17Calculus - Green's Theorem
Relationships Between Green's Theorem, Stokes' Theorem and the Divergence Theorem | |||
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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) |
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Greens' Theorem (divergence form) |
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If you want a complete lecture on this topic, here is a good video. | |||
Prof Leonard - Green's Theorem [1hr-45min-53secs]
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Green's Theorem |
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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 |
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 |
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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 ) | |
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\(\displaystyle{ \oint_{\mathcal{C}}{ \vec{F} \cdot \vec{T} ~ds} = }\) \(\displaystyle{ \iint_{\mathcal{R}}{(curl ~\vec{F}) \cdot \hat{k} ~ dA} }\) |
\(\displaystyle{ \oint_{\mathcal{C}}{\vec{F} \cdot d\vec{r} } = }\) \(\displaystyle{ \iint_{\mathcal{R}}{ \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) dA} }\) |
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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 |
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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]
video by Michael Hutchings |
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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]
video by Michael Hutchings |
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David Metzler - Greens Theorem Proof [16mins-22secs]
video by David Metzler |
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Khan Academy - Green's Theorem Proof (1) [14mins-25secs]
video by Khan Academy |
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Khan Academy - Green's Theorem Proof (2) [19mins-26secs]
video by Khan Academy |
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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 |
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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 |
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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 |
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1569 video
video by Dr Chris Tisdell |
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close solution
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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 |
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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.
Solution |
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1570 video
video by Dr Chris Tisdell |
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close solution
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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 |
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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 \( x=0, x+y=1, y=0 \)
Solution |
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1571 video
video by Dr Chris Tisdell |
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close solution
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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 |
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Let \( \mathcal{C} \) be the circle \( x^2 + y^2 = 4. \) Evaluate \(\displaystyle{ \oint_{\mathcal{C}}{ \left[ (3y-e^{\tan^{-1}x})~dx + (7x + \sqrt{y^4+1}) ~dy \right] } }\)
Solution |
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1572 video
video by Dr Chris Tisdell |
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close solution
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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 |
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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 \((1,1)\).
Solution |
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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 |
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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 \( Q(x,y) = y + x^2 \).
Solution |
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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 |
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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 \(Q(x,y)=2x^2+\cos y\).
Solution |
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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 |
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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 \(\vec{F}=\langle -y, x \rangle\).
Solution |
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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 |
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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 \(\vec{F}=\langle y^2, x^2 \rangle\).
Solution |
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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 |
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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. \( y=\sqrt{x}, 0 \leq x \leq 4\)
Solution |
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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 |
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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 \( \vec{F} = \langle e^x+2xy, 4x^2+\sin y \rangle \).
Solution |
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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 |
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Determine the flux of the field \( \vec{F} = \langle -x, -y \rangle \) across \(x^2+y^2=16\).
Solution |
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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 |
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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.
Final Answer |
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4
Problem Statement |
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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.
Solution |
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1984 video
video by Dr Chris Tisdell |
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Final Answer |
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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 |
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Evaluate \(\displaystyle{ \oint_{\mathcal{C}}{ 2xy~dx + (x^2+y^2)~dy } }\) where \(\mathcal{C}\) is \( 4x^2 + 9y^2 = 36 \).
Solution |
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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 |
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Evaluate \(\displaystyle{ \oint_{\mathcal{C}}{(x^2-y^2)~dx + 2xy~dy} }\) on the curve which bounds the region \(\mathcal{R}: \{(x,y): 0 \leq x \leq 1, 2x^2 \leq y \leq 2x \}\)
Solution |
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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 |
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Evaluate \(\displaystyle{ \oint_{\mathcal{C}}{ 2y~dx-3x~dy } }\) on the unit circle.
Solution |
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1582 video
video by Khan Academy |
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Intermediate |
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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 |
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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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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
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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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Page Sections
How to Ace the Rest of Calculus: The Streetwise Guide, Including MultiVariable Calculus |
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Practice Instructions
Unless otherwise instructed, use Green's Theorem to evaluate these integrals.
