1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $F$?

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – Granger Obliviate Jan 3 '17 at 11:15
  • 1
    $\begingroup$ 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. $\endgroup$ – StackTD Jan 3 '17 at 11:29
  • 1
    $\begingroup$ 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. $\endgroup$ – StackTD Jan 3 '17 at 11:31
  • $\begingroup$ 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. $\endgroup$ – Granger Obliviate Jan 3 '17 at 11:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.