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 threedimensional form of the circulation form of Green's Theorem. Similarly, the Divergence Theorem is the threedimensional form of the divergence form of Green's Theorem. The following table summarizes these relationships.  
2dim 
3dim  
Green's Theorem (circulation form) 

Greens' Theorem (divergence form) 

If you want a complete lecture on this topic, here is a good video.  
Prof Leonard  Green's Theorem [1hr45min53secs]

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Green's Theorem 

IF WE HAVE 
 a simplyconnected region \(\mathcal{R}\) which is bounded by a smooth curve \(\mathcal{C}\), oriented anticlockwise (counterclockwise) and traversed once, and 
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 simplyconnected, 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 simplyconnected regions as well.
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 3dim 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 3dim 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.
video by Dr Chris Tisdell 

Another byproduct 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.
video by Michael Hutchings 

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.
video by Michael Hutchings 

video by David Metzler 

video by Khan Academy 

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 

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 outwardpointing.
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 outwardpointing.
Solution 

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 

video by Dr Chris Tisdell 

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

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

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 

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 

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 

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 

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 

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 

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 

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 

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 outwardpointing.
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 outwardpointing.
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 outwardpointing.
Solution 

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 

video by Michel vanBiezen 

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

video by Khan Academy 

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Evaluate \(\displaystyle{ \oint_{\mathcal{C}}{ 2y~dx3x~dy } }\) on the unit circle.
Problem Statement 

Evaluate \(\displaystyle{ \oint_{\mathcal{C}}{ 2y~dx3x~dy } }\) on the unit circle.
Solution 

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 

video by Dr Chris Tisdell 

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You CAN Ace Calculus
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  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle 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
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
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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Practice Instructions
Unless otherwise instructed, use Green's Theorem to evaluate these integrals.