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Green's Theorem

Relationships Between Green's Theorem, Stokes' Theorem and the Divergence Theorem

Green's Theorem has two forms, the circulation form and the divergence form. Green's Theorem is in two dimensions, While Stokes' Theorem is the three-dimensional form of the circulation form of Green's Theorem. Similarly, the Divergence Theorem is the three-dimensional form of the divergence form of Green's Theorem. The following table summarizes these relationships.

2-dim

3-dim

Green's Theorem (circulation form)

Stokes' Theorem

Greens' Theorem (divergence form)

Divergence Theorem

Green's Theorem

IF WE HAVE

- a simply-connected region \(\mathcal{R}\) which is bounded by a smooth curve \(\mathcal{C}\), oriented anticlockwise (counter-clockwise) and traversed once, and
- a vector field \(\vec{F} = M\hat{i}+N\hat{j}\) which is continuously differentiable on the region \(\mathcal{R}\).

THEN THIS EQUATION HOLDS

\(\displaystyle{ \oint_{\mathcal{C}}{M~dx+N~dy} = \iint\limits_{\mathcal{R}}{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} dA } }\)

Notation - - The circle on the integral sign clarifies that the curve \(\mathcal{C}\) is a closed curve. It is not necessary and not all books and instructors use it.

If you want a complete lecture on this topic, here is a video.

Prof Leonard - Green's Theorem [1hr-45min-53secs]

Green's Theorem relates the calculation of line integrals to double integrals over area in the special case when the line integral completely encloses a single region in the plane. This enclosed region must be simply-connected, i.e. the enclosing line must not cross itself and it must consist of one simple closed curve. This theorem is very powerful and allows us to evaluate line integrals using double integrals.

Let's get started with a video. This video is excellent in explaining Green's Theorem, including several examples. It covers the idea of simply-connected regions as well.

Dr Chris Tisdell - What is Green's Theorem?

Alternative Forms of Green's Theorem

Okay, so the theorem at the top of the page is the basic form of Green's Theorem. There are two other alternative forms that are equivalent to this basic form. These forms were touched on in the previous video. The next video, below the table, explains them in more detail.

1. circulation or curl ( in 3-dim called Stokes' Theorem )

\(\displaystyle{ \oint_{\mathcal{C}}{ \vec{F} \cdot \vec{T} ~ds} = }\) \(\displaystyle{ \iint_{\mathcal{R}}{(curl ~\vec{F}) \cdot \hat{k} ~ dA} }\)

\(\displaystyle{ \oint_{\mathcal{C}}{\vec{F} \cdot d\vec{r} } = }\) \(\displaystyle{ \iint_{\mathcal{R}}{ \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) dA} }\)

 

2. flux or divergence ( in 3-dim called the Divergence Theorem )

\(\displaystyle{ \oint_{\mathcal{C}}{ \vec{F} \cdot \vec{n} ~ds} = }\) \(\displaystyle{ \iint_{\mathcal{R}}{div~\vec{F}~dA} }\)

\(\displaystyle{ \iint_{\mathcal{R}}{ \left( \frac{\partial N}{\partial y} + \frac{\partial M}{\partial x}\right) dA} }\)

In the table above, \(\vec{T}\) is the unit tangent vector and \(\vec{n}\) is the outward unit normal vector.

This next video explains these alternative forms in more detail and has some examples.

Dr Chris Tisdell - Green's Theorem

Another by-product of Green's Theorem is that we can now use a line integral to calculate area. Here is the equation.

\(\displaystyle{ A = \frac{1}{2} \oint_{\mathcal{C}}{x~dy - y~dx} }\)

The previous video shows the derivation of this equation. If you skipped it, now is the time to go back and watch it.

Meaning of Green's Theorem

So, now that you know how to use Green's Theorem, what is really going on here? This next video gives a quick review of the Theorem and then explains the interpretation and meaning of it.

Michael Hutchings - Multivariable calculus 4.4.1: Review and interpretation of Green's theorem

Proofs of Green's Theorem

Here are several video proofs of Green's Theorem. Each instructor proves Green's Theorem differently. So it will help you to understand the theorem if you watch all of these videos.

Khan Academy - Green's Theorem Proof (2 videos)

Michael Hutchings - Multivariable calculus 4.3.4: Proof of Green's theorem

David Metzler - Greens Theorem Proof

Okay, you are now ready for some practice problems. Next, you need to understand how to parameterize surfaces before you go on to surface integrals and the three dimensional versions of Green's Theorem, Stokes' Theorem and the Divergence Theorem.

parametric surfaces →

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Practice Problems

Instructions - - Unless otherwise instructed, use Green's Theorem to evaluate these integrals.

