Confusion regarding orientation of curves in Green's Theorem I am in the middle of helping some friends out with their vector calculus assignment (I don't do this course). Now in their assignment they have the following question:


Consider the integral $$\oint_{C_1} \frac{-y^3 dx + xy^2 dy}{(x^2 + y^2)^2}$$
    where $C_1$ is the ellipse $x^2 + 4y^2 = 4$. Assuming that Green's Theorem can be applied to the region $D$ between the circle $C_2 : x^2 + y^2 = 9$ and the ellipse $C_1$, show that $$\oint_{C_1}\frac{-y^3 dx + xy^2 dy}{(x^2 + y^2)^2} = \oint_{C_2} \frac{-y^3 dx + xy^2 dy}{(x^2 + y^2)^2}$$
    where $C_1$ and $C_2$ are both traversed in the anti - clockwise direction sense as viewed from the positive $z$ - direction.


Now let us call $M = \frac{-y^3}{(x^2 + y^2)^2}$ and $N = \frac{xy^2}{(x^2 + y^2)^2}$. I want to invoke something about Green's Theorem that tells me something like 
$$\oint_{C_1} M dx + N dy - \oint_{C_2} M dx + N dy = \int\int_D (N_x - M_y) dxdy.$$
If this holds since the right hand side is zero, we would have our desired equality.
However what is causing the confusion is the left hand side. Is this even valid? In all the problems we have encountered, the curves $C_1$ and $C_2$ are traversed in opposite directions. For example $C_2$ is traversed clockwise and $C_1$ is traversed counter - clockwise. However now both curves are traversed anti - clockwise so how do we get around this? Is there a mistake in the assignment?
Thanks.
 A: The region in question:

The boundary of $D$ with the natural orientation (in preparation for the use of Green's Theorem) is the union $(−C_1) \cup C_2$, where $C_1$ and $C_2$ are traversed in the counterclockwise direction.  See oenamen's answer for the proper justification.  Writing $−C_1$ means "traverse $C_1$ in the opposite direction". So, Green's theorem gives us
$$
\begin{align}
\iint_{D} N_x - M_y \,dx\,dy &= \oint_{(−C_1) \cup C_2} M\,dx + N\,dy \\
  &= \oint_{−C_1} M\,dx + N\,dy + \oint_{C_2} M\,dx + N\,dy \\
  &= -\oint_{C_1} M\,dx + N\,dy + \oint_{C_2} M\,dx + N\,dy.
\end{align}
$$
Just show that the integral over $D$ is zero to get your equality.
A: The boundary of $D$ is the curve $-C_1$ and $C_2$ (that is, $C_1$ is traversed in the opposite direction). 
To see why the orientation of $C_1$ is flipped, have a look at the diagram below. 
Imagine the curves $l$ and $-l$ are on top of one another, so their contribution to the line integral vanishes. 
Notice that $N_x - M_y = 0$ inside $D$, so the surface integral vanishes.
(Here it is important that we have avoided the singularity at the origin.)

A: Green's Theorem says
$$
\oint_{C_1}M dx+Ndy = \iint_{D_1} (N_x-M_y)dxdy
$$
not
$$
\oint_{C_1}Mdx+N dy = \iint_{D_1} (M_y-N_x)dxdy.
$$
Then we have
$$
\oint_{C_1}M dx+Ndy
$$
$$
\oint_{C_1}M dx+Ndy -\oint_{C_2}M dx+Ndy + \oint_{C_2}M dx+Ndy
$$
$$
=\left(\oint_{C_1}M dx+Ndy -\oint_{C_2}M dx+Ndy \right) + \oint_{C_2}M dx+Ndy
$$
$$
= \iint_D(N_x-M_y)dxdy + \oint_{C_2}M dx+Ndy
$$
$$
= 0 + \oint_{C_2}M dx+Ndy
$$
$$
= \oint_{C_2}M dx+Ndy
$$
