So I'm supposed to use Green' theorem to calculate the line integral
$$
\int_{C_1} \frac{x^2-1}{x^2+4y^2}dx +\frac{x}{x^2+4y^2}dy
$$
Where $C_1$ is the part of the parabola $y=1-x^2$ from point $(1,0)$ to $(-1,0)$
My first problem: I was able to calculate $\partial P/ \partial y$ and $\partial Q/ \partial x$ but it was indeed very tedious. Is there another way to calculate it? I thought about considering the derivatives only evaluated at the parabola, in a way I could write $x^2-1=-y$ but I don't know if I can do this because when we calculate the surface integral at Green's theorem we are considering the whole surface, right?
Anyways, I found out $\partial Q/ \partial x – \partial P/ \partial y = 0$. Then, since the theorem requires me to have a closed path, I chose my "second path" as the ellipsis $x^2+4y^2=1$. So then I can write:
$$
\int_{C_1} \frac{x^2-1}{x^2+4y^2}dx +\frac{x}{x^2+4y^2}dy= -\int_{C_2} x^2-1dx +xdy
$$
After this we just need to parametrize the ellipsis as $y=\frac{1}{2} \sin(t); \enspace x=\cos(t)$.
Sadly enough the answer I get is not the correct one (which is $\pi/2$). What am I doing wrong/ is there a better way to proceed?
I appreciate any tips/corrections.
Best Answer
We have that
$$\int_{C_1} \frac{x^2-1}{x^2+4y^2}dx + \frac{x}{x^2+4y^2}dy = \int_{C_1} \frac{-y}{x^2+4y^2}dx+\frac{x}{x^2+4y^2}dy$$
since on $C_1$ $y=1-x^2$. This new vector field is conservative on any region that doesn't contain the origin. To use Green's theorem, close the loop with the upper half of the ellipse $x^2+4y^2=1$ which means that
$$I = \int_{x^2+4y^2=1\:\cap\:y\geq 0} -ydx + xdy = \frac{\pi}{2}$$