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.

  • $\begingroup$ the two derivatives are not the equal $\endgroup$ Sep 14, 2020 at 16:25
  • $\begingroup$ 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) $\endgroup$ Sep 14, 2020 at 16:28
  • $\begingroup$ I can't see how you got $\;Q_x-P_y=0\;$ ...In fact I get a rather ugly expression $\endgroup$
    – DonAntonio
    Sep 14, 2020 at 16:42
  • $\begingroup$ 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 $C_1$. $\endgroup$
    – Maxim
    Sep 21, 2020 at 15:45

1 Answer 1


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

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

  • $\begingroup$ clever ${}{}{}$ $\endgroup$ Sep 14, 2020 at 16:44
  • $\begingroup$ 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. $\endgroup$
    – H44S
    Sep 14, 2020 at 17:41
  • $\begingroup$ 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 $$y=0$$ $\endgroup$
    – H44S
    Mar 2, 2021 at 19:21
  • $\begingroup$ @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 $0$ $\endgroup$ Mar 2, 2021 at 21:20
  • $\begingroup$ But my curve is no longer a simple closed curve (It has intersections) so I'm not sure green theorem holds $\endgroup$
    – H44S
    Mar 3, 2021 at 2:07

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