Level A - Basic

Practice A01

Evaluate \(\displaystyle{\oint_{\mathcal{C}}{(y^2~dx+3xy~dy)}}\) where \(\mathcal{C}\) is the boundary of the semiannular region D in the upper half plane between the circles \(x^2+y^2=1\) and \(x^2+y^2=4\).

answer

solution

Practice A02

Calculate the outward flux of \(\vec{F}=-x\hat{i}+2y\hat{j}\) over the square with corners \((\pm 1, \pm 1)\) where the unit normal is outward-pointing.

solution

Practice A03

Evaluate \(\displaystyle{\oint_{\mathcal{C}}{y^2~dx+x^2~dy}}\) where \(\mathcal{C}\) is the triangle bounded by the lines \(x=0, x+y=1, y=0 \)

solution

Practice A04

Let \(\mathcal{C}\) be the circle \(x^2+y^2=4\).
Evaluate \(\displaystyle{ \oint_{\mathcal{C}}{ \left[ (3y-e^{\tan^{-1}x})~dx + \right. }}\) \(\displaystyle{ \left. (7x+\sqrt{y^4+1})~dy \right] }\)

solution

Practice A05

Evaluate \(\displaystyle{ \int_{\mathcal{C}}{x~dx-x^2y^2~dy}}\) where \(\mathcal{C}\) is the triangle with vertices \((0,0)\), \((0,1)\) and \((1,1)\).

solution

Practice A06

Evaluate \(\displaystyle{\oint_{\mathcal{C}}{P~dx+Q~dy}}\) where \(\mathcal{C}\) is the square with corners \((\pm 1, \pm 1)\); \(P(x,y)=x+y^2\) and \(Q(x,y)=y+x^2\).

solution

Practice A07

Evaluate \(\displaystyle{\oint_{\mathcal{C}}{P~dx+Q~dy}}\) where \(\mathcal{C}\) is the triangle with corners \((0,0)\), \((1,1)\), \((2,0)\), \(P(x,y)=y+e^x\) and \(Q(x,y)=2x^2+\cos y\).

solution

Practice A08

Evaluate \(\displaystyle{\int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}}}\) where \(\mathcal{C}\) is the circle \(x^2+y^2=4\) and \(\vec{F}=\langle -y, x \rangle\).

solution

Practice A09

Evaluate \(\displaystyle{\int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}}}\) where \(\mathcal{C}\) is along \(y=x^2/2\) and \(y=x\) and \(\vec{F}=\langle y^2, x^2 \rangle\).

solution

Practice A10

Evaluate \(\displaystyle{\int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}}}\) where \( \vec{F}=\langle y^3, 6xy^2 \rangle\) and \(\mathcal{C}\) is the curve with the 3 segments:
1. \( y=0, 0 \leq x \leq 4\), 2. \( x=4, 0 \leq y \leq 2\), 3. \( y=\sqrt{x}, 0 \leq x \leq 4\)

solution

Practice A11

Evaluate \(\displaystyle{\int_{\mathcal{C}}{\vec{F} \cdot d\vec{r}}}\) where \(\mathcal{C}\) is boundary between the two circles \(x^2+y^2=1\) and \(x^2+y^2=3\) that lies in the first quadrant and \(\vec{F}=\langle e^x+2xy, 4x^2+\sin y \rangle\).

solution

Practice A12

Determine the flux of the field \(\vec{F}=\langle -x, -y \rangle\) across \(x^2+y^2=16\).

solution

Practice A13

Evaluate \(\displaystyle{\oint_{\mathcal{C}}{(x^2-y^2)~dx + 2xy~dy}}\) on the curve which bounds the region \(\mathcal{R}: \{(x,y): 0 \leq x \leq 1, 2x^2 \leq y \leq 2x \}\)

solution

Practice A14

Evaluate \(\displaystyle{\oint_{\mathcal{C}}{2y~dx-3x~dy}}\) on the unit circle.

solution

Practice A15

Calculate the outward flux of \(\vec{F}=-x\hat{i}+2y\hat{j}\) over the square with corners \((\pm 1, \pm 1)\) where the unit normal is outward-pointing.

answer

solution


Level B - Intermediate

Practice B01

Verify Green's theorem for the line integral \(\displaystyle{\oint_{\mathcal{C}}{xy~dx+x~dy}}\) where \(\mathcal{C}\) is the unit circle.

solution

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