# Green's Theorem and Line Integrals

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.

• the two derivatives are not the equal Sep 14, 2020 at 16:25
• That vector field is not conservative. The trick is to say the integral of that vector field over the parabola is equivalent to the integral of a different vector field which is conservative. You get that different vector field by changing the numerator of $P$ to $-y$. Then you may close the loop with the upper half of that ellipse, not the lower half. (I'm just writing this as a reminder, you might have already done this but I can't tell because your second integral doesn't have limits) Sep 14, 2020 at 16:28
• I can't see how you got $\;Q_x-P_y=0\;$ ...In fact I get a rather ugly expression Sep 14, 2020 at 16:42
• Can also be solved without Green's theorem: $$\int_{C_1} \boldsymbol F \cdot d\boldsymbol s = \int_{C_1} \frac {(-y, x)} {x^2 + 4 y^2} \cdot d\boldsymbol s = \frac 1 2 \arg(x + 2 i y) \bigg\rvert_{(x, y) = (1, 0)}^{(-1, 0)},$$ since, with a suitable choice of $\arg$, $\arg(x + 2 i y)/2$ is a potential for $(-y, x)/(x^2 + 4 y^2)$ in a region containing $C_1$. Sep 21, 2020 at 15:45

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

$$I = \int_{x^2+4y^2=1\:\cap\:y\geq 0} -ydx + xdy = \frac{\pi}{2}$$

• clever ${}{}{}$ Sep 14, 2020 at 16:44
• Thank you very much. The detail I was not understanding was the fact I could calculate the integral of the ellipsis on this new vector field which is conservative. I was using the initial vector field for the integral of the ellipsis.
– H44S
Sep 14, 2020 at 17:41
• After a long time, I was checking this problem again and I realised this would not work, since closing the loop with the upper half of $$x^2+4y^2=1$$ actually gives us 3 regions. I thought the intersection of the parabola and the elipse would occour only at $$y=0$$
– H44S
Mar 2, 2021 at 19:21
• @H44S It doesn't actually matter because the vector field is conservative on any region that doesn't enclose the origin. Any minus signs due to a flip in orientation don't matter because the double integrals are all $0$ Mar 2, 2021 at 21:20
• But my curve is no longer a simple closed curve (It has intersections) so I'm not sure green theorem holds
– H44S
Mar 3, 2021 at 2:07