Gradient fields with holes and the fundamental theorem of line integrals

Question

I'm learning vector calculus for the GRE, so this is my first time encountering these concepts. To my knowledge, a gradient field $\mathbf{F}(x,y) : D\to \mathbb{R}^2$ is one where there exists a scalar function $f(x,y) : D\to \mathbb{R}$ such that $$ \mathbf{F}=\nabla f $$ Here, I'm not specifying $D$ to be simply-connected, or even open. I believe this is a valid definition for any subset $D\subset \mathbb{R}^2$.

The fundamental theorem for line integral states that if $\mathbf{F}$ is a gradient field, i.e. $\mathbf{F}=\nabla f$ for some $f$, then along any path $C$ given by the parameterization $\mathbf{r}(t)$ for $a\leq t \leq b$ $$ \int_C \mathbf{F}\cdot d\mathbf{r}= f(\mathbf{r}(b))- f(\mathbf{r}(a)) $$

This might be a dumb question, but what if the domain has holes?

For example, take the scalar function $$f(x,y) = \frac{xy}{x^2+y^2}$$ then the corresponding gradient field is $\mathbf{F}(x,y) = \left( \frac{y(y^2-x^2)}{(x^2+y^2)^2}, \frac{x(x^2-y^2)}{(x^2+y^2)^2} \right)$.

I tried to take the line integral over the upper half circle, and the lower half circle, and it turns out they are both zero. Could I have applied the fundamental theorem of line integrals to this vector field $\mathbf{F}$?

Answer 1

Yes you could have use the theorem. If you take a look at the short proof of the theorem,

\begin{align*} \int_C \mathbf{F}\cdot d\mathbf{r} &= \int_a^b \mathbf{F} (\mathbf{r} (t)) \mathbf{r}'(t) dt\\ &= \int_a^b\nabla f (\mathbf{r} (t)) \mathbf{r}'(t) dt \\ &= \int_a^b \frac{d}{dt} (f\circ \mathbf{r})(t) dt\\ &= (f\circ \mathbf{r})(b)-(f\circ \mathbf{r})(a). \end{align*}

We used only chain rule and the standard fundamental theorem of calculus. No assumptions about the domain $D$ (besides that the curve $C$ lies inside $D$).

Answer 2

No, unfortunately you cannot apply FTLI if the gradient field has holes. I believe the (almost) gradient field $$\vec{F} (x, y) = \left(\frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}\right)$$ which is (almost) the gradient of $f = \arctan(y/x)$ (except that $f$ is not defined at $(0, 0)$). It has the property that $$\int_C \vec{F} \cdot \vec{dr} = 2 \pi$$ for any closed curve $C$ which surrounds the origin. What you can do for such problematic vector fields, however, is to notice that $\vec{F}$ is a gradient field away from its discontinuities, so you can always evaluate the above integral by deforming $C$ into any nice closed loop you like (and here, a particularly nice one is the unit circle). In other words, $$\int_C \vec{F} \cdot \vec{dr} = \int_{C'} \vec{F} \cdot \vec{dr}$$ for any $C'$ also containing the origin (which is the only problematic point) and preserving orientation. The reason that such a deformation is possible is due to Stokes' Theorem, which notes that $$\int_{\partial A} \vec{F} \cdot \vec{dr} = \iint_{A} (\vec{\nabla} \times \vec{F}) \cdot \vec{dA}$$ for any $A$, as long as the function is $\mathcal{C}^2$ over $A$. Choosing $A$ to be $C \setminus C'$, we can see that $$\int_{\partial A} \vec{F} \cdot \vec{dr} = \int_C \vec{F} \cdot \vec{dr} - \int_{C'} \vec{F} \cdot \vec{dr} = \iint_{A} (\vec{\nabla} \times \vec{F}) \cdot \vec{dA} = 0$$