# Can Green's Theorem disagree with itself sometimes?

• kostoglotov
In summary, Green's theorem can be extended to arbitrary regions, but the region containing the origin cannot be included.
kostoglotov

## Homework Statement

Firstly, I was seeking any clarification on whether I've made any mistakes. Secondly, further insight into Green's Theorem, if my working is all good.

Regarding the vector field $\vec{F} = \frac{1}{x^2+y^2} \langle -y,x,0 \rangle$ I decided to test Green's Theorem out.

It would seem that for this vector field $\oint_C \vec{F}\cdot d\vec{r} \neq \int\int_D (\nabla \times \vec{F})\cdot \hat{k} \ dA$

## The Attempt at a Solution

So $curl \ \vec{F} = \vec{0}$ so $\int\int_D (\vec{0})\cdot \hat{k} \ dA = 0$ regardless of our path C or domain D.

So let's consider the unit circle as our path

$C: \ x^2+y^2=1$ for $\oint_C \vec{F} \cdot d\vec{r}$

Standard polar form conversion $\vec{r}(t) = \langle cos(t), sin(t), 0 \rangle$

So $\oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi}\vec{F}(\vec{r}(t)) \cdot \vec{r}'(t) dt = 2\pi$

So Green's Theorem doesn't work for all vector fields, even if it's a simple enclosed region...is this due to a discontinuity at (x,y) = (0,0)?

The point is that the computation of curl F is only valid away from r = 0, where the vector field is singular. If you include the appropriate delta function at r = 0, you will find that Green's theorem holds.

No one (including Green) has ever claimed that Green's theorem works for all vector fields. A correct statement of Green's theorem is
"If L and M have continuous partial derivatives inside a simple closed curve, then $$\oint_C (Ldx+ Mdy)= \int\int_D\left(\frac{\partial M}{\partial x}- \frac{\partial L}{\partial y}\right) dxdy$$.

Your example does not have "continuous partial derivatives" at the origin. (And if you try to exclude (0, 0) by adding a small circle around it as part of the boundary, the boundary is no longer "simple".)

kostoglotov
Orodruin said:
The point is that the computation of curl F is only valid away from r = 0, where the vector field is singular. If you include the appropriate delta function at r = 0, you will find that Green's theorem holds.

... :) So what does all that mean?

Orodruin said:
The point is that the computation of curl F is only valid away from r = 0, where the vector field is singular. If you include the appropriate delta function at r = 0, you will find that Green's theorem holds.

Does this mean a region that does not include the origin will be fine for Green's Theorem?

HallsofIvy said:
"If L and M have continuous partial derivatives inside a simple closed curve, then

This seems too restrictive. It should be straight forward to generalise it to arbitrary distributions.

Orodruin said:
This seems too restrictive. It should be straight forward to generalise it to arbitrary distributions.
Yes, if you are willing to go to "homotopies", "homologies", and "algebraic topology".

Green's theorem as you know it has been proven when the region ##D## is simple (type I or type II):

$$\oint_C \vec F \cdot d \vec r = \iint_D Q_x - P_y \space dA$$

The vector field is given as:

$$\vec F = \frac{-y \hat i + x \hat j}{x^2 + y^2}$$

For the given vector field, there is an issue at the origin because the vector field is undefined at the origin and it doesn't have continuous partial derivatives there either. Any attempt to apply Green's theorem directly will fail for any region containing the origin.

This poses a problem for the general version of Green's theorem, but Green's theorem can be extended to apply to regions that are not simple, i.e the region will have a hole somewhere. We do this by dividing the larger region into two simpler regions where we can apply Green's theorem and superimpose the results.

So taking the curve as the positively oriented unit circle ##C_1: x^2 + y^2 = 1##, it encloses the region ##D: x^2 + y^2 \leq 1##. This is going to cause a problem because it contains the origin, and as mentioned before, we can't do that for this particular vector field.

By enclosing the origin in a negatively oriented, smaller circle ##C_2: x^2 + y^2 = r, \space 0 < r < 1##, we can extend Green's theorem to show that:

$$\oint_C \vec F \cdot d \vec r = 2 \pi$$

Where ##C = C_1 \cup C_2##.

We do this by dividing ##D## into two simple regions ##D'## and ##D''## such that ##D = D' \cup D''## and we apply Green's theorem to the two simple regions like so:

$$\iint_D Q_x - P_y \space dA = \iint_{D'} Q_x - P_y \space dA + \iint_{D''} Q_x - P_y \space dA = \oint_{\partial D'} Pdx + Qdy + \oint_{\partial D''} Pdx + Qdy$$

The line integrals are along common boundary lines and are opposite in direction, so they cancel and we get:

$$\oint_{\partial D'} Pdx + Qdy + \oint_{\partial D''} Pdx + Qdy = \oint_{C_1} P dx + Q dy + \oint_{C_2} P dx + Q dy = \oint_C P dx + Q dy$$

This is Green's theorem as we know it because we have shown:

$$\iint_D Q_x - P_y \space dA = \oint_C P dx + Q dy$$

Even though we divided the region ##D## into two simpler regions before applying Green's theorem. This allows you to get around that pesky origin problem.

So for the problem at hand, we have specifically shown:

$$\oint_{C_1} \vec F \cdot d \vec r = \oint_{C_2} \vec F \cdot d \vec r = \int_0^{2 \pi} d \theta = 2 \pi$$

kostoglotov and jim mcnamara
You're trying to use Green's Theorem without the field being of C1 class, it's not continuously differentiable.
As the previous poster mentioned you need to use the second form of Green's theorem if you wish to use it at all in this problem, you'll end up integrating by definition regardless, just in a smaller circle

## 1. What is Green's Theorem and how does it relate to vector fields?

Green's Theorem is a mathematical theorem that relates a line integral around a simple closed curve to a double integral over the region enclosed by the curve. It is commonly used in vector calculus to evaluate the work done by a vector field along a closed curve.

## 2. Can Green's Theorem be applied to any type of vector field?

Green's Theorem can be applied to any type of vector field as long as it satisfies certain conditions, such as being continuously differentiable and having a continuous first derivative. However, it is most commonly used for conservative vector fields, which have the property that the line integral is independent of the path taken.

## 3. What does it mean for Green's Theorem to "disagree with itself"?

When we say that Green's Theorem can disagree with itself, we mean that the result obtained from the line integral along a closed curve does not match the result obtained from the double integral over the enclosed region. This can happen in certain cases when the vector field is not conservative or when the curve is not simple.

## 4. How can Green's Theorem be used to solve practical problems?

Green's Theorem can be used in various practical applications, such as calculating the work done by a force field on a moving object, calculating the flux of a vector field through a surface, or finding the area of a region bounded by a curve. It provides a powerful tool for evaluating line integrals and double integrals, which are commonly used in physics, engineering, and other fields.

## 5. Can Green's Theorem be extended to higher dimensions?

Yes, Green's Theorem can be extended to higher dimensions through the use of generalizations such as Stokes' Theorem and the Divergence Theorem. These theorems relate surface integrals and volume integrals, respectively, to higher-dimensional versions of line integrals. They are essential tools in the study of vector calculus and have many important applications in mathematics and physics.

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