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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.

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3 Answers 3

up vote 2 down vote accepted

The region in question:

enter image description here

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.

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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.)

Green's theorem on the annulus

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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 $$ \oint_{C_1}M dx+Ndy $$ $$ \oint_{C_1}M dx+Ndy -\oint_{C_2}M dx+Ndy + \oint_{C_2}M dx+Ndy $$ $$ =\left(\oint_{C_1}M dx+Ndy -\oint_{C_2}M dx+Ndy \right) + \oint_{C_2}M dx+Ndy $$ $$ = \iint_D(N_x-M_y)dxdy + \oint_{C_2}M dx+Ndy $$ $$ = 0 + \oint_{C_2}M dx+Ndy $$ $$ = \oint_{C_2}M dx+Ndy $$

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How did you get that $\oint_{C_1} M dx + Ndy - \oint_{C_2} Mdx + Ndy = \int\int_D ( N_x - M_y) dxdy$? –  user38268 May 6 '12 at 17:03
    
Greens Theorem applies to regions which are not simply connected. All I did was apply it the the region bounded by $C_1$ and $C_2$. If you are in an introductory multivariable calculus course, this should have been proved for you. The idea behind the proof is to split up the region into simply connected regions and then apply Greens Theorem on each piece, noting the the "cuts" that you introduce appear with both orientations and cancel out. –  nullUser May 6 '12 at 17:10
    
I can see how it's done in $C_1$ and $C_2$ are traversed in the opposite direction, but now they are traversed in the same direction?? –  user38268 May 6 '12 at 17:11
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That is why we need the minus sign. $-\oint_{C_2} = \oint_{-C_2}$ where $-C_2$ is $C_2$ traversed in the opposite orientation. –  nullUser May 6 '12 at 17:12
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