as

$$l_1: x^2+(y-1)^2=1\text{, } l_2: x^2+(y-4)^2=1$$

According to the answer key given by my TA, the first integral can't be evaluated using the fundamental theorem of line integrals. He mentions the fact that $V$ is conservative, except at (0,1) where it is not defined, but the first curve doesn't pass through this point. Using a verbose method which isn't really the focus here, he finds that the integral evaluates to 2$\pi$, not zero, which is what conservativity would imply.

However, he evaluates the second integral using the fundamental theorem and gets zero.

I believe both of these answers are correct, as I didn't find any errors in his evaluation of the first integral. My question is, why can the fundamental theorem be applied to the second integral but not the first one?

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    $\begingroup$ The first curve loops around a bad point, whereas the second does not. Your fundamental theorem - I assume! - says something like: the line integral of $V= (A,B)$ over a closed curve is zero if $\partial A/ \partial y = \partial B/ \partial x$ for all points of the enclosed region - is that correct? So if the curve loops around a point where $V$ is 'bad', the theorem is not applicable. N.B. I am being somewhat imprecise in this, but it's the idea... $\endgroup$ – peter a g Dec 14 '15 at 19:40
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V is conservative, except at (0,1) where it is not defined, but the first curve doesn't pass through this point

The issue is whether the curve surrounds the point, not whether it passes through.

V contributes a fixed amount ($2\pi$) to the integral for every time the integration path winds around the singular point $(0,1)$. The number of times winding occurs is measured taking orientation into account, so that clockwise and anticlockwise loops cancel each other out.

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The problem is that $V$ isn't actually conservative on the open set $\mathbb{R}^2 - \{(0,1)\}$. (If it were, as you say, by the gradient theorem the integral would have to be zero.) By integration, we find that a potential function $f$ would have to be $f(x,y) = \operatorname{arctan}(\frac{x}{1-y}) + c$, which isn't defined on $\mathbb{R} \times \{ y = 1 \}$.

On the other hand, that potential function is defined on $\mathbb{R} \times \{y > 1\}$, so that $V$ is conservative there; in particular, by the gradient theorem the integral over $l_2$ vanishes.

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The fundamental theorem say that: if $V=\nabla f$ for some scalar function$f$ than the path integral of $V$ on a smooth line depends only on the initial and final points. As a consequence, for a close path we have: $$ V=\nabla f \Rightarrow \oint_C V\cdot dr=0 $$ In this case we say that the vector field $V$ is conservative, and we can show that the converse is also true, i.e. a field $V$ is conservative only if its line integral is independent of the path.

So the question becomes: whan a field $V$ is the gradient of some scalar field?

A necessary condition is that $$\frac{\partial V_x}{\partial y}=\frac{\partial V_y}{\partial x}$$ but in general such condition is not sufficient.

A very beautiful result is that this condition is sufficient only if the region where the vector field $V$ is defined is simply connected, i.e.such that any closed loop in this region can be continuously contracted to a point. This is the Poincaré Lemma.

So, In your case the closed path $l_1$ enclose a region that is not simply connected since the field $V$ is not defined in the point $(0,1)$ contained in the region ( the region has a hole) , but the region enclosed by $l_2$ does not contains this point, so it is simply connected, and this is the cause of the difference.

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Notice that $$ V = \nabla \left ( \arctan \left (\frac{y-1}{x} \right )\right) = \nabla f(x,y) $$ and is continuous on $(x,y)\in \mathbb{R}^2 \setminus ( 0,1)$ since the derivative isn't defined at $(0,1)$. Regarding the fundamental theorem, notice for closed curves $\gamma$ we have that $$ \int_{\gamma} \nabla f \cdot ds = \int_D \nabla \times \nabla f \, dxdy $$ where $\partial D = \gamma$ by green's theorem (where it makes sense, $f \in C^2$). Thus if $f$ is $C^2$ on $D$, we have that $\nabla \times \nabla f =0$ on $D$, so $$ \int_{\gamma} \nabla f \cdot ds = 0$$ For the case of $I_2$, $f$ is $C^2$ on $D$, thus we have $$ \int_{I_2} V \cdot ds = 0 $$ For the case of $I_1$, $f$ is not $C^2$ on $D$, thus we need to compute the integral directly. $$ \int_{I_1} V \cdot ds = \int_{I_1} (1-y)dx + xdy $$ Let $(1-y) = -\sin \theta$ and $x = \cos \theta$ with $\theta \in [0, 2\pi)$, then $$\int_{I_1} (1-y)dx + xdy =\int_0^{2 \pi } (\sin^2 \theta + \cos^2 \theta) d\theta = \int_0^{2 \pi } d \theta = 2 \pi$$

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