Compute $\int_C \textbf{F} \cdot \textbf{ds}, \;\textbf{F}(x,y)=\frac{2}{x^2+y^2}(y\textbf{i}-x\textbf{j})$ Let
$$
\textbf{F}(x,y)=P(x,y)\textbf{i}+Q(x,y)\textbf{j}=
\begin{cases} 
\frac{2}{x^2+y^2}(y\textbf{i}-x\textbf{j}) &\text{if } x^2+y^2\neq 0\\
\textbf{0} &\text{if } x=y=0,
\end{cases}
$$
and let C be the curve given by $3x^{8}+2y^{6}=1$, oriented counter-clockwise.
Compute $\int_C \textbf{F} \cdot \textbf{ds}$.
It would be nice if I could use Greene's theorem because (substituting with polar coordinates)
$$
\frac{\partial P}{\partial y}= 
\frac{\partial Q}{\partial x}=
\frac{2(x^2-y^2)}{(x^2+y^2)^2}=
\frac{2\cos2\theta}{r^2}.
$$
But Greene's theorem requires that P and Q have continuous partial derivatives in the region contained by C, and
$
\lim_{r \to 0^+}\frac{2\cos2\theta}{r^2}=\pm\infty.  
$
(except where $\cos 2\theta=0)$.
Is there a way to ignore the fact that the partial derivatives are not continuous at (0,0)?
 A: No, in general we can't ignore the singularity at $(0, 0)$, and in fact the nonzero answer to this problem illustrates that in general we can't do this.
The problem with computing the line integral directly is that the curve $C$ is annoying to parameterize. We can, however, exploit Green's Theorem and the fact that our vector field $\bf F$ satisfies $$\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 0:$$
Consider the curve $$C' = C \amalg -\mathbb{S}^1,$$ which is the disjoint union of the given curve $C$ and unit circle $- \mathbb{S}^1$, where $-$ indicates that the curve is oriented clockwise. Now, the components $P, Q$ of $\bf F$ do have continuous partial derivatives on the (distorted annular) region $R$ bounded by $C'$, so applying Green's Theorem gives that
$$\int_{C'} {\bf F} \cdot d{\bf s} = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA = \iint_R 0 \, dA = 0.$$
On the other hand, the line integral over $C'$ is just the sum of the line integrals over the constituent curves $C$, $-\mathbb{S}^1$:
$$0 = \int_{C'} {\bf F} \cdot d{\bf s} = \int_{C \amalg -\mathbb{S}^1} {\bf F} \cdot d{\bf s} = \int_{C} {\bf F} \cdot d{\bf s} + \int_{- \mathbb{S}^1} {\bf F} \cdot d{\bf s} = \int_{C} {\bf F} \cdot d{\bf s} - \int_{ \mathbb{S}^1} {\bf F} \cdot d{\bf s},$$
where in the last equality we use the fact that reversing the orientation of a curve reverses the value of the line integral of a vector field along that curve. (In particular, the last line integral is taken over the unit circle evaluated anticlockwise.) Rearranging gives that
$$\int_{C} {\bf F} \cdot d{\bf s} = \int_{ \mathbb{S}^1} {\bf F} \cdot d{\bf s},$$
that is, we can compute the original integral by replacing the curve $C$ by the curve $\mathbb{S}^1$, which is simpler to parameterize.

 The standard parameterization ${\bf r}(t) := \cos t \cdot {\bf i} + \sin t \cdot {\bf j}$, $t \in [0, 2\pi]$, of the unit circle produces an especially nice integral, as $${\bf F}({\bf r}(t)) = 2 \sin t \cdot {\bf i} - 2 \cos t \cdot {\bf j}, \qquad d{\bf s} = -\sin t \cdot {\bf i} + \cos t \cdot {\bf j}.$$ So, ${\bf F}({\bf r}(t)) \cdot {\bf r}'(t) = -2$ and hence $$\int_{C'} {\bf F} \cdot d{\bf s} = \int_{\mathbb{S}^1} {\bf F} \cdot d{\bf s} = \int_0^{2 \pi} {\bf F}({\bf r}(t)) \cdot {\bf r}'(t) \, dt = \int_0^{2 \pi} -2 \,dt = -4 \pi.$$ Note in particular this shows that we cannot ignore the singularity in $\bf F$ when applying Green's Theorem; if we had, we would have (hence) incorrectly concluded that the line integral has value $0$.

