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Let $C$ be parametrization $\mathbf{r}=\space5\cos(t) \mathbf{i}+\space4\sin(t) \mathbf{j}\space, t \in [0, 2\pi]$. Calculate $\oint_C \mathbf{F}\cdot d \mathbf{r}$ where $F$ is vector field

$$F(x,y)= \frac{-4x^3y}{(x^4+y^4)^2} \mathbf{i}+\frac{x^4-3y^4}{(x^4+y^4)^2} \mathbf{j}$$

Using Green's theorem:

\begin{align} &\oint_C \bigg(\frac{-4x^3y}{(x^4+y^4)^2} \mathbf{i}+\frac{x^4-3y^4}{(x^4+y^4)^2} \mathbf{j} \bigg) \cdot d \mathbf{r} \\\\ &= \iint_R \bigg[\frac{\partial}{\partial x} \bigg(\frac{x^4-3y^4}{(x^4+y^4)^2 } \bigg)- \frac{\partial}{\partial y} \bigg(\frac{-4x^3y}{(x^4+y^4)^2} \bigg)\bigg] dA \end{align}

But the vector field presents a singularity on $(0,0)$ so I cannot directly apply Green's. I use Extended Green’s Theorem instead.

$$\oint_CF-\oint_LF=\iint_R (\nabla×F) ·k\space dA = 0$$

$$\oint_CF=\oint_LF$$ Which $L$ will let me calculate the line integral in the easiest way possible?

PD: I've already tried with a square $(|x|=1\space |y|=1)$ but it just gives me more problems.

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  • $\begingroup$ Perhaps the Dirac delta function is necessary at the singular point. $\endgroup$ – marshal craft Nov 21 '15 at 22:10
  • $\begingroup$ How does it work? $\endgroup$ – encoded Nov 22 '15 at 0:05
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I don't know what problems the square gave you. It works excellently:
Define $$\mathbf r(t) = \begin{cases}(t+3)\mathbf{i} - \mathbf{j}& t \in [-4,-2)\\ (\mathbf{i} + (t+1)\mathbf{j}& t\in [-2, 0)\\ (1 - t)\mathbf{i} + \mathbf{j}& t \in [0,2)\\ -\mathbf{i} + (3-t)\mathbf{j}& t\in [2, 4)\end{cases}$$ Then $$\dot{\mathbf r}(t) = \begin{cases}\mathbf{i}& t \in [-4,-2)\\ \mathbf{j}& t\in [-2, 0)\\ -\mathbf{i}& t \in [0,2)\\ -\mathbf{j}& t\in [2, 4)\end{cases}$$ So $$\begin{align}\oint_LF\cdot d\mathbf r = &\int_{-4}^{-2} \frac{4(t+3)^3}{((t+3)^4+1)^2} dt\\ &+\int_{-2}^{0}\frac{1-3(t+1)^4}{(1+(t+1)^4)^2}dt\\ &+\int_{0}^{2} \frac{4(1-t)^3}{((1-t)^4+1)^2} dt\\ &+\int_{2}^{4}\frac{3(3-t)^4-1}{(1+(3-t)^4)^2}dt\\=&\int_{-1}^{1} \frac{4t^3}{(t^4+1)^2} dt\\ &+\int_{-1}^{1}\frac{1-3t^4}{(1+t^4)^2}dt\\ &+\int_{-1}^{1} \frac{4t^3}{(t^4+1)^2} dt\\ &+\int_{-1}^{1}\frac{3t^4-1}{(1+t^4)^2}dt\end{align}$$

For the first and third integrals, the integrand is odd, so the integrals are $0$. The second and fourth integrals are opposites, so their sum is $0$.

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  • $\begingroup$ I think the problem arrise when instead of simply computing the line integral as you have for the closed path, instead he used "Green's" theorem or sometimes known as "Stokes" theorem or generalized fundamental law of calculus relating line integral of closed path to the area integral over surface. In his case I think he said he ran into issue with singular point that only would arise when computing the surface integral. Due to "Green's" theorem one can relate the line integral you computed with a surface integral. $\endgroup$ – marshal craft Nov 22 '15 at 1:39
  • $\begingroup$ @marshalcraft - sorry, but you've misunderstood the problem. Because of the singularity at the origin, he is looking at an annular region with $C$ being the outer boundary and $L$ being the inner boundary. Because the component of the curl normal to the plane vanishes on the interior of this region, Green's theorem says the integral along the boundary is also $0$. Because you must traverse them in opposite directions, this means $\oint_C - \oint_L = 0$, or $\oint_C = \oint_L$. So the question was to find an $L$ whose integral could be calculated, as I did. $\endgroup$ – Paul Sinclair Nov 22 '15 at 1:59
  • $\begingroup$ I guess I misunderstood the part where encoded is attempting to solve a double integral and claims a singularity at the origin. I guess I further misunderstand how you, while solving a line integral questioned why you did not see such a singularity? $\endgroup$ – marshal craft Nov 22 '15 at 2:23
  • $\begingroup$ He isn't solving a double integral, he is evaluating a line integral. The singularity at the origin is the reason he cannot just apply Green's theorem to get a double integral over the entire interior of the curve. Instead, the idea is to select a second curve $L$ inside the original whose line integral is easy to evaluate, and such that it includes the singular point in its own interior so that the region between the curves has no singularity. Then the Green's theorem argument I gave above shows that the two curves have the same integral. So evaluating the easy one evaluates the original. $\endgroup$ – Paul Sinclair Nov 22 '15 at 4:01
  • $\begingroup$ I didn't see a singularity because my line integral doesn't get any closer than a distance of $1$ to the singularity. I evaluated it as a line integral, not by any variant of Stoke's theorem, so the fact that the function is singular some distance away is immaterial to this calculation. It is immaterial to the use of Green's theorem to equate the two boundaries because Green's theorem is being applied only to the other side of $L$ where there is no singularity. $\endgroup$ – Paul Sinclair Nov 22 '15 at 4:05

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