# Green's theorem details

I'm studying Green theorem. My textbook gives me two examples in which I have a few doubts

1st example

Imagine the annulus $D=\{(x,y) \in \mathbb{R^2}: r^2 < x^2 + y^2 < R^2\}$

$F$ the a vector field of class $C^1$ defined in an open set of $\mathbb{R^2}$. D is an almost regular (union of elementary domains) domain. The boundary of the set is $\partial D=\Gamma_1 \cup \Gamma_2$ with $\Gamma_1$ the circle of radius R anticlockwise and $\Gamma_2$ the circle of radius r clockwise, both centered in the origin.

By Green's theorem

$\int\int_D (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dx dy= \int_{\Gamma_1} Pdx + Qdy + \int_{\Gamma_2} Pdx + Qdy$

If F is a conservative field then $\int\int_D (\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y})dx dy= 0$ so $\int_{\Gamma_1} Pdx + Qdy = - \int_{\Gamma_2} Pdx + Qdy$

If instead we had $\Gamma _1$ and $\Gamma _2$ both clockwise or anticlockwise we would have $\int_{\Gamma_1} Pdx + Qdy = \int_{\Gamma_2} Pdx + Qdy$

I'm with everything except my last paragraph. How can we conclude that? Is it because in $\int_{\Gamma_1} Pdx + Qdy = - \int_{\Gamma_2} Pdx + Qdy$ the minus signal is also because $\Gamma _1$ and $\Gamma _2$ have different directions? Also if F is a closed field shouldn't we have $\int_{\Gamma_1} Pdx + Qdy = \int_{\Gamma_2} Pdx + Qdy = 0$?

2nd example

$F$ the a vector field of class $C^1$ defined in $\mathbb{R^2} except \{0,0\}$.

$F(x,y) = (P,Q) = (\frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2})$

We immediately conclude that F is a conservative field.

D is an almost regular (union of elementary domains) domain. The boundary of the set is $\partial D=\Gamma \cup C$ with C the circle of radius R anticlockwise and $\Gamma$ other closed line (exterior to C) also anticlockwise, both centered in the origin (it's a similar set to the one in the 1st example but now instead of two circles we have a circle closer to the origin and other closed path less close to the origin)

C can be parameterized as $\gamma (t) = (R\cos t, R \sin t)$ $0<t<2\pi$

By Green's theorem

$\oint _C F d\gamma = \int_0^{2\pi} F(\gamma (t)) \gamma ´(t) dt = 2\pi$

By the conclusions of the first example we have $\int_{\Gamma} Pdx + Qdy = \int_{C} Pdx + Qdy = 2\pi$

If we had another closed line $\Gamma _2$ limiting a almost regular domain that did not include the origin in the interior then: $\int_{\Gamma} P dx + Q dy = 0$

Ok first of all: why isn't the integral of F in C zero? Is it because F is not defined in the origin (does this mean it's not a $C^1$ class function?). Therefore in any other path that doesn't include the origin that integral must be zero?

Hope my questions are clear.

Thanks!

For your first question:

I'm with everything except my last paragraph. How can we conclude that? Is it because in $\int_{\Gamma_1} Pdx + Qdy = - \int_{\Gamma_2} Pdx + Qdy$ the minus signal is also because $\Gamma _1$ and $\Gamma _2$ have different directions? Also if F is a closed field shouldn't we have $\int_{\Gamma_1} Pdx + Qdy = \int_{\Gamma_2} Pdx + Qdy = 0$?

To apply Green's theorem, $\Gamma_1$ and $\Gamma_2$ are oriented as stated (anticlockwise and clockwise respectively; take a look at how you would divide region $D$ into elementary domains to see why this is the case). To make this explicit, write $\Gamma_1^+$ and $\Gamma_2^-$ for the paths with these orientations. Now if the integral is $0$, then you have: $$\int_{\Gamma_1^+} Pdx + Qdy + \int_{\Gamma_2^-} Pdx + Qdy = 0 \iff \int_{\Gamma_1^+} Pdx + Qdy =- \int_{\Gamma_2^-} Pdx + Qdy$$ But changing the orientation of a path simply changes the sign of the integral, so you also have: $$\int_{\Gamma_1^+} Pdx + Qdy = \int_{\Gamma_2^+} Pdx + Qdy \quad\mbox{or}\quad \int_{\Gamma_1^-} Pdx + Qdy = \int_{\Gamma_2^-}Pdx + Qdy$$ which means you get a simple equality if you choose the same orientation.

Note: I think they mean $F$ is a conservative field. You could talk about a closed path, but a closed field doesn't make sense to me and from the context, I expect they mean $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=0$, which is the case if $F$ is conservative (i.e. the gradient of some function).

The second question isn't completely clear to me, is there information missing? Is $D$ as in example 1? Now $\Gamma$ is just a path and $\Gamma_2$ is (another?) line...? Is anything given about $F$?

• Hi StackTD! About the first question I think I get it now. The only thing I'm still questioning is "if F is a closed field shouldn't we have $\int_{\Gamma_1} Pdx + Qdy = \int_{\Gamma_2} Pdx + Qdy = 0$"? Shouldn't the integral of a closed field in a closed line be zero? About the second question you're right there is information missing that I already added (hope the question is clear now). Thanks! – Granger Obliviate Jan 3 '17 at 11:15
• I added a note about question 1, I assume they mean a conservative field. I'll take a look at the updated question 2 later. – StackTD Jan 3 '17 at 11:29
• Due to "We immediately conclude that F is a closed field.", I suppose you use the term closed field for what most would call a conservative field. – StackTD Jan 3 '17 at 11:31
• Yes! I meant when $\frac{\partial Q}{\partial x} = \frac {\partial P}{\partial y}$. I saw the term in my professor notes, I thought it was used but apparently it's not... So yes it's conservative what I mean. I don't understand how this integrals are not zero when F is a conservative field. – Granger Obliviate Jan 3 '17 at 11:39