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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

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

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

video by Prof Leonard

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.

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? [47mins-49secs]

The last example in this video has an incorrect answer. The final answer should be \(14/3\).

video by Dr Chris Tisdell

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.

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 [39mins-13secs]

video by Dr Chris Tisdell

Another by-product of Green's Theorem is that we can now use a line integral to calculate area. Here is the equation. \[ A = \frac{1}{2} \oint_{\mathcal{C}}{x~dy - y~dx} \] 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 [7mins-25secs]

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.

Michael Hutchings - Multivariable calculus 4.3.4: Proof of Green's theorem [18mins-2secs]

David Metzler - Greens Theorem Proof [16mins-22secs]

video by David Metzler

Khan Academy - Green's Theorem Proof (1) [14mins-25secs]

video by Khan Academy

Khan Academy - Green's Theorem Proof (2) [19mins-26secs]

video by Khan Academy

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.

Practice

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

Basic

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 \( x^2 + y^2 = 4 \).

Problem Statement

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 \( x^2 + y^2 = 4 \).

Solution

1569 video

video by Dr Chris Tisdell

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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.

Problem Statement

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

1570 video

video by Dr Chris Tisdell

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

Problem Statement

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

1571 video

video by Dr Chris Tisdell

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

Problem Statement

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

Solution

1572 video

video by Dr Chris Tisdell

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

Problem Statement

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

1573 video

video by PatrickJMT

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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 \( Q(x,y) = y + x^2 \).

Problem Statement

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

1574 video

video by Krista King Math

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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 \(Q(x,y)=2x^2+\cos y\).

Problem Statement

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

1575 video

video by Krista King Math

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

Problem Statement

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

1576 video

video by MIP4U

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

Problem Statement

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

1577 video

video by MIP4U

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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. \( y=\sqrt{x}, 0 \leq x \leq 4\)

Problem Statement

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

1578 video

video by MIP4U

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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 \( \vec{F} = \langle e^x+2xy, 4x^2+\sin y \rangle \).

Problem Statement

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

1579 video

video by MIP4U

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

Problem Statement

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

Solution

1580 video

video by MIP4U

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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.

Problem Statement

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.

Final Answer

4

Problem Statement

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

1984 video

video by Dr Chris Tisdell

Final Answer

4

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Evaluate \(\displaystyle{ \oint_{\mathcal{C}}{ 2xy~dx + (x^2+y^2)~dy } }\) where \(\mathcal{C}\) is \( 4x^2 + 9y^2 = 36 \).

Problem Statement

Evaluate \(\displaystyle{ \oint_{\mathcal{C}}{ 2xy~dx + (x^2+y^2)~dy } }\) where \(\mathcal{C}\) is \( 4x^2 + 9y^2 = 36 \).

Solution

2200 video

video by Michel vanBiezen

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

Problem Statement

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

1581 video

video by Khan Academy

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

Problem Statement

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

Solution

1582 video

video by Khan Academy

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Intermediate

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

Problem Statement

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

Solution

1568 video

video by Dr Chris Tisdell

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You CAN Ace Calculus

Topics You Need To Understand For This Page

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

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Theorem Relationships

Green's Theorem

Explanation

Alternative Forms

Meaning

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Practice

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Practice Instructions

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

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