Line integral for vector field $F(x,y)=\left(-\frac{y}{x^2 + y^2} , \frac{x}{x^2+y^2}\right)$ with the curve $\vert x \vert + \vert y \vert = 5$

Question

$Q)$ For a vector field $F(x,y)=\left(-\frac{y}{x^2 + y^2} , \frac{x}{x^2+y^2}\right)$ in $\mathbb{R}^2$

Find the $\int_{\vert x \vert + \vert y \vert = 5 } -\frac{y}{x^2 + y^2} dx + \frac{x}{x^2 + y^2}dy $

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In my lecture's note he claim that $\int_{\vert x \vert + \vert y \vert = 5 } -\frac{y}{x^2 + y^2} dx + \frac{x}{x^2 + y^2}dy $ = $\int_{x^2 + y^2 = 1 } -\frac{y}{x^2 + y^2} dx + \frac{x}{x^2 + y^2}dy $

And then, he solved by parameterizing curve $x^2+y^2=1$ to $(cost, sint), [0 \leq t \leq 2\pi]$

But my doubt is firstly the vector field, $F$ is not defined at $(0,0)$, Hence it is not simply connected. So we can't say the not conservative of the $F$ though $curl F = (0,0)$. Therefore we can't guarantee the path independence, we can't claim that $\int_{\vert x \vert + \vert y \vert = 5 } = \int_{x^2 + y^2 = 1 } $.

Is my opinion right? If the lecture is right what is his ground justifying $\int_{\vert x \vert + \vert y \vert = 5 } = \int_{x^2 + y^2 = 1 } $ ? If the lecture's solution is false, How to Solve it?

Any help would be appreciated. thanks.

Answer 1

The integral along the boundary of a C shaped contour centered at $(0,0)$ is zero because the plane without the positive $x$ axis is a simply connected domain.

Now close the letter C to form an O. By taking the limit, you see that the integral along the inside boundary of the O is the same as the integral along the outside boundary. Finaly do the same thing with the circle $x^2 + y^2=1$ and the square $|x|+|y|=5$ as internal and external boundaries.

Answer 2

The unit circle $\partial D$ and your rhombus boundary $\partial R$ together bound an annular region $$S:=\bigl\{(x,y)\in{\mathbb R}^2\bigm|1\leq x^2+y^2, \ |x|+|y|\leq5\bigr\}\ .$$ Apply now Greens theorem to $S$ and obtain $$0=\int_S{\rm curl}(F)\>{\rm d}(x,y)=\int_{\partial S}F\cdot d{\bf r}=\int_{\partial R}F\cdot d{\bf r}-\int_{\partial D}F\cdot d{\bf r}\ .$$ As you have computed $\int_{\partial D}F\cdot d{\bf r}=2\pi$ you then know that $$\int_{\partial R}F\cdot d{\bf r}=2\pi$$ as well.

By the way: Note that your $F$ is the same as $$F(x,y)=\nabla{\rm arg}(x,y)\qquad\bigl((x,y)\ne(0,0)\bigr)\ ,$$ so that the line integral of $F$ along any curve is just the total increment of the polar angle along this curve.