What happens when C is only partly within the vector field?

Question

So I have the following question where $$\vec{G}=(x,y)/\sqrt{x^2+y^2-1}$$ is a vector field. I have to calculate the line integral for $$\int_{\mathrm{C}}\vec{G}\, d\vec{r}\, $$ where $C$ is the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$.
I know that $\vec{G}$ has the domain $x^2+y^2 \gt 1,$ (meaning that the circle with radius 1 is not part of the domain) and the ellipse has radius 2 with x-component and 3 for y-component. I have calculated param $x=2cost$ and $y=3sint$ where $0\le t\le 2\pi$. So does this mean that only a part of the ellipse is within $\vec{G}$?

How do I calculate the line integral in this case?

Answer 1

The ellipse $C: x^2/4+y^2/9=1$ is well within the field $\vec{G}=\frac{x\hat i+y\hat j}{\sqrt{x^2+y^2-1}}$ which is defined for the exterior of circle $x^2+y^2=1$ so you can easily evaluate the line integral using Green's theorem(after making it simply connected using a two-way cut).

$\int_C\vec{G}.\vec{dr}=\int\int_R curl(\vec{G})dR$

where $R$ is the region bounded by ellipse on $xy-$plane.

Answer 2

You do not need to parametrize this, nor do you need Green's theorem. By usual integration tricks, your function $\vec{G}$ has a potential: $$ g(x,y) = \sqrt{x^2+y^2-1}. $$ You can check that $\vec{G} = \nabla g$, and this is valid over a domain that "encloses" your ellipse in a nice, legal way. Now use the fundamental theorem: $$ \int_C \vec{G} \cdot d\vec{r} = g(2,0) -g(2,0) = 0 $$ as $C$ is a closed curve